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MyStatLab™ for Business Statistics

MyStatLab is a course management system that provides engaging learning experiences and delivers proven results while helping students succeed. Tools are embedded which make it easy to integrate statistical software into the course. And, MyStatLab comes from an experienced partner with educational expertise and an eye on the future.

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StatCrunch MyStatLab includes web-based statistical software, StatCrunch, within the online assessment platform so that students can analyze data sets from exercises and the text. In addition, MyStatLab includes access to www.StatCrunch.com, the full web-based program where users can access thousands of shared data sets, create and conduct online surveys, perform complex analyses using the powerful statistical software, and generate compelling reports.

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MyStatLab™ for Business Statistics

Help Me Solve This breaks the problem into manageable steps. Students enter answers along the way.

View an Example walks students through a problem similar to the one assigned.

Textbook links to the appropriate section in the etext.

Tech Help is a suite of Technology Tutorial videos that show how to perform statistical calculations using popular software.

Engaging Video Resources • Business Insight Videos are 10 engaging videos showing managers at top companies using statistics in

their everyday work. Assignable questions encourage discussion.

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Business Statistics

E d i t i o n

n d

R o b e R t A . D o n n e l ly, J R .

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ISBN 10: 0-321-92512-2 ISBN 13: 978-0-321-92512-1

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Library of Congress Cataloging-in-Publication Data Donnelly, Robert A. Business statistics / Robert A. Donnelly, Jr.—2nd ed. p. cm. ISBN 978-0-321-92512-1 1. Commercial statistics. 2. Industrial management—Statistical methods. I. Title. HF1017.S74 2013 519.5024'65—dc23

2012005942 10 9 8 7 6 5 4 3 2 1—CRK—16 15 14 13

Editor in Chief: Deirdre Lynch Acquisitions Editor: Marianne Stepanian Development Editor: Dana Bettez Assistant Editor: Sonia Ashraf Senior Managing Editor: Karen Wernholm Associate Managing Editor: Tamela Ambush Senior Production Project Manager: Peggy McMahon Associate Director of Design, USHE EMSS/HSC/EDU:

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To my wife, Debbie, who supported and encouraged me every step of the way. I could not have done this without you, Babe.

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Robert (Bob) A. Donnelly, Jr. is a professor at Goldey-Beacom College in Wilmington,

Delaware, with more than 25 years of teaching experience. He teaches classes in statistics,

operations management, spreadsheet modeling and project management at both the un-

dergraduate and graduate level. Bob earned an undergraduate de-

gree in chemical engineering from the University of Delaware,

after which he worked for several years as an engineer with the

Diamond Shamrock Corporation in a chlorine plant. Despite success

in this field, Bob felt drawn to pursue a career in education. It was

his desire to teach that took him back to school to earn his MBA and

Ph.D. in Operations Research, also from the University of Delaware.

Bob also teaches in the MBA program at the International School

of Management in Paris, France. He thoroughly enjoys discussing

research methods and business statistics with both his French and

American students.

Bob’s working experience gathered prior to his teaching career

has provided him with many opportunities to incorporate real-life ex-

amples into classroom learning. His students appreciate his knowledge of the business world

as well as his mastery of the course subject matter. Many former students seek Bob’s assis-

tance in work-related issues that deal with his expertise. Typical student comments focus on

his genuine concern for their welfare and his desire to help them succeed in reaching their

goals.

R o b e R t A . D o n n e l ly, J R .

About the Author

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Preface | xi

Acknowledgments | xvi

Dear Students | xx

Index of Applications | xxi

Chapter 1 An Introduction to Business Statistics | 1

Chapter 2 Displaying Descriptive Statistics | 21

Chapter 3 Calculating Descriptive Statistics | 77

Chapter 4 Introduction to Probabilities | 147

Chapter 5 Discrete Probability Distributions | 193

Chapter 6 Continuous Probability Distributions | 247

Chapter 7 Sampling and Sampling Distributions | 289

Chapter 8 Confidence Intervals | 333

Chapter 9 Hypothesis Testing for a Single Population | 383

Chapter 10 Hypothesis Tests Comparing Two Populations | 437

Chapter 11 Analysis of Variance (ANOVA) Procedures | 501

Chapter 12 Chi-Square Tests | 575

Chapter 13 Hypothesis Tests for the Population Variance | 625

Chapter 14 Correlation and Simple Linear Regression | 651

Chapter 15 Multiple Regression and Model Building | 709

Chapter 16 Forecasting | 767

Chapter 17 Decision Analysis (online) | 17-1

Chapter 18 Nonparametric Statistics (online) | 18-1

Appendix A: Tables | 827

Appendix B: Answers to Selected Even-Numbered Problems | 865

Credits | 933

Index | 935

bRief Contents

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vii

An Introduction to Business Statistics | 1 1.1 Business Statistics and Their Uses 2 1.2 Data 4 1.3 Descriptive and Inferential Statistics 11 1.4 Ethics and Statistics—It’s a Dangerous World of Data Out There 14 Summary 16 | Key Terms 17 | Problems 17 | Solutions to YoUR tURn 19

Displaying Descriptive Statistics | 21 2.1 The Role Technology Plays in Statistics 23 2.2 Displaying Quantitative Data 25 2.3 Displaying Qualitative Data 42 2.4 Contingency Tables 55 2.5 Stem and Leaf Display 59 2.6 Scatter Plots 61 Summary 66 | Key Terms 67 | Problems 67 | Solutions to YoUR tURn 73

Calculating Descriptive Statistics | 77 3.1 Measures of Central Tendency 78 3.2 Measures of Variability 91 3.3 Using the Mean and Standard Deviation Together 101 3.4 Working with Grouped Data 112 3.5 Measures of Relative Position 116 3.6 Measures of Association Between Two Variables 126 Summary 132 | Key Terms 133 | Equations 134 | Problems 137 |

Solutions to YoUR tURn 141

Solutions to YoUR tURn 189

Preface | xi Acknowledgments | xvi Dear Students | xx Index of Applications | xxi

tAble of Contents

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viii t A B L E o F C o n t E n t s

Discrete Probability Distributions | 193 5.1 Introduction to Discrete Probability

Distributions 195

5.2 Binomial Distributions 207 5.3 Poisson Distributions 218 5.4 The Hypergeometric

Distribution 229 Summary 235 | Key Terms 236 |

Equations 236 | Problems 237 | Solutions to YoUR tURn 242

Continuous Probability Distributions | 247

6.1 Continuous Random Variables 248 6.2 Normal Probability Distributions 250 6.3 Exponential Probability

Distributions 269

6.4 Uniform Probability Distributions 273 Summary 279 | Key Terms 280 |

Equations 280 | Problems 281 | Solutions to YoUR tURn 286

Sampling and Sampling Distributions | 289

7.1 Why Sample? 290 7.2 Types of Sampling 291 7.3 Sampling and Nonsampling

Errors 298

7.4 The Central Limit Theorem 301 7.5 The Sampling Distribution of the

Proportion 317 Summary 323 | Key Terms 324 |

Equations 325 | Problems 325 | Solutions to YoUR tURn 329

Confidence Intervals | 333 8.1 Point Estimates 334 8.2 Calculating Confidence Intervals

for the Mean when the Standard Deviation (S) of a Population Is Known 335

8.3 Calculating Confidence Intervals for the Mean when the Standard Deviation (S) of a Population Is Unknown 347

8.4 Calculating Confidence Intervals for Proportions 354

8.5 Determining the Sample Size 359 8.6 Calculating Confidence Intervals for

Finite Populations 366 Summary 371 | Key Terms 372 |

Equations 372 | Problems 375 | Solutions to YoUR tURn 378

Hypothesis Testing for a Single Population | 383

9.1 An Introduction to Hypothesis Testing 384

9.2 Hypothesis Testing for the Population Mean When S Is Known 390

9.3 Hypothesis Testing for the Population Mean when S Is Unknown 403

9.4 Hypothesis Testing for the Proportion of a Population 412

9.5 Type II Errors 416 Summary 425 | Key Terms 426 |

Equations 426 | Problems 427 | Solutions to YoUR tURn 432

Hypothesis Tests Comparing Two Populations | 437

10.1 Comparing Two Population Means with 439

• Independent Samples

• Known Population Standard Deviations 1S1 and S22

10.2 Comparing Two Population Means with 448

• Independent Samples

• Unknown Population Standard Deviations

1S1 and S22 10.3 Hypothesis Testing With Dependent

Samples 464

10.4 Comparing Two Population Proportions with Independent Samples 474

Summary 484 | Key Terms 485 | Equations 485 | Problems 489 | Solutions to YoUR tURn 494

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ix C O N T E N T Six C O N T E N T S t A B L E o F C o n t E n t s ix

Analysis of Variance (ANOVA) Procedures | 501

11.1 One-Way ANOVA: Examining the Effect a Single Factor Has on the Means of Populations 503

11.2 Randomized Block ANOVA: Examining the Effects of a Single Factor by Blocking a Second Factor 523

11.3 Two-Way ANOVA: Examining the Effects Two Factors Have on the Means of Populations 540

Summary 557 | Key Terms 559 | Equations 560 | Problems 563 | Solutions to YoUR tURn 568

Chi-Square Tests | 575 12.1 Comparing Two or More Population

Proportions 577

12.2 Determining If Observed Frequencies Follow a Known Probability Distribution 585

12.3 Testing the Independence of Two Variables 605

Summary 612 | Key Terms 613 | Equations 613 | Problems 614 | Solutions to YoUR tURn 619

Hypothesis Tests for the Population Variance | 625

13.1 Testing the Variance of a Single Population 626

13.2 Comparing the Variances of Two Populations 634

Summary 644 | Equations 644 | Problems 645 | Solutions to YoUR tURn 648

Correlation and Simple Linear Regression | 651

14.1 Dependent and Independent Variables 653

14.2 Correlation Analysis 653 14.3 Simple Linear Regression

Analysis 660

14.4 Using a Regression to Make a Prediction 672

14.5 Testing the Significance of the Slope of the Regression Equation 679

14.6 Assumptions for Regression Analysis 684

14.7 A Simple Regression Example with a Negative Correlation 687

14.8 Some Final (but Very Important) Thoughts 696

Summary 697 | Key Terms 698 | Equations 699 | Problems 701 | Solutions to YoUR tURn 704

Multiple Regression and Model Building | 709

15.1 Developing the Multiple Regression Model 710

15.2 Explaining the Variation of the Dependent Variable 717

15.3 Inferences about the Independent Variables 726

15.4 Using Qualitative Independent Variables 732

15.5 Model Building 739 Summary 756 | Key Terms 757 |

Equations 758 | Problems 759 | Solutions to YoUR tURn 761

Forecasting | 767 16.1 Introduction to Forecasting 768 16.2 Smoothing Forecasting

Methods 770

16.3 Forecasting with Regression Analysis 783

16.4 Forecasting with Seasonality 797 Summary 814 | Key Terms 815 |

Equations 816 | Problems 817 | Solutions to YoUR tURn 820

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x t A B L E o F C o n t E n t s

Decision Analysis (online) | 17-1 17.1 Introduction to Decision

Analysis 17-2

17.2 Constructing a Decision Table 17-3 17.3 Decision Making Under

Uncertainty 17-6

17.4 Decision Making Under Risk 17-12 17.5 Decision Making Using Decision

Trees 17-16

17.6 Using Bayes’ Theorem to Calculate Posterior Probabilities 17-27

Summary 17-32 | Key Terms 17-33 | Equations 17-34 | Problems 17-34 | Solutions to YoUR tURn 17-37

Nonparametric Statistics (online) | 18-1

18.1 Introduction to Nonparametric Statistics 18-2

18.2 The Sign Test 18-4 18.3 The Wilcoxon Rank-Sum Test for

Two Independent Samples 18-14

18.4 The Wilcoxon Signed-Rank Test for Two Dependent Samples 18-22

18.5 The Kruskal-Wallis One-Way ANOVA 18-27

18.6 The Spearman Rank-Order Correlation Coefficient 18-31

Appendix C: Tables for Chapter 18 18-48 Summary 18-37 | Key Terms 18-38 |

Equations 18-39 | Problems 18-40 | Solutions to YoUR tURn 18-44

Appendix A | 827 table 1 Binomial Probabilities | 827 table 2 Poisson Probabilities | 838 table 3 Cumulative Probabilities for the Standard

Normal Distribution | 843

table 4 Cumulative Probabilities for the Standard Normal Distribution | 844

table 5 Student’s t-distribution | 845 table 6 F-distribution | 848 table 7 Critical Values of the Studentized

Range, Q | 858

table 8 Chi–Square Distribution | 861 table 9 Critical Values for the Durbin-Watson

Statistic | 862

Appendix B: Answers to Selected Even-Numbered Problems | 865

Credits | 933

Index | 935

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b u s i n e s s s tA t i s t i C s , second edition, is a one- or two-semester textbook written in a conversational tone designed to reduce the level of anxiety that many business students experience when taking a statistics course.

Many of today’s business students are intimidated by their statistics textbook. These stu- dents often view their textbook as an obstacle to overcome rather than a tool to help them succeed. To address this issue, I have written Business Statistics in a straightforward, conver- sational tone that helps to reduce the anxiety many students experience with this course. My experience as both a writer and a teacher has taught me that students learn more effectively when they feel a personal connection with their instructor. Many traditional textbooks tend to “talk down” to students in a manner that many find difficult to understand. I prefer a textbook that “talks to” the students as I do in the classroom providing them a sense that I’m on their side, encouraging them every step of the way.

I strongly believe that students learn most effectively when they solve statistics prob- lems as they learn new concepts rather than later (often right before the next exam). To facili- tate this philosophy, I provide the student with a parallel problem that I have labeled “Your Turn,” which allows them to work alongside the example that I am demonstrating in the chap- ter. I attempt to motivate them to do these exercises with a little levity but it’s not beneath me to downright beg them to give it a try. I show the entire solution at the end of the chapter, so they can quickly check if their answer is correct. I call this my “learn it, do it, check it” cycle, where students learn by reading an example, doing a similar problem on their own, and finally checking their answer to confirm they understand. In effect, the textbook also plays the role of a workbook for the student, keeping them actively engaged in the learning process. Too often, students skim through an example that is completely solved for them in the text and convince themselves they understand the concept—that is, until they are trying to solve a similar problem in an exam for the first time. My approach encourages students to work through examples and confirm they grasp the concept before moving on to the next topic.

I have inserted many author’s comments in the margins throughout each chapter, that provide useful insights along the way. This feature is analogous to the side comments you would make to your students during a lecture to help them better understand the material. I have found this to be an effective technique to help keep students focused on material that they may find confusing or overwhelming.

To help students be successful in your course, Business Statistics, second edition, has the following attributes:

Is written in a straightforward, conversational tone—to help reduce the anxiety that many business students experience with the topic of statistics.

Utilizes a “learn it, do it, check it” cycle—by providing parallel Your Turn problems throughout each chapter, the textbook essentially serves as a workbook allowing students to convince themselves they really understand a concept before moving on to the next topic.

Incorporates author’s comments in the margins—which are analogous to the side comments that an instructor would make during a lecture to help students better under- stand the material.

n e w t o t h i s e D i t i o n I am very excited to offer several new features to the second edition of Business Statistics. I have

Added two new online chapters: Chapter 17, Decision Analysis, and Chapter 18, Non- parametric Statistics. Chapter 17, Decision Analysis, provides a detailed discussion of decision making under uncertainty and decision making under risk along with a step- by-step description on the construction and analysis of decision trees. Chapter 18, Non- parametric Statistics, provides a detailed description of the following procedures: Sign Test, Wilcoxon Rank-Sum Test, Wilcoxon Signed-Rank Test, Kruskal-Wallis One-Way ANOVA, and Spearman Rank-Order Correlation Coefficient. These chapters can be found on the textbook’s Web site www.pearsonhighered.com/donnelly.

Preface

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xii P R E F A C E

Updated technology coverage to Microsoft Excel 2013, with instructions for Excel 2011 for Mac and Excel 2010 for Windows provided online as needed. Through my experi- ence in the classroom, I have been aware of the increasing number of Mac users who have been frustrated with software compatibility issues. The version of the Excel Add-in PHStat that is utilized in this edition is compatible with Excel 2011 for Mac. These in- structions can be found on the textbook’s Web site www.pearsonhighered.com/donnelly.

Increased the number of problems by 25%, totaling over 1,110 business-related prob- lems. Additionally, 35% of the problems in the text are new or updated.

Doubled the number of data sets included in problems, examples, and Your Turns, total- ing over 340 data sets.

Introduced topics of covariance and the correlation coefficient at the end of Chapter 3, Calculating Descriptive Statistics. The correlation coefficient is also covered in Chapter 14, Correlation and Simple Linear Regression.

Used Excel functions to determine p-values and critical scores for hypothesis tests that use the normal, student’s t, F, and chi-square distributions. This feature provides students with more options for this type of analysis.

Removed critical sample mean and critical sample proportion as optional steps to hy- pothesis testing in Chapters 9 and 10 to streamline these procedures. These two topics are now included in the section describing Type II Errors at the end of Chapter 9.

Added learning objectives to each chapter opener, which describe the skills that the stu- dent is expected to acquire after studying this material.

Added an Index of Applications that allows faculty and students to conveniently locate specific types of problems and examples.

t e x t b o o k f e At u R e s

Current business examples that keep the students’ interest—Statistical procedures are applied to products and services that students can relate to such as the following:

• Approximating the probability of an accident similar to the BP oil spill in the Gulf of Mexico occurring again in the near future

• Liberty Mutual Insurance Company comparing the proportion of auto accident claims for clients with and without good student discounts

• Comparing satisfaction scores for various smartphone brands • The shortage of Internet protocol (IP) addresses using the original IPv4 format

C h a p t e r 7 | 7.4 The Central Limit Theorem 315

As the calculations show, for an infinite population where n>N is less than 5%, the correction factor is close to 1.0, essentially dropping out of the standard error calculation. The next Your Turn problem demonstrates the finite population concept in an effort to save my job.

My college requires instructors to have an average approval rating of 9.0 on a scale of 1–10 from stu- dent evaluations as a condition for employment. This current semester, I have 120 students, of whom 30 completed the evaluation. My average score on the evaluation was 8.8. Historical data have indicated that the standard deviation for student evaluations in the college is 3.2. My dean has scheduled a meet- ing with me today to discuss my future employment with the college. Can you help me save my job?

Y o u r t u r n # 7N

Answer can be found on u page 331

Rating data used in the Your Turn #7 example is technically ordinal data which we defined in Chapter 1 as data that are rank-ordered with no assurance that the differences be- tween the ranks are uniform. We have no certainty that the difference between a rating score of 5 and 6 is identical to the difference between a rating score of 8 and 9. Because of the ambiguity of the intervals, it’s not proper to calculate the mean and standard deviation of or- dinal data. However, it is common practice to assume that rating data used in this Your Turn are close enough to interval data which allows calculations of the mean and standard devia- tion (which we do perform in this example). Just be aware that some people in the world of statistics take offense at such liberties!

7.4 Section Problems Basic Skills 7.7 For a population with a mean equal to 150 and a standard

deviation equal to 30, calculate the standard error of the mean for a sample size of

a. 10 b. 30 c. 50

7.8 For a normal population with a mean equal to 80 and a standard deviation equal to 15, determine the probability of observing a sample mean of 85 or less from a sample of size 10.

7.9 For a population that is left skewed with a mean of 24 and a standard deviation equal to 15, determine the probability of observing a sample mean of 23 or more from a sample of size 32.

7.10 Identify the symmetrical interval that includes 95% of the sample means for a population with a mean equal to 45 and a standard deviation equal to 9 using a sample size of 36.

7.11 Consider a population of 250 with a mean of 60 and a standard deviation equal to 23. What is the probability of obtaining a sample mean of 62 or less with a sample size of 45?

applications 7.12 According to the National Association of Theater Owners,

the average price for a movie in the United States in 2012 was $7.96. Assume the population standard deviation is $0.50 and that a sample of 30 theaters was randomly selected.

a. Calculate the standard error of the mean. b. What is the probability that the sample mean will be

less than $7.75? c. What is the probability that the sample mean will be

less than $8.10? d. What is the probability that the sample mean will be

more than $8.20?

7.13 Iron Hill Café is a full-service restaurant in Columbus, Ohio, which is owned and managed by John Stevens. The number of minutes that diners spend at the table has a major impact on the profitability of the restaurant. Sup- pose the average number of minutes that diners spend at a table for dinner at the Iron Hill Café is 93 minutes with a standard deviation of 17 minutes. Assume the number of minutes diners spend at their table follows the normal probability distribution. Calculate the probability that the average number of minutes that diners spend at their table for dinner will be less than 100 minutes

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272 C h a p t e r 6 | Continuous Probability Distributions

So, for exponential distributions, there are no nasty standard deviation calculations that we dealt with in previous chapters. For our Piggly Wiggly example, the standard deviation is simply as follows:

s = m = 4 minutes

As with our previous distributions, the standard deviation of the exponential distribution measures the average variability around the mean of the distribution.

Calculating exponential probabilities Using excel We can confirm the probability that the next customer at a Piggly Wiggly will arrive at the deli counter within the next two minutes using EXPON.DIST, Excel’s built-in function, which has the following characteristics:

=EXPON.DIST1x, l, cumulative2 where

cumulative = FALSE if you want the probability density function cumulative = TRUE if you want the cumulative probability We choose the TRUE option for cumulative because we want the probability that the

next customer will arrive in 2 minutes or less. The parameter l refers to the average rate of arrivals over a specific interval. In our Piggly Wiggly example, customers arrive, on average, every four minutes 1m2. We can convert this average value into a rate 1l2 as follows:

m = 4 minutes per customer

l = 1 m

= 1 customer

4 minutes = 0.25 customers per minute

An average of 0.25 customers each minute is equivalent to

a 0.25 customers minute

b a 60 minutes hour

b = 15 customers per hour

In other words, if customers are arriving every 4 minutes, 15 customers arrive in 1 hour (there are 15 four-minute intervals in an hour). It’s simply a matter of explaining the same in- formation two different ways. However, the terms m and l must be based on the same units. If m is expressed in minutes per customer, then l must be expressed in customers per minute.

We’re now ready to use the EXPON.DIST function to calculate P1x … 22 with l = 0.25. =EXPON.DIST12, 0.25, TRUE2 = 0.3935

As you can see, our result, P1x … 22 = 0.3935 matches what we obtained using Equation 6.4.

Take a shot (no pun intended) on the next Your Turn problem that involves using the exponential distribution in the National Hockey League.

During the 2011–2012 National Hockey League season, the Detroit Red Wings led the league in shots on goal, averaging 32.2 per 60-minute game. Assume that the number of minutes between shots on goal follows the exponential distribution.

a. Calculate the following probabilities using Equations 6.4 and 6.5 and verify them using Excel.

1. The probability that the next shot on goal will occur during the next minute.

2. The probability that the next shot on goal will occur between 2 and 3 minutes from now. b. Verify your answers from part a using Excel. c. What is the standard deviation for this distribution?

(Because I’m such a nice guy, I’ll give you the following hint: You need to convert 32.2 shots per hour into minutes between shots.)

Y o U r t U r n # 6N

Answers can be found on u page 287

M06_DONN5121_01_SE_C06.3.indd 272 08/08/13 6:47 PM

Your Turn problems after every major section—These prob- lems are strategically placed throughout the chapter and are de- signed to reinforce key concepts. Complete solutions to these problems can be found at the end of the chapter. I feel that students learn more effectively when they actively solve problems rather than skim through examples that are completely solved for them.

Step-by-step approach to complicated statistical procedures— Many students tend to get “lost in the forest” when facing compli- cated procedures such as hypothesis testing, analysis of variance (ANOVA), and regression. My approach is to break these procedures down into bite-size, repeatable steps that can be applied to solve a variety of problems. As a result, the student has a consistent road map to follow when deciding how to proceed with a specific technique.

Highlighted text—Throughout the textbook, I have highlighted text to draw the student’s attention to a key point in the chapter. This will help to reinforce important concepts that could other- wise be overlooked by the student.

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xiii C O N T E N T Sxiii C O N T E N T S P R E F A C E xiii

Author’s comments—In the margins, you will find comments that help clarify specific topics. These comments often point to an appropriate location in the chapter and are analogous to the comments you as an instructor might make in class to provide your stu- dents with some additional insight to the material.

Statistics in Practice—Throughout the textbook, examples of how statistics is used in today’s business environment are described in specially marked sections within the chap- ter. Examples include the following:

• Government reports of unemployment figures using confidence intervals and the interpretation of these results

• Comparison of customer feedback for different snack food products from Herr’s Food Company

• A statistical technique that health care insurance companies can rely on to investigate unusual billing practices from doctors’ offices

• Comparison of performances of Olympic athletes across different sporting events

92 C h a p t e r 3 | Calculating Descriptive Statistics

The following data set shows five online donations recently received by the American Red Cross:

$35 $10 $50 $20 $60

The range for these data is shown as follows:

Range = $60 - $10 = $50

The benefit of the range is that it is easy to calculate and understand. However, this measure of variability has the following two drawbacks: (1) It is based only on two numbers in the data set, and (2) it is highly affected by outliers. For instance, if the last donation were changed from $60 to $1,000, the range would be $1,000 - $10 = $990, instead of $50. The $990 range does not accurately reflect the overall variability of our donations because most of the values lie between $10 and $50.

Formula 3.5 for the range

Range = Highest value - Lowest value

The range, which is one measure of variability, can found by subtracting the lowest value from the highest value in the data set.

the range The range is the difference between the highest value and the lowest value in a data set. The range, which is the simplest measure of a distribution’s variability, can be found using Equation 3.5.

The range is the easiest measure of variability to calculate, but it has the most limitations, one being that it is highly affected by outliers.

Stats in practice: Using the range to Measure Volatility

Although the range as a measurement of variability has its limi- tations, it can be useful when measuring volatility of the stock market. If a volatile day for the stock market is defined as a swing of more than 200 points, we can use the daily range of the stock market index to identify volatile days. Figure 3.9 shows the number of volatile days in the stock market from 2000 until 2012.

The surge in volatile days from 2008 to 2011 reflects the uncer- tainty in the financial markets as the economy struggled through a significant recession. The quiet period seen in 2012 represents a more stable market as investors showed more confidence in the economy. In this example, we’ve demonstrated that we can use the range to conclude that the economy is showing signs of recovering from the recent financial turmoil.

{ figure 3.9 Number of Volatile Days in the Stock Market

the Variance and Standard Deviation A much more useful measure of variability is the variance, which is determined by using all of the data values rather than only the highest and lowest. The variance measures the vari- ability, or spread, of the data points in a set around the set’s mean. The formula for calculat- ing a variance will depend on whether your data set is a sample or a population. Next, we look at each scenario.

The variance measures the variability, or spread, of the data points in a set around the set’s mean.

M03_DONN5121_01_SE_C03.1-3.2.indd 92 8/14/13 9:31 AM

In quality control settings, businesses prefer a smaller standard deviation, which is an indication of more consistency in the process.

The integration of Microsoft Excel® 2013—I utilize Excel to demonstrate the use of technology in business statistics, but not at the expense of understanding the underly- ing concepts. I have spoken to students who tell me that they know how to perform a procedure such as ANOVA on the computer, but do not feel comfortable interpreting the results. By letting Excel do all the work they miss the opportunity to understand the underlying concepts of the technique. The philosophy employed in this textbook avoids this unfortunate situation.

C h a p t e r 11 | 11.1 One-Way ANOVA: Examining the Effect a Single Factor Has on the Means of Populations 515

Figure 11.8C shows the ANOVA output for the smartphone example, which matches our results from the previous section.

4. Fill in the ANOVA: Single Factor dialog box as shown in Figure 11.8B. Click OK.

{ figure 11.8a Conducting a One-Way ANOVA

Test Using Excel (Steps 1–3)

{ figure 11.8b Conducting a One-Way ANOVA

Test Using Excel (Step 4)

{ figure 11.8c Conducting a One-Way ANOVA

Test Using Excel (Final Result)

Notice that the p-value reported in Cell K13 shown in Figure 11.8C equals 0.001887. This value represents the probability of obtaining a critical F-score of 9.86 or more if the population means are truly equal. Because this p-value is less than a = 0.05, we reject the null hypothesis and conclude that the average scores in these populations are not equal.

M11_DONN5121_01_SE_C11-1.indd 515 9/6/13 11:12 AM

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s t u D e n t R e s o u R C e s Student Solutions Manual, by Bob Donnelly and Roman Erenshteyn, provides detailed, worked-out solutions to all even-numbered problems (ISBN-10: 0-321-93070-3; ISBN-13: 978-0-321-93070-5).

Study Cards for Business Statistics Software. This series of study cards, available for Excel® with XLSTAT™, Excel® 2013, Minitab®, JMP®, SPSS®, R®, StatCrunch™, and TI-83/84 graphing calculators, provides students with easy, step-by-step guides to the most common business statistics software. Avail- able at www.myPearsonStore.com.

i n s t R u C t o R R e s o u R C e s Instructor’s Solutions Manual, by Bob Donnelly and

Roman Erenshteyn, provides detailed, worked-out solu- tions for all problems. Available for download at www. pearsonhighered.com/irc.

Test Bank, by Bob Donnelly, includes true/false, multiple- choice, fill-in, and problem-solving questions based on the definitions, concepts, and ideas developed in each chapter of the text. Available for download at www .pearsonhighered.com/irc.

TestGen® (www.pearsonhighered.com/testgen) enables in- structors to build, edit, print, and administer tests, using a computerized bank of questions developed to cover all the objectives in the text. TestGen is algorithmically based, allow- ing instructors to create multiple but equivalent versions of the same question or test with the click of a button. Instructors can also modify test bank questions or add new questions. Tests can be printed or administered online. The software and test bank are available for download from Pearson Education’s online catalog (www.pearsonhighered.com/irc).

t e C h n o l o g y R e s o u R C e s O n l i n e R e s o u r c e s c a n b e d o w n l o a d e d f r o m w w w .pearsonhighered.com/donnelly and include the following:

Chapters 17, Decision Analysis, and Chapter 18, Nonpara- metric Statistics.

Data sets in Excel format for chapter problems, examples, and Your Turns.

Excel for Mac 2011/Excel 2010 for Windows instruc- tions for students to refer to when Excel instructions shown in the textbook are not compatible with the Excel for Mac 2011 or Excel 2010 for Windows versions of Excel.

Online case files and answers, by Bob Donnelly, can be used in conjunction with the chapters to further class discus- sions or to use as class projects or homework assignments.

PHStat, the latest version of PHStat, the Pearson statistical add-in for Windows-based Excel 2003, 2007, and 2010. This version is also compatible with Excel 2011 for Mac. PHStat is available for download with an access code. If your book was packaged with

an access code, download PHStat from www.pearsonhighered. com/phstat. If your book was not packaged with an access code, visit myPearsonStore.com to purchase a PHStat access code.

MyStatLab™ Online Course (access code required) MyStat- Lab is a course management systems that delivers proven results in helping individual students succeed. MyStatLab can be suc- cessfully implemented in any environment—lab-based, hybrid, fully online, traditional—and demonstrates the quantifiable dif- ference that integrated usage has on student retention, subsequent success, and overall achievement. MyStatLab’s comprehensive online gradebook automatically tracks students’ results on tests, quizzes, homework, and in the study plan. Instructors can use the gradebook to provide positive feedback or intervene if students have trouble. Gradebook data can be easily exported to a variety of spreadsheet programs, such as Microsoft Excel.

MyStatLab provides engaging experi- ences that personalize, stimulate, and

measure learning for each student.

Tutorial Exercises with Multimedia Learning Aids: The homework and practice exercises in MyStatLab align with the exercises in the textbook, and they regenerate algorith- mically to give students unlimited opportunity for practice and mastery. Exercises offer immediate helpful feedback, guided solutions, sample problems, animations, videos, and eText clips for extra help at point-of-use.

Adaptive Study Plan: Pearson now offers an optional fo- cus on adaptive learning in the study plan to allow students to work on just what they need to learn when it makes the most sense to learn it.

Additional Statistics Question Libraries: In addition to questions that are aligned with your textbook, MyStatLab courses come with two additional question libraries. Four hundred and fifty Getting Ready for Statistics questions offer the developmental math topics students need for the course. One thousand Conceptual Question Library re- quire students to apply their statistical understanding.

StatCrunch™: MyStatLab includes a Web-based statis- tical software, StatCrunch, within the online assessment platform so that students can easily analyze data sets from exercises and the text. In addition, MyStatLab includes ac- cess to www.StatCrunch.com, a Web site where users can access shared data sets, conduct online surveys, perform complex analyses, and generate compelling reports.

Integration of Statistical Software: Knowing that students often use external statistical software, we make it easy to copy our data sets, both from the ebook and the MyStatLab questions, into software programs. Students also have access to a variety of support tools—Technology Instruction Videos, Technology Study Cards, and Manuals for select titles.

Business Insight Videos: Ten engaging videos show man- agers at top companies using statistics in their everyday work. Assignable questions available.

Supplements

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MyStatLab™

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StatTalk Videos: Fun-loving statistician Andrew Vickers takes to the streets of Brooklyn, New York, to demonstrate important statistical concepts through interesting stories and real-life events. A series of 24 videos with accompany- ing assessment questions and instructor’s guide available.

Expert Tutoring: Although many students describe the whole of MyStatLab as “like having your own personal tu- tor,” students also have access to live tutoring with quali- fied statistics instructors via MyStatLab.

And, MyStatLab comes from a trusted partner with educa- tional expertise and an eye on the future.

Knowing that you are using a Pearson product means know- ing that you are using quality content. That means that our eTexts are accurate, that our assessment tools work, and that our questions are error free. And whether you are just getting started with MyStatLab, or have a question along the way, we’re here to help you learn about our technolo- gies and how to incorporate them into your course.

To learn more about how MyStatLab combines proven learning applications with powerful assessment, visit www.mystatlab. com or contact your Pearson representative.

MyStatLab™ Ready to Go Course (access code required): These new Ready to Go courses provide students with all the same great MyStatLab features that you’re used to, but make it easier for instructors to get started. Each course includes pre- assigned homework and quizzes to make creating your course even simpler. Ask your Pearson representative about the details for this particular course or to see a copy of this course.

MathXL® for Statistics Online Course (access code re- quired): MathXL® is the homework and assessment engine that runs MyStatLab. (MyStatLab is MathXL plus a learning man- agement system.)

With MathXL for Statistics, instructors can

Create, edit, and assign online homework and tests using algorithmically generated exercises correlated at the objec- tive level to the textbook.

Create and assign their own online exercises and import TestGen tests for added flexibility.

Maintain records of all student work, tracked in MathXL’s online gradebook.

With MathXL for Statistics, students can

Take chapter tests in MathXL and receive personalized study plans and/or personalized homework assignments based on their test results.

Use the study plan and/or the homework to link directly to tutorial exercises for the objectives they need to study.

Knowing that students often use external statistical soft- ware, we make it easy to copy our data sets, both from the ebook and the MyStatLab questions, into software like StatCrunch™, Minitab, Excel, and more.

MathXL for Statistics is available to qualified adopters. For more information, visit our Web site at www.mathxl.com or contact your Pearson representative.

StatCrunch™ StatCrunch is powerful Web-based statistical software that allows users to perform complex analyses, share data sets, and generate compelling reports of their data.

Collect. Users can upload their own data to StatCrunch or search a large library of publicly shared data sets, spanning almost any topic of interest. An online survey tool allows users to quickly collect data via Web-based surveys.

Crunch. A full range of numerical and graphical methods allow users to analyze and gain insights from any data set. Interactive graphics help users understand statistical con- cepts, and are available for export to enrich reports with vi- sual representations of data.

Communicate. Reporting options help users create a wide variety of visually appealing representations of their data.

Full access to StatCrunch is available with a MyStatLab kit, and StatCrunch is available by itself to qualified adopters. For more information, visit our Web site at www.StatCrunch.com or con- tact your Pearson representative.

PowerPoint® Lecture Slides. These presentations are avail- able for each chapter. The PowerPoint slides provide an instruc- tor with individual lecture outlines to accompany the text. The slides include many of the figures and tables from the text. In- structors can use these lecture notes as is or can easily modify the notes to reflect specific presentation needs.

The Student Edition of Minitab® is a condensed edition of the professional release of Minitab statistical software. It offers the full range of statistical methods and graphical capabilities, along with worksheets that can include up to 10,000 data points. Individual copies of the software can be bundled with the text (ISBN-10: 0-13-143661-9; ISBN-13: 978-0-13-143661-9).

JMP Student Edition is an easy-to-use, streamlined version of JMP desktop statistical discovery software from AS Insti- tute, Inc., and is available for bundling with this text (ISBN-10: 0-321-89164-3; ISBN-13: 978-0-321-89164-8).

s p e C i A l t h A n k s I would like especially to thank Deirdre Lynch and Marianne Ste- panian for their support and guidance with this second edition, Dana Bettez for keeping the entire project on schedule, Erin Lane for promoting this edition with her marketing skills, Joe Vetere for producing professional quality figures and screenshots in the textbook, Peggy McMahon for coordinating the production process with PreMediaGlobal, Sonia Ashraf for coordinating the reviewers, Amy Ray for improving my writing with her edito- rial skills, Cathy Zucco-Teveloff and Dirk Tempelaar for their accuracy checking to ensure an error-free textbook, and David Stephan for the development of the Excel Add-in, PHStat.

And most importantly, I would like to thank my wife, Debbie, for her constant love and support throughout the entire project.

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This textbook was shaped and contributed to by dozens of class testers, consultants, focus group participants, and reviewers who gave generously of their time, expertise, and creativity.

R e v i e w e R s f o R t h e s e C o n D e D i t i o n

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Acknowledgments

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Student Reviewers/Focus Group Participants for the First Edition This textbook was shaped and contributed to by hundreds of students who gave generously of their time, expertise, and creative unique perspective as students. This is who this textbook is for and we appreciate your input. Of particular note, are the following students:

Jay Bhatt, Goldey-Beacom College Rafie K. Boghozian, East Tennessee State University Lisa Bosman, College of Menominee Nation Anne Drougas, Dominican University Ashraf El-Houbi, Lamar University Douglas Evans, The Ohio State University Kevin Foster, City College of New York Fred Howington, Washtenaw Community College Molly Jensen, University of Arkansas Constance McLaren, Indiana State University Lan Ma Nygren, Rider University Eric Ozsen, San Francisco State Universtiy

Gopi Parimi, Goldey Beacom College Robert Patterson, Penn State–Erie, The Behrend College Nora Perrone, Newmann University Phillip Pfaff, Canisius College Hamid Pourmohammadi, California State University–

Dominquez Hills Alex Riker, Boston College Sanjani Shah, Goldey-Beacom College Tim Sullivan, Southern Illinois University Edwardsville Minghe Sun, The University of Texas at San Antonio Cori Thompson, Madison Area Technical College Linda Williams, Tidewater Community College

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D e A R s t u D e n t s

Information overload is a recognized mark of our time, and the daily data tsunami isn’t going to recede any time soon. There’s no doubt you will be exposed to an unprecedented volume of data and information throughout your career. While many around you are crying “I’m overwhelmed!” or even “man overboard!” you can take a different, much better approach. Developing the skills needed to organize and interpret important information is a key to success, especially in business.

The business statistics course you are beginning will give you invaluable tools that enable you to use data to make good business decisions. For example, if you were a manager for AT&T or Verizon, could you conclude that the dropped-call rates for the two companies are significantly different based on a sample of customers from each company? Would that data alone be enough to affect how you manage your business? Or, would you want to break it down further and analyze it by some additional other factor, the locations of dropped calls perhaps? After you have completed Chapter 10, you’ll be able to analyze data such as this and answer these questions with a high degree of reliability.

Throughout your business career you will find people at every level making decisions that have real consequences affecting business profitability and the jobs of real people. A skilled data analysis can give you an amazing window on aspects of a business that are simply not apparent without it.

I have written this textbook with you, the student, as my focus and the overall goal of helping you succeed both in this course and later in your career. I developed my approach over many years of teaching, and on the basis of that experience, I offer the following advice to help you achieve your own goals:

Make it to class regularly. If you don’t, you’ll miss the details that help you master the subject. No textbook, no matter how well written, can take the place of your instructor and the classroom interaction. Seriously, go to class!

Take advantage of the “Your Turn” problems placed throughout the chapters. Solving them will reinforce key concepts and let you know if you fully understand the material. The solutions to the problems at the end of each chapter give you immediate feedback, so I encourage you not to peek at the answers before you solve the problems or you won’t really know how you are doing!

Solve as many of the chapter problems as you can before exams. Business statistics is not a subject that rewards cramming or winging it. The saying “practice makes perfect” holds true in this field, and working through a variety of problems will build your confidence during the semester. I have provided solutions to all the even-numbered problems in Appendix B at the end of the book.

I hope that you come to share my enthusiasm for the value of business statistics this semester and that what you learn in this course contributes to your future success in the business world.

b o b D o n n e l ly

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Index of Applications

A Automobiles

Ages of cars, 617 Auto financing market, 376 Automotive dealerships, 283 Average age of a Scion driver, 347 Average gasoline consumption, 429 Average rental price of car, 447–448 Average score of car, 137 Avis rental car, 139, 301, 522 BMW owners, 71 Brand loyalty, 492 Car battery, 285 Choosing cars, 189 Drive-through lane, 71 Engine size of sedans, 18-35, 130–131, 695 Explorer, 704 General Motors (GM), 617, 782–783 Hybridcars, 241, 617 Interstate speeds, 645 Inventory, 241 Latitude Tour tire in, 366 Maintenance and repairs, 416 Miles per gallon (MPG), 18–35, 695 Selling price of new domestic cars, 18–41 Snow plow controls, 17-15–17-16 Speed of cars traveling, 282 U.S. market share, 603–604 Variability of the tire life, 18-43, 281, 491, 643, 645 Volkswagen, 376 Warranty for a particular SUV, 268

B Business and Economics

Acme Markets, 18-41, 131, 702 Advertisers, 563 AOL Internet business, 812–813 AutoTrader, 18-14 Average monthly revenue, 68 Average price of television, 268 Average prices for flat-panel televisions, 783 Avon Products, 539–540 Bagel Boys, 540 Banking fees, 346 Bank of America stock prices, 90 Best Buy, 483, 659 Bookstore, 238 Buy.com, 18-35, 683 Car dealership, 174 Cell phone companies, 431 Chlorine shipping, 813 Christmas cards, 17-34 Christmas trees, 17-5–17-6, 283 Clear Bags, 377 Consumers product demand, 206 Crisis in the real estate market, 423–424 Currency exchange rates, 91 Daily demand for birthday cakes, 17-16 Demand for bananas, 17-11 Downturn in the housing market, 69 E-commerce Web site, 284 Euro currency crisis, 376 European debt crisis, 416 Fannie Mae, 411 Fitness apparel, 17-10–17-11 “Focus on the economy,” 585

French pension system, 423 Full-time small-business owners, 429, 431 Gasoline prices, 54, 73, 326, 354 GDP, 54 GM market share, 782 GNC, 702–703 Gold prices, 796 Homeownership, 358 Housing market, 284, 321, 430–431, 565 Inventory, 782 iPhone sales, 101 Island Art, 619, 759 Laptop demand, 187 Laptop prices, 111, 761 Laptop sales, 91 Light bulbs production records, 228 Market cap, 139 Market shares, 69, 70 New product manufacturing, 17-31 Number of days homes were on the market, 137 Printed T-shirt, 17-11, 238 Product demand, 17-15 Purchasing DVDs, 175 Recession, 283, 431, 483, 489 Revenues of cell phone accessories, 301 Sales calls, 616 Selling cameras, 228 Selling prices, 18-31, 18-41, 18-44, 111, 563–564 Small Business Administration (SBA), 429,

431, 490 Small-business owners, 71 Soccer equipment, 17-6 Software sales, 53–54 Spain unemployment, 818–819 Stock markets, 18 Stone Harbor, 18 Supply order, 17-6 Toys Unlimited, 819 US Housing Starts, 73 Variability of the salaries, 659 Video game profits, 65 Warehouse clubs, 540 Weekly purchases of printers, 604

C Construction

Baseball stadium, 67–68 Beach properties, 702 Building full-scale facility, 17-30 Building new facility, 17-35 Crisis in the real estate market, 423–424 Downturn in the housing market, 69 First-time home buyers, 18-40 Handyman, 567 Homeownership, 17-30–17-31, 358, 492 Housing prices, 18-14, 281 Jersey Shore Realtors, 716, 725, 730, 738, 755 Maintenance and repairs, 326–327 Monthly apartment rent, 18-43 Myrtle Beach Homes, 761 New housing, 18-42, 71, 72, 411, 489 Occupancy rate of the complex, 17-15 Precision Doors, 285 Prudential Realtors, 300 RAD Construction Company, 283 US Housing Starts, 73

Consumers AT&T wireless, 140 Average age of, 18–14, 402 Average delivery time for orders, 402 Complaints, 67, 69 Consumer reports, 564 Credit scores, 416 Customer arrivals, 604 Customer education levels, 154 Customer satisfaction, 58–59, 69, 101, 138, 139,

175, 206, 217, 328, 370, 377, 424, 428, 522, 617, 646, 695, 701–702

Customer service, 130, 139, 182 Food orders, 174 Forecasting, 818 Hair salon, 565 At a jewelry kiosk, 126 Music, 564 Optical department, 285 Product demand, 206 Purchasing through e-commerce Web site, 217–218 Shopping experiences, 175 Spending expectations, 817 Waiting time, 115–116, 140, 240, 322, 347, 402, 430 Wireless, 68

D Demographics. See Statistics/ Demographics

E Economics. See Business and Economics Education

Absence, 71, 101, 139, 238, 702 Age of an MBA student, 112 Average back-to-school spending, 447 Average score for the class in exam, 317 Bellevue University, 812 Class sizes, 73, 402, 464, 647 College students’ salaries, 101 Customer survey, 155 Degree-granting institutions, 175 E-book reading, 240 Education levels, 603 Final exam scores, 140, 716, 725, 737–738, 755 For-profit schools, 489 GMAT, 672 Goldey-Beacom College, 41 Grade distribution, 58, 189, 610 Grade inflation, 683, 818 Grade point average (GPA), 187–188, 300,

365–366, 659, 683, 818 Graduate Management Admission Council

(GMAC), 18-41, 376 High school salaries, 567 100-point test score, 633–634 Independence School, 783 Kindergarten, 563 MBA programs, 18-41, 53, 240, 327, 618 MCAT, 474 Missed classes, 18-41–18-42, 139, 702 Multiple-choice quiz, 187, 189, 604 Number of graduating students, 126 Pearson Education, 18-40–18-41

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Question answering, 217 Salary information for graduates, 283 SAT scores, 18-43, 68, 140 Spring breaks, 321 Statistics grades, 68, 326 Student population, 187 Students enrolment, 140, 328, 358 Sussex College, 54 Trends in International Mathematics and Science

Study (TIMSS), 70 U.S. Department of Education, 18-41

Engineering Camera warranty, 619 Compact fluorescent light bulbs (CFL), 375, 647 Electronic traffic counters, 327 Laptop battery, 18-36, 140, 273, 430, 696 Quality control, 328 Rechargeable batteries, 463–464 Smart meters, 429 Sources of electricity, 70 Texas Oil drilling, 17–36

Entertainment. See Sports and Entertainment Environment

Carbon emissions, 91 Engine size of sedans and, 130–131, 695 Environmental Protection Agency (EPA), 18–35,

539, 695 Fuel efficiency, 491 Grass fertilizers, 564 Hurricanes, 240–241 Miles per gallon (MPG), 18-35, 695 Oil spill, 326 Organisation for Economic Co-operation and

Development (OECD), 282 Snowfall, 616

F Finance

Account balance, 70, 429, 647–648 Apple stock price, 796 Auto financing market, 376 Average cash bonus, 818 Average hotel price, 316 Average price for a movie, 315 Baggage fees, 489 Bank of America, 377 Bank of Delaware, 738, 755 Car-loan payments, 241 Citibank, 18-44, 493 Consumer Price Index (CPI), 782 Cost of ink cartridges, 347 Credit cards, 18-21, 41, 175, 268, 402, 567, 610 Credit scores, 18-42, 140, 416, 555–556, 659 Creditworthiness, 18 Earnings per share (EPS), 716, 725, 731, 754–755 Electronic tax filing, 782 Faculty salaries, 760 Fair Isaac, 18-42, 140 Financial aid, 239 Flexible spending accounts (FSA), 17-35 Food costs, 71, 759, 783 French pension system, 423 Gold prices, 796 Health care costs, 54 Hotel room rates, 18-26, 281, 427 Housing prices, 18–14 H&R Block, 18-14 Income of states, 301 Income tax, 175, 241, 267–268, 321–322, 346, 358 iPhone sales, 101 Microloan program, 18-42

MLB payroll, 73, 704 Money saved for retirement, 70 Mortgage rates, 18-40, 238–239, 796 National Foundation for Credit Counseling

(NFCC), 18-21 Negative equity, 327–328, 493 Pay for college, 71 Paying annual dues to play golf, 322 PGA money list, 61 Portfolios, 238 Recession, 283, 431, 483, 489 Sallie Mae, 18-41 “Sandwich generation,” 327 Soffrittos, 759 Starting salaries, 554 Stock markets, 18, 819 Student-loan debt, 18-40, 492, 614 Tax prep costs, 375 Tennis pay out, 17-30 Travel insurance, 238 TurboTax, 18-31 U.S. households, 217

Food and drinks Annual chocolate consumption, 267 Average beer consumption, 354 Average number of minutes of diners, 315–316 Baking loaves, 206 BBQ potato chips, 366 Brands and types of soda, 72 Cafeteria, 813 Coffee industry, 402 Coffee vending machine, 278 Consumption of bottled water, 326 Creamery, 18-26 Customer dining experience, 53, 54 Customers’ orders, 174 Daily demand for birthday cakes, 17-16 Diet cola, 241 Drive-through, 61, 111 Drive through service, 565 Entrée types, 54 Food expenditures, 71, 759, 783 Gatorade, 538–539 Herr Foods, 646 Hot dogs, 672 McDonald’s, 490 Number of cups of coffee, 156 Number of meals eaten in restaurants, 411 Number of pounds of bananas, 522 Order takeout for dinner, 615 Pepsi products, 326 Pizza home delivery time, 188–189, 285, 316, 523 Sugar consumption, 282 Taste test, 494, 648

G Government

Age group of registered voters, 614 Federal government, 584–585 For-profit schools, 489 French government, 492–493 Health care plan, 490 Ohio voters, 615 Union membership, 614 Wall Street Journal poll, 174 White House staff, 428

H Health

Alcohol consumption, 326 American Board of Family Medicine, 131, 701 Aston, 645

BMI, 328, 566 Caloric consumption, 347 CDC guidelines, 614–615 Child-care center, 281 City Hospital, 716, 726, 730, 755 Consumer-Directed Health Plan (CDHP), 327 Exercise, 327, 555 Fertility rate, 138 Flu shot, 239 Health care plan, 490 Health insurance, 175, 317, 322, 375, 584–585 Hours of sleep per night, 281 Low-density lipoprotein (LDL), 490, 678, 679 Neonatal intensive care unit (NICU), 18-40 Never events, 72 Number of hours of sick time, 539 Obamacare, 375 Obesity, 328, 483 Opinion Research Corporation, 54 Patient records, 115, 240, 284, 366, 645 Patient satisfaction, 370 Radon levels, 370 Resident hours, 645 Safety awareness, 18-43 Skin cancer cases, 585 Smoking, 359 Staffing of emergency room (ER) personnel,

18-42–18-43, 563 Traffic accident, 284 Weight-loss program, 474, 565–566, 616 Weight of NFL players, 316

L Labor

Adecco lay off, 217 Age discrimination, 41–42 Age group, 566, 584 American Association of Retired Persons

(AARP), 567 Average hourly wage rates, 375, 377, 463 Average retirement age, 447 Average salary for a federal employee, 325 Average sleep time, 424 Average weekly wages, 282 Average workweek, 327 Citibank interview, 604 Commute time to work for, 278 Earnings, 18 Employee hours, 565 Executive officer positions, 217, 328 Faculty members, 241, 683 Faculty salaries, 760 Flexible spending accounts (FSA), 17-35 Forced-ranking system, 237 Golf cart attendants, 347 Hiring seasonal employees, 69 Long hours resident doctors work, 428 Monthly salary, 18-43, 448 Neumann, 18-35 Organisation for Economic Co-operation and

Development (OECD), 316 Paid vacation days, 354 Safety awareness, 18-43 Satisfaction, 238 Sick days, 615, 618 Spain unemployment, 818–819 Staffing of emergency room (ER) personnel, 563 Unemployment rate, 90, 138, 283, 316, 385, 416 Union, 430 Variability of the salaries, 643 Years of sales experience, 703 Younger employees retirement plans, 429

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M Miscellaneous

Age of an MBA student, 112 American Idol viewers, 782 Annual snowfall, 72 Apartments.com, 217 Average daily rate for a luxury hotel, 267 Average hotel price, 316 Average of holiday shoppers spent, 411 Average of hours a day watching TV, 316, 354 Average utility bills, 411, 447, 567 Calls received by customer support center, 228 Cats living in individual households, 238 Cell phone minutes, 796 Cheez-It crackers, 72 Church’s congregation, 377 Comcast monthly cable bill, 18-35–18-36, 430,

717, 726, 731 Contact rate, 424 Delmarva Power utility bill, 717, 726, 731 Entertainment magazine, 17-34–17-35 Executives’ salary ranges, 156 Favorite national park survey, 156 Flipping a single coin, 174 Generation Y, 321, 326 G&R Publishing, 17-11 Kayak rental company, 17-5 Later-born and first-born one-year-old

children, 464 Lowes home improvement chain, 377 Manuscript typos, 240 Number of bedrooms per house, 206 Number of births in a local hospital, 228 Office occupancy, 483 Participating groups, 182 People living within their households, 101,

155–156, 186, 206, 217, 241, 283, 424 pH readings, 647 Pretzels, 634, 645 Proportion of tips on bills, 366 Purina dog food, 645 Quadrophobia, 619 Retirement ages, 556 Rolling a pair of dice, 174 Scheduling visit, 189 Seating arrangement, 182 Selection of juries, 182 Squirt Squad cleaning service, 725, 730, 738, 755 Standard 52-card deck, 174, 175, 187, 242 Texting while driving, 283–284 Time required for housekeeping, 278 Time required to pass through security, 278 U.S. Postal Service, 819–820 Wedding location, 17-11

P Politics

Campaign strategy, 377 Federal government, 584–585 Political poll, 616 Wall Street Journal poll, 174

S Sports and Entertainment

Actor and actress roles, 189 Average number of viewers, 70 Average price for a movie, 315 Baseball, 69, 91, 116, 125–126, 267, 375, 424,

492, 614

Basketball games, 18–14, 188, 239, 284, 354, 369, 703–704

Casino resort industry, 522–523 Children spent with each media, 70 Christmas M&M’s, 234 Curling, 268 Data analyzing, 428 Driving distances of various golf balls, 539 Fan Cost Index, 138 Fantasy-football, 18-26 Five-card poker, 182 Golf cart attendants, 347 Golf clubs, 72, 645–646 Golf scores, 139, 239, 474, 493, 616, 648, 796 Great Smoky Mountain National Park, 819 Green Bay Packers, 241–242 Heights of the players in NBA, 317 Holiday sales, 493 Lead time, 285 Left-handed golfers, 326 Lottery game, 182, 189 Median ticket prices, 18–31 MLB payroll, 73 MLB pitchers, 760 Movie rankings, 189 Movie theater ticket prices, 556, 795 National Football League games (NFL), 18–31,

111, 138, 301, 316, 563 National Park Service (NPS), 646–647 Netflix subscribers, 65 NFC North Division teams, 182 Orange Lake Resort, 285 Paying annual dues to play golf, 322 PGA money list, 61 Points per game, 61 Regifting, 239 Running play or a passing play, 174, 483 Soccer goals, 615 Spring breaks, 321 2012 Summer Olympics medal count, 610 Summer travelers, 411 Super Bowl, 189, 238, 796, 820 Swimming, 18–43, 140, 182 Tennis pay out, 17–30 Touchdown passes, 111–112, 604 Vacation properties, 618 Video game profits, 65 Villanova basketball, 240, 716–717, 730–731,

738 Weight of NFL players, 316 Whale watching tours, 241 Winter Olympic sport, 268

Statistics/Demographics Annual incomes, 115 City populations, 125 Contributions, 376 Dr. Gregg’s office, 604 Eastern Time Zone, 284 High school graduates, 268 High school salaries, 567 Spending on Valentine’s Day gifts, 316 Wall Street Journal poll, 174 ZoneDiet.com, 186–187

T Technology

AOL Internet business, 812–813 Apple, 239–240, 428 Average usage of mobile devices, 268 Blackberry devices, 228, 322

Canon copier, 17-16 Cost of ink cartridges, 347 Customizing a desktop computer online, 182 Data usage and speeds, 138, 647 E-commerce Web site, 284 Establishing four-digit pin number, 189 Facebook, 321, 489 Federal Communications Commission, 618 Google Maps, 239–240 Google’s Android operating system, 218,

228, 269 HDTV, 283 Internet access, 174 Malicious attacks, 430 Owning a tablet, 268 Owning cell phone, 429 Research in Motion (RIM), 228 Revenues of cell phone accessories, 301 Search engine, 53 Smart meters, 429 Social network usage, 137, 321, 416, 428, 489,

492, 611 Spam e-mails, 615 Swype technique, 473–474 Text messages, 138, 429–430 T-Mobile, 73 TurboTax software, 18–31 Use of smartphones and laptops, 72 Verizon Wireless, 18-40, 68, 139, 376, 429,

647, 701 Wi-Fi service, 416 Work-related emails, 618

Temperature Department of Energy, 679 Freezer temperatures, 634 Hawaii ocean temperatures, 42, 377

Transportation Aircraft turnaround time, 285 Airfare benchmarking, 18-26 Air France, 617, 646 Airline arrivals, 68 Airline performance, 73 Airlines incentives, 375 American Airlines, 268, 646 Average gasoline consumption, 429 Average number of miles driven, 267 Average speed of drivers, 316 Commuting to school, 642–643 Cruising to different destinations, 464 Delta Airlines, 138, 237 Domestic airfares, 377 Driver’s license, 328 Driving distance, 521–522 Federal Aviation Administration (FAA), 521 Fuel efficiency, 491 GMAC, 376 No-shows, 17-35–17-36, 41 Number of traffic tickets, 228 On-time and late flights, 187, 490, 616 Orbitz, 18-30–18-31 Package weight, 522 Pizza delivery time, 316, 523, 634 Southwest Airlines, 239 Spirit Airlines, 17-35–17-36 Travel insurance, 238, 647 TSA times, 73, 463, 642 United Airlines, 240 Urban traffic congestion, 555 U.S. air traffic control system, 354 Wi-Fi service, 416

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I n t H I s c H A P t e R , Y o U w I l l l e A R n :

■ How statistics is used in the business world.

■ About the sources and data and the methods for collecting it.

■ How to classify data by the level of measurement.

■ To distinguish between time series and cross-sectional data.

■ To distinguish between descriptive and inferential statistics.

■ About the ethical implications of misusing statistics.

An Introduction to Business Statistics

C H A P t E R 1 m A P

1.1 Business Statistics and Their Uses | 2

1.2 Data | 4

1.3 Descriptive and Inferential Statistics | 11

1.4 Ethics and Statistics—It’s a Dangerous World of Data Out There | 14

C H A P t E R

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2 C h a p t e r 1 | An Introduction to Business Statistics

Welcome to the world of statistics! Although some of you might be excited about learning statistics, others of you are probably less than thrilled about it. Perhaps you have to take “sadistics” because your major requires it. However, before you write off the value of this learning opportunity, let’s discuss the role that statistics can play in your life. In today’s world, everyone is a consumer of statistics. By this, I mean that you are continually sur- rounded by data and statements about those data in an effort to influence you to purchase something, vote for someone, or change your opinion about an issue. Consider the follow- ing examples:

• When CBS Sports announces that 109 million viewers tuned in to the 2013 Super Bowl, do you understand the method used to determine this number? (How does CBS know that I, you, or anyone else watched the game?)

• When we hear on TV that President Barack Obama’s approval rating is 56% and in small print see {4%, do you understand the significance of this percentage?

• When you read an advertisement claiming that a new product is recommended by four out of five doctors, do you question the validity of the claim? (For instance, were the doctors paid for their endorsements?)

• When an online survey reports that Canon digital cameras are preferred to Nikon, does it concern you that the majority of the respondents could be Canon loyalists who repeat- edly voted, skewing the results, or that the survey was conducted on a Canon user’s forum?

Never before in the history of humankind have people had more data and information at their fingertips than you do at this moment. Statistics can have a powerful effect on our feel- ings, our opinions, and the decisions that we make in our personal and professional lives. As a result, it’s very important that the statistics we report are both accurate and unbiased to ensure that they are properly utilized.

1.1 Business Statistics and Their Uses Statistics is the mathematical science that deals with the collection, analysis, and presenta- tion of data—data that can then be used as a basis for inference and induction. Business statistics are statistics applied to the business world in an effort to improve people’s decision making in fields such as marketing, operations, finance, and human resources, to name a few. Let’s look at a few examples of how business statistics can help an organization’s decision makers.

Marketing research Organizations rely heavily on business statistics when they conduct marketing research to determine what consumers want. For example, Kellogg’s could perform a taste test to determine if consumers prefer the company’s Cheez-It crackers to Nabisco’s Cheese Nips. (Being a life-long Cheez-It addict, I know where my vote will go.) Kellogg’s could also gather information about each consumer participating in the test in an effort to determine if the people who prefer one brand to the other share similar characteristics. This would

be useful information for Kellogg’s future marketing efforts.

advertising Television networks set their advertising rates for commercials based on the sizes of viewing audiences. The networks receive the information from Nielsen Media Research, which col- lects data from approximately 25,000 U.S. households. The sample of households surveyed has been carefully selected so that the results can be used to infer the viewing habits of the entire country. Statistics are used to ensure the sample is properly chosen and to process the data into meaningful information for the networks. Based on this information, CBS was able to charge a record $3.8 million for a 30-second Super Bowl ad in 2013.

Statistics is the mathematical science that deals with the collection, analysis, and presentation of data, which can then be used as a basis for inference and induction.

In the 1980s, Marriott conducted an extensive survey to see how potential customers felt about the company’s current hotel offerings. Based on the results, Marriott designed a new hotel chain known as Courtyard by Marriott, which has been a huge success.

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C h a p t e r 1 | 1.1 Business Statistics and Their Uses 3

Operations Statistics can also be used to help businesses operate better. Quality control, for example, is a vital concern for all successful organizations. The combination of business statistics and quality control is a marriage made in heaven. If properly implemented, statistics can help manufacturing and service organizations monitor their processes and determine when quality problems begin to occur. For instance, Kellogg’s can use statistics to determine if my Cheez- Its are overbaked or too salty. (Based on the box I’m sampling from at this moment, I’d say the company is doing an excellent job with its statistical quality control.)

Finance When I began writing this textbook, the U.S. economy was in the middle of a deep reces- sion. Part of the economic downturn was due to the poor lending practices of banks, particu- larly in the mortgage industry. If used properly, business statistics is an excellent tool to help banks identify consumers who are good credit risks and those who are not based on charac- teristics such as income, education, and home-ownership. For example, Fair Isaac Corpora- tion (FICO) developed the credit scoring system (FICO score) currently used by the industry and which is based on a variety of statistical techniques.

Business statistics are statistics applied to the business world to help improve decision making.

Stats in practice: Careers in Statistics

It’s a great time to be in the job market if you have an interest and aptitude for statistics. This is due to an increase in demand for the skills of individuals who have developed a level of statistical literacy. The biggest driver for this demand is a global society that is both data-rich and data-dependent as technology advances at an explosive rate. In recognition of this development more than 150 professional organizations, including the American Statistical Asso- ciation, have designated 2013 as the International Year of Statistics.

There are abundant employment opportunities in today’s business world that require expertise in statistics. The following jobs are just a small sampling of the type of jobs available to people who have mastered this skill set.

• Transportation statistician—Our daily lives are surrounded by transportation systems that require ongoing attention by people with statistical talents. Examples include:

° Delivery companies such as FedEx and UPS, which are al- ways seeking opportunities to improve the efficiency of their delivery systems by analyzing data.

° Airlines such as American Airlines rely on statistics for their yield management system which determines airfares on a continuous basis. You can thank these systems when you experience unexpected swings in airfare as you book your next vacation!

• Financial analyst—Using a variety of statistical tools, financial analysts provide investment advice to businesses and individu- als. Banks and investment firms are examples of organizations seeking this type of position.

• Actuary—In order to maintain profitability, insurance companies need actuaries to analyze risk factors for their customers in

order to establish appropriate premiums for their service. Statistical techniques play a major role in this analysis. Hospitals, banks, and government agencies are other examples of organiza- tions that rely on these skills.

• Sports statistician—Many professional baseball, basketball, and football teams have hired statisticians in an effort to gain a competitive advantage. Examples include:

° Assistant General Manager Peter Brand, who was drama- tized in the 2011 film Moneyball, used statistical analysis to help the Oakland Athletics assemble a competitive Major League Baseball team during the 2002 season. The Boston Red Sox won the World Series in 2004 (their first since 1918!) relying on some of the same statistical modeling used by the Oakland Athletics.

° The Memphis Grizzlies, an NBA team, recently hired a vice president of basketball operations because of his statistical expertise.

• Political analyst—A great deal of attention has been paid to predicting political outcomes, such as elections, using statistical tools.

° Nate Silver is an American statistician who has developed an impressive reputation for his accurate predictions in the political arena. I recommend visiting his Web site, fivethirtyeight.blogs.nytimes.com, to gain some insight into the benefits of statistics in this field.

And of course I’d be remiss not mentioning one last type of employment in the field of statistics—education. I can personally testify that teaching statistics to students and writing books to help them learn has been a very rewarding experience.

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4 C h a p t e r 1 | An Introduction to Business Statistics

1.2 Data Data are the foundation of the field of statistics and can be defined as the values assigned to specific observations or measurements. If I’m collecting data on my wife’s snoring behavior, I can do so in different ways. I can measure how many times Deb snores over a 10-minute period. I can measure the length of each snore in seconds. I could also measure how loud each snore is using a descriptive phrase like “That one sounded like a bear just waking up from hibernation” or “Wow! That one sounded like a sea lion calling for its young.” (How a sound like that can come from a person who can fit into a pair of size 2 jeans is beyond me.)

In each instance, I would be recording data on the same event but in a different form. In the first instance, I would be measuring a frequency, or number of occurrences. In the second instance, I would be measuring duration, or length of time. And in the final instance, I would be measuring the event by describing its volume using words rather than numbers.

However, data all by themselves are not particularly useful. By definition, data are just the raw facts and figures that pertain to a measurement of interest. Information, on the other hand, is derived from the facts for the purpose of making decisions. One of the major rea- sons to use statistics is to transform data into information. For example, Table 1.1 shows my golf scores over a two-month period. (For those of you who are non-golfers, lower scores are better.)

Information is data that are transformed into useful facts that can be used for a specific purpose, such as making a decision.

Secondary data are data collected by someone else that you are “borrowing.”

The Internet has become a rich source of data for statistics published by various industries. I once found a Japanese study on the effect of fluoride on toad embryos. Before this discovery, I had no idea toads even had teeth, much less a cavity problem!

Data are values assigned to observations or measurements and are the building blocks of statistical analysis.

Each individual golf score would be considered a data point. By themselves, the data points have limited value, other than to suggest that I should not quit my day job. (To quit my day job and become a professional golfer, I would need a score in the 65–70 range.) To these data, let’s add the fact that in my desperate efforts to buy a better golf game, I quietly purchased a brand new driver on July 1, thinking my wife would not notice a new club in my golf bag (she did). We may be able to conclude, with the help of statistics, that the purchase helped improve my game (imagine a doubtful look on Deb’s face). In order to answer this question, we will need to employ a statistical analysis that will be covered later in this text. Stay tuned as I use statistics hopefully to convince my wife that I really needed this new club!

the Sources of Data We classify the sources of data into two broad categories: primary data and secondary data. Secondary data are data somebody else has collected and made available for others to use. The U.S. government collects and publishes a variety of data that are readily available on- line. The U.S. Department of Labor collects mountains of data on topics such as consumer prices, inflation, unemployment, and productivity. The home page for the department’s Web site is shown in Figure 1.1.

Every 10 years the U.S. Department of Commerce conducts a nationwide census to gather a wide variety of data related to the country’s population. The data are used by Con- gress to make decisions about the funding for community services throughout the United States. The data are used to adjust the number of representatives each state is allowed to elect to Congress. You can find census data on the Department of Commerce’s Web site.

{ table 1.1 | Golf-Score Data

date score

June 13 94

June 20 96

June 27 93

July 10 89

July 16 86

July 24 89

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C h a p t e r 1 | 1.2 Data 5

The U.S. Geological Survey (USGS) provides an impressive assortment of scientific information that is used to manage water, energy, biological, and mineral resources across the Earth. For instance, did you know there are 250 species of squirrels in the world? If you don’t believe me, you can look the information up at the USGS’s Web site and become the local squirrel “expert” in your area.

The main drawback of using secondary data is that you have no control over how the data were collected. People tend to believe anything that’s in print, even if it’s not true. (You believe me, don’t you?) Some of it is wrong, and, as you will learn later in the chapter, some of it is deliberately biased. The advantage to secondary data is that they are cheap (some- times free) and that they are immediately available. For someone looking for data quickly, secondary data provide instant gratification (assuming the data are accurate, of course).

Primary data, on the other hand, are data collected by the person or organization that eventually uses the data. This type of data can be expensive to acquire, but the main advan- tage is that primary data are your data, and you have nobody else to blame but yourself if you make a mess of it. You can obtain primary data in many ways, such as by direct observa- tion, via experiments, and through surveys.

Direct observation is a method of gathering data while the subjects of interest are in their natural environment, often unaware they are being watched. Observing wild animals stalk- ing their prey in the forest or teenagers at the mall on Friday night are two examples. (Or are they the same example?) The advantage of this method is that the subjects will unlikely be influenced by the data collection process.

A focus group is a direct observational technique whereby individuals are often paid to discuss their attitudes toward products or services in a group setting controlled by a mod- erator. For example, Fisher Price heavily relies on focus groups of both adults and children to obtain valuable feedback on new toy ideas. The participants are aware they are being observed.

In an experiment, subjects are exposed to certain treatments and the data of interest are recorded. An experiment that tests the use of a new medical drug is an example. Two dif- ferent groups would be established: The first group would receive the new drug; the second group would be the control group. People in the control group would be told they are getting the new drug but would in fact get a placebo with no medication. The reactions from each group would be measured and compared to determine whether the new drug is effective.

The benefit of experiments is that they allow statisticians to control factors that could influence the results, such as the gender, age, and education of a participant. One major con- cern about collecting data through experiments is that the response of the subjects might be influenced by the fact they are participating in a study.

A survey involves directly asking people a series of questions. Surveys can be adminis- tered by e-mail, via the Web, through snail mail, face to face, or over the telephone. (It’s the telephone survey I’m most fond of, especially when I get the call just as I’m sitting down to dinner, getting into the shower, or finally making some progress on the chapter I’m writ- ing.) The questionnaire needs to be carefully designed to avoid any bias that could affect

Primary data are data that you have collected for your own use.

Direct observation is a method of gathering data while the subjects of interest are in their natural environment, often unaware they are being watched.

A focus group is a direct observational technique whereby individuals are often paid to discuss their attitudes toward products or services in a group setting controlled by a moderator.

{ figure 1.1 The Home Page of the Bureau

of Labor Statistics

In an experiment, subjects are exposed to certain treatments and the data of interest are recorded.

A survey involves directly asking people a series of questions and can be administered by e-mail, via the Web, through snail mail, face to face, or over the telephone.

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6 C h a p t e r 1 | An Introduction to Business Statistics

participants’ responses or confuse them. Bias can occur when a question is stated in a way that encourages or leads a respondent to a particular answer. For example, “Wouldn’t you agree that all drivers should wear a seat belt?” is a biased question. The influence the survey itself has on the responses of participants can also affect the quality of the data collected. Some participants will respond in a way they feel the survey would like them to. Figure 1.2 shows a portion of a survey I developed for users of the Claymont Community Center in Delaware. To encourage respondents to participate, an effective survey will state its purpose in the beginning, ask questions in a clear and concise manner, and place the more personal demographic questions last—when the respondents feel more comfortable with the process.

We relied on SurveyMonkey to develop this textbook. Faculty and students tested the book before publication and provided valuable feedback through this Web site. I’m a satisfied customer!

Research has shown that a question posed in a positive tone will tend to evoke a more positive response. A question posed in a negative tone will tend to evoke a more negative response. A good strategy is to pre-test your questionnaire before releasing it to the actual participants.

Claymont Community Center Health Clinic

Customer Service Survey

Dear Customer,

Our goal is to provide you with the best service possible. You can help us by taking a few minutes to answer the following questions about the Claymont Community Center (CCC). This valuable information will be used to help us to improve our services in the near future. The information gathered with this survey will be kept strictly confidential. Thank you for helping us. (Please place this form in the Customer Survey Box in the Health Clinic. If you need more time, you may return the form on your next visit or ask for a stamped envelope to return it by mail.)

1. How long have you been using the Health Clinic? (Check one) _____ Less than one month _____ 1 to 6 months _____ 6 months to a year _____ 1 to 3 years

2. How satisfied are you with the quality of service being provided by the medical staff at the Health Clinic? _____ Very Satisfied _____ Satisfied _____ Dissatisfied _____ Very Dissatisfied Comment__________________________________ __________________________________________

3. Please provide the following personal information. Again, all information will remain confidential.

Zip code of your residence:________________

Gender: _____ Male _____ Female Age: _____

Marital Status: _____ Married _____ Single _____ Separated _____ Divorced _____ Widowed

Are you the head of your household? _____ Yes _____ No

What is your annual gross income? _____ Less than $10,000 _____ $30–40 _____ $10–20 _____ $40 _____ $20–30

{ figure 1.2 An Example of a Survey

Online surveys are a convenient way to acquire data. Companies such as SurveyMonkey provide people with a low-cost way to design surveys, collect responses, and analyze the data. The SurveyMonkey Web site claims that 80% of the Fortune 100 companies are users of the service. However, there are challenges to Internet surveys, which will be discussed later in this chapter.

To test your understanding of data sources, I encourage you to spend a few minutes answering the questions in the following Your Turn section.

Identify if the data required for each example are primary or secondary. For primary data, determine the best way in which the data should be collected. In other words, should the data be collected via obser- vation, experiment, or survey?

1. Apple would like to measure the satisfaction levels of customers who purchased its new iPad product.

2. Pepsi would like to determine if consumers prefer the taste of Diet Pepsi to Diet Coke. 3. Cleveland State University needs to determine the current inflation rate to determine the annual

salary increases for its staff for the upcoming year. 4. McDonald’s would like to determine the average wait time for customers who use its drive-through

windows during the lunch hour.

Y O u r t u r n # 1N

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C h a p t e r 1 | 1.2 Data 7

the two Main types of Data Another way to classify data is by whether they are quantitative or qualitative:

• Quantitative data use numerical values to describe something of interest either by measuring it (such as its weight, height, or distance) or by counting it (such as the number of customers or repeat customers a business receives).

• Qualitative data use descriptive terms to measure or classify something of interest. One example of qualitative data is the name of a respondent in a survey and his or her level of education. Mathematical operations such as addition, subtraction, multiplication, and division cannot be performed on this type of data.

Classifying Data by their Level of Measurement Another important way to classify data is by the way the data are measured. This distinction is critical because it affects which statistical techniques we can use in our analysis of the data. The four levels of measurement are nominal, ordinal, interval, and ratio.

A nominal level of measurement deals strictly with qualitative data assigned to prede- termined categories. One example is the gender of a survey respondent, with the catego- ries being male and female. This type of data is referred to as nominal data, or categorical data. It does not allow us to perform any mathematical operations on it, such as adding or multiplying. We can only give the data names and categorize them. (The word nominal ac- tually means “pertaining to names.”) We also cannot rank-order the data in any way from highest to lowest. An example is the state in which the survey respondent resides, such as Delaware or New Jersey (although I would try to rank my home state of Delaware on top.)

Other examples of nominal data are zip codes and telephone numbers, which can’t be added, subtracted, or placed in a meaningful order of greater than or less than. Even though the data consist of numbers, they are handled just like qualitative data. Nominal data are considered the lowest level of data, and, as a result, the statistical techniques used to analyze them are the most restrictive.

An ordinal level of measurement can be conducted on data that are on the next level up on the food chain. Ordinal data has all the properties of nominal data, with the added feature that we can rank-order the values from highest to lowest. The following example explains ordinal measurement: Recently, I felt my manhood challenged by two neighbors who claimed their lawnmowers were faster than mine. Naturally, this had to be settled by a lawnmower race down our street, Gaebel Lane. Sadly, I present Figure 1.3, which shows that I lost the race. My neighbor Tom came in first (1), my neighbor Scott came in second (2), and I came in third (3).

Quantitative data use numerical values while qualitative data relies on descriptive terms to describe something of interest.

Nominal data are data that are described as a category or a label. Examples are gender (male or female), marital status (married, single, divorced, widowed), or yes/no responses.

Ordinal data have all the properties of nominal data, with the added feature that we can rank-order the values from highest to lowest.

{ figure 1.3 An Example of Ordinal

Measurement: Tom, Me, and Scott (from left to right) on Our Mowers

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8 C h a p t e r 1 | An Introduction to Business Statistics

We still can’t perform mathematical operations on this type of data, but we can say that Tom’s lawnmower was faster than mine and Scott’s. However, we cannot say how much faster because we didn’t record the times of the lawnmowers. We just noted who came in first, second, and third. Ordinal data that have been collected do not allow us to make mea- surements between the categories or to say, for instance, that Scott’s lawnmower is twice as fast as Bob’s. For that, we need a different type of data. (In case you were wondering, I have been unsuccessful at convincing Deb that I need more horsepower to restore our family honor at the Second Annual Gaebel Lane Lawnmower Race.)

Education level is another example of ordinal data. A master’s degree is ranked higher than a bachelor’s degree, which in turn is ranked higher than a high school diploma. How- ever, we are unable to measure the difference between these degrees in a meaningful, mathematical way. For instance, it would not be accurate to claim the difference between a master’s and a bachelor’s degree is more than the difference between a bachelor’s and a high school degree. A property of ordinal data is that the differences between categories are not meaningful and, therefore, cannot be measured.

Ordinal data can also be numerical. One example of numerical ordinal data is when we rate movies with one, two, three, or four stars. Although we can order the movies by their ratings, we can’t, for example, claim that a four-star movie is four times as good as a one- star movie.

The interval measurement level is yet a higher level of measurement. It measures inter- val data, which are strictly quantitative. Temperature measurements in degrees Fahrenheit are an example of interval data. With this level, we can measure the differences between the categories with actual numbers in a meaningful way. For instance, 70°F is 5 degrees warmer than 65°F. However, multiplication and division can’t be performed on this level of data. Why not? Simply because we cannot argue that 100°F is twice as warm as 50°F. The logic of this claim becomes more obvious when we convert the temperatures to the Celsius scale. The same two temperatures convert to 38°C and 10°C, respectively, so the twice-as-warm argument does not hold true. To help explain this, try baking a cake at twice the recommended temperature in half the recommended time. Yuck!

Another characteristic of interval data is that they do not have a true zero point. The term true zero point means that a zero data value indicates the absence of the object be- ing measured. For instance, 0°F and 0°C do not represent the absence of temperature, even though it may feel like it.

Your grade point average (GPA) is another example of interval data. We can measure the difference between a 4.0 and a 2.0 GPA by simply subtracting the two values. However, it would not be an accurate statement to claim that a 4.0 student is twice as smart as a 2.0 student. Also, GPA has no true zero point because a 0.0 GPA does not indicate the absence of a grade point average.

The most versatile of data types is the ratio level of measurement. Ratio data are as good as it gets as far as data are concerned. Examples of this type of data are ages, weights, prices, and salaries. Ratio data have all the features of interval data, with the added benefit of a true zero point. For instance, a zero salary indicates the absence of any salary. With a true zero point, we can say that a person who is six feet in height is twice as tall as a three-foot person or that a 20-year-old person is half the age of a 40-year-old.

The distinction between interval and ratio data is a fine line. To help identify the proper scale, use the “twice as much” rule. If the phrase “twice as much” accurately describes the relationship between two values that differ by a multiple of two, then the data can be consid- ered to be ratio level.

Table 1.2 summarizes the properties of the four levels of data measurement, and Figure 1.4 shows the relationship between the levels of data measurement and the two main types of data,

For example, dollars are considered to be ratio data because $20 is twice as much as $10.

Interval data, which are strictly quantitative, allow us to measure the differences between the categories with actual numbers in a meaningful way.

Types of Data

Qualitative Quantitative

Nominal Ordinal Interval Ratio

{ figure 1.4 The Two Main Types of Data

and Their Corresponding Levels

Ratio data have all the features of interval data, with the added benefit of having a true zero point.

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C h a p t e r 1 | 1.2 Data 9

which creates data measurement scales. As we explore different statistical techniques later in this book, we will revisit these different measurement scales. You will discover that specific techniques require certain types of data.

We can summarize classifying data by the level of measurement with the following baseball analogy:

• Nominal data. The numbers on the players’ uniforms would be nominal data because these values are simply a label and are not used for any type of measurement. A player with the number 30 is not necessarily better than the player wearing number 15.

• Ordinal data. The batting order of the players would be considered ordinal data because we can rank-order the players going to the plate to hit. However, there is no useful or meaningful way to measure the differences between a player hitting 5th and a player hitting 6th.

• Interval data. The birth years of the players would be considered interval data. We can say that a player born in 1984 is three years older than a player born in 1987, but the “twice-as-much rule” does not hold true for calendar years. A person born in the year 1000 is not twice as old as a person born in the year 2000. This is because there is no true zero point with calendar years. The year 0 does not indicate the ab- sence of age or time; it is merely an arbitrary reference point.

• Ratio data. The number of home runs hit by the best hitter on the team is an example of ratio data. A player with 20 home runs has twice as many as a player with 10 home runs. There is a true zero point for these data because zero home runs represent the absence of home-run hits.

Because this is such an important concept, which will be used later in the course, take a few minutes to answer the questions in the following Your Turn section before moving on in the chapter.

{ table 1.2 | The Four Levels of Data Measurement: A Summary

level description example

Nominal Arbitrary labels for data No ranking allowed

Zip Codes (19808, 76137)

Ordinal Ranking allowed No measurable meaning to the

number differences

Education level (master’s degree, doctorate degree)

Interval Meaningful differences No true zero point

Calendar year (2014, 2015)

Ratio Meaningful differences True zero point

Income ($48,000, $0)

Identify the type of data (qualitative/quantitative) and the level of measurement for each of the following data sources:

a. Your IQ score b. The price for one gallon of gasoline c. The letter grade earned in your statistics course d. The number of boxes of Frosted Flakes on the shelf of a grocery store e. The types of cars driven by students in your class

Y O u r t u r n # 2N

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10 C h a p t e r 1 | An Introduction to Business Statistics

time Series vs. Cross-Sectional Data Data can also be classified as time series or cross-sectional. Table 1.3 shows the average unemployment rates for the Unites States, California, Delaware, Michigan, and Texas dur- ing the month of January for indicated years. Time series data are values that correspond to specific measurements taken over a range of time periods. Each of the last five columns in Table 1.3 represents a time series of unemployment rates. Figure 1.5 shows a line graph of one of the columns, U.S. unemployment rates from 2008 until 2012. The upward trend in unemployment in 2008 and 2009 reflects the economic recession that occurred during that time period.

Time series data are values that correspond to specific measurements taken over a range of time periods.

{ table 1.3 | Unemployment Rate Data, 2008–2012

year united states % ca % de % mi % tx %

2008 4.9 5.9 3.8 7.1 4.3

2009 7.6 10.1 6.7 11.6 6.4

2010 9.7 12.3 8.8 13.7 8.2

2011 9.0 12.4 8.5 10.7 8.3

2012 8.3 10.9 7.0 9.0 7.3

Based on: http://www.bls.gov.

Cross-sectional data are values collected from a number of subjects (firms, individual, states, regions, and so forth) during a single time period.

Unlike time series data, which measure values over a range of time, cross-sectional data are values collected from a number of subjects (firms, individual, states, regions, and so forth) during a single time period. Each of the five rows labeled 2008–2012 in Table 1.3 represents a cross-section of unemployment rates for the entire country and the various states during a single year. Figure 1.6 shows a column graph of the 2012 U.S. un- employment rates for the country and the states in the table. Michigan’s high unemploy- ment rate reflects the impact that the recession had on the automotive industry. Both time series and cross-sectional data told their own stories from the same data set—the data shown in Table 1.3.

4 5 6 7 8 9

10 U.S. Unemployment Rates

Pe rc

en t

2008 2009 2010 2011 2012 Year

{ figure 1.5 A Time Series Graph of U.S.

Unemployment Rates, 2008–2012

0 US CA DE

Location

2012 Unemployment Rates

MI TX

2 4 6 8

Pe rc

en t

12 { figure 1.6 A Cross-Sectional Graph of 2012

Unemployment Rates

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C h a p t e r 1 | 1.3 Descriptive and Inferential Statistics 11

Try the next Your Turn problems to test your understanding of time series and cross- sectional data.

Identify the following data as either time series data or cross-sectional data.

1. The following table shows the final win/loss records of the National League Central Division for Major League Baseball at the end of the 2012 season.

Team Wins Losses

Cincinnati Reds 91 65

St. Louis Cardinals 88 74

Milwaukee Brewers 83 79

Pittsburgh Pirates 79 83

Chicago Cubs 61 107

Houston Astros 55 107

2. The following table shows the final win/loss records of my favorite Major League Baseball team that I love to hate, the Pittsburgh Pirates, for the 2005–2012 seasons. (This is so painful to list.)

Year Wins Losses

2005 67 95

2006 67 95

2007 68 94

2008 67 95

2009 62 99

2010 57 105

2011 72 90

2012 79 83

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Y O u r t u r n # 3N

1.3 Descriptive and Inferential Statistics The two main branches of statistics are descriptive and inferential. The main focus of descrip- tive statistics is to summarize and display data. Descriptive statistics plays an important role today because of the vast amount of data readily available at our fingertips. With a basic com- puter and an Internet connection, we can access volumes of data in no time at all. Being able to summarize all of these data accurately in order to get a look at the “big picture,” either graphically or numerically, is the job of descriptive statistics.

The majority of statistics that you encounter in your day-to-day life are descriptive. Figures 1.5 and 1.6 are examples of descriptive statistics for unemployment rates. An average is another popular descriptive statistic, such as the average exam score in your statistics class. The statistic provides your instructor with an overview of how well your class performed on the test.

Descriptive statistics are useful, but have limitations. By summarizing large quantities of data, you lose information. For instance, your current GPA is a descriptive statistic sum- marizing your academic performance. However, this value alone does not provide informa- tion about the level of difficulty of the courses you have taken or how your GPA compares to your classmates’.

The purpose of descriptive statistics is simply to summarize, or display, data so we can quickly get an overview of the information.

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12 C h a p t e r 1 | An Introduction to Business Statistics

As important as descriptive statistics are to us number crunchers, we really get excited about inferential statistics. This category covers a large variety of techniques that allow us to make actual claims about a population based on a sample of data. We use the term popu- lation in statistics to represent all possible subjects that are of interest to us in a particular study. The term sample refers to a portion of the population that is representative of the population from which it was selected. Figure 1.7 illustrates this important concept.

Inferential statistics allows us to make claims or conclusions about the data based on a sample of them.

A population represents all possible subjects of interest. A sample is a subset of a population.

Because much of Alaska is a rainforest, mosquitoes grow large—so large that Alaskans refer to them as the state’s unofficial bird!

{ figure 1.7 The Relationship Between a

Population and a Sample Population

Sample

Many populations are too large or inaccessible to measure completely, so looking at a sample of the population of interest is the only feasible way to estimate it. For instance, it would not be practical to measure the average life span of the mosquito population in Alaska. We would need to measure the life span of a random sample of mosquitoes and use these data to estimate the average life span of the entire population.

Suppose I would like to estimate the average weight per box of my favorite cereal, Kellogg’s Frosted Flakes, in the population. (No, I don’t own stock in Kellogg’s, but I really like Frosted Flakes, so I want the company to stay in business.) For Kellogg’s sake, I hope the population average is very close to the target weight of 18 ounces per box. Filling boxes with less than 18 ounces could result in dissatisfied customers (me included), whereas overfill- ing boxes is giving away free cereal and will reduce profits. To accomplish this task, I would randomly select a sample of cereal boxes and measure their average weight. Let’s say the average weight is 18.2 ounces. With inferential statistics, I could determine the likelihood that the average weight of the population of boxes is actually 18 ounces if my sample aver- age were 18.2 ounces. Even if the likelihood were very low, say less than 1%, I would still have to conclude that the average weight of the population of boxes is more than 18 ounces. I would then proceed to check Kellogg’s filling process to see why the company is giving away free Frosted Flakes. This example demonstrates the usefulness of statistics in quality control programs. Figure 1.8 summarizes the process of inferential statistics.

{ figure 1.8 Using Inferential Statistics for

Quality Control Purposes: An Example

Assume the unknown average weight per box of cereal in the population is 18.0 oz.

Select a random sample of cereal boxes from the population.

Calculate the sample average—18.2 oz.

Using inferential statistics, determine that the probability

of the actual population average being 18.0 oz is less than 1%

when the sample average is 18.2 oz.

Conclude that the actual population average is greater than 18.0 oz.

The following are other examples of inferential statistics:

• Based on a recent sample, I am 95% certain that the average age of my customers is between 32 and 35 years.

• Based on a random survey, the average salary for male employees in a particular job cat- egory across the country is higher than for comparable female employees.

Descriptive statistics describe a sample while inferential statistics draw a conclusion about a population.

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C h a p t e r 1 | 1.3 Descriptive and Inferential Statistics 13

In summary, descriptive statistics reports only on the observations at hand and nothing more. Inferential statistics makes a statement about a population based solely on results from a sample taken from that population.

Two more terms need to be mentioned here regarding populations and samples. Data used to describe something of interest about a population are known as parameters. An example of a parameter is the average weight of the population of Frosted Flake boxes (assumed to be 18 ounces). Data that describe a sample from that population are known as statistics. In our previous example, the sample average of 18.2 ounces would be considered a statistic.

Let’s check your understanding of this important concept with the following questions.

Data that describe a characteristic about a population are known as parameters.

Data that describe a characteristic about a sample are known as statistics.

Identify each of the following as either descriptive or inferential statistics.

1. Households with children under the age of 18 are more likely to have access to the Internet (77%) than family households with no children (68%).

2. Julie, who cuts and styles hair in her home salon, had 23 customers last week. 3. The average exam score for my statistics exam was an 88. 4. A recent poll showed that 57% of Americans had a favorable opinion of the president of the United

States.

Y O u r t u r n # 4N

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Stats in practice: using Smartphone Data to Monitor health

With today’s technology, our smartphones have become a rich data source of our day-to-day activities. Researchers at the Massachusetts Institute of Technology used this data from student volunteers and found they could reliably conclude when they were ill. Sick students moved around much less than nor- mal and depressed students had fewer calls and text messages. As a result of this research, a company called Ginger.io was formed whose mission is to provide tools to monitor the health

of individuals using their smartphone data. When there is a disruption in someone’s normal daily pattern, Ginger.io can send a message similar to a car’s “check engine light” to that person’s friends or doctors. In order to accomplish this task, Ginger.io must analyze massive amounts of smartphone data and link it to health patterns. Sounds like this company can really use the talent of a few good statisticians! Anybody out there interested?

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14 C h a p t e r 1 | An Introduction to Business Statistics

{ figure 1.9 Misusing Statistics: A Graphical

Example, U.S. Unemployment Rates, 2008–2012

0 5

10 15 20 25 30 35 40

U.S. Unemployment Rates

Pe rc

en t

2008 Year

2011 20122009 2010

A biased sample is a sample that does not represent the intended population and that can lead to distorted findings. Biased sampling can occur either intentionally or unintentionally.

1.4 Ethics and Statistics—It’s a Dangerous World of Data Out There

People often use statistics when attempting to persuade you to accept their point of view. Perhaps they are motivated to convince you to purchase something or to support a cause they are advocating. This motivation can lead to the misuse of statistics in several ways. One of the most common misuses is choosing a sample that ensures results consistent with the desired outcome rather than choosing a sample representative of the population of interest. This is known as having a biased sample.

Suppose I’m a lobbyist for the golf industry, and my task is to convince Congress to es- tablish a national golf holiday. During this honored day, all government and business offices would be closed so that we all could run out to chase a little white ball into a hole that’s way too small with sticks purposely designed by the evil golf club companies to make the task impossible. Sounds like fun to me! Somehow, I need to demonstrate to Congress that the average level-headed American is in favor of this. Here is where the genius part of my plan lies: Rather than survey the general American public, I pass out my survey form only at golf courses. But wait—it only gets better. I design the survey to read as follows:

We would like to propose a national golf holiday. Everybody gets the day off from work and can play golf all day. (You would not need permission from your spouse to play.) Are you in favor of this proposal?

A. Yes, most definitely.

B. Sure, why not?

C. No, I would rather spend the entire day at work.

P.S. If you choose C, your golfing privileges will be permanently revoked everywhere in the country for the rest of your life. We are dead serious.

Granted, this is an extreme example, but my point is that survey results can easily be manipulated by how we ask questions and who is responding to them. And often, major decisions rest on these results.

Another way to misuse statistics is to make differences seem greater or lesser depending on the way the data are graphed. Earlier in the chapter, I presented Figure 1.5, which showed a rather dramatic increase in the U.S. unemployment rate in 2008 and 2009. Figure 1.9 graphs the same data, but the increase in unemployment appears much less dramatic. The trick I employed (no pun intended) was to flatten the graph by simply changing the scale of the y-axis, which shows the unemployment rate. In Figure 1.5, the scale ranges from 4% to 10%, whereas in Figure 1.9, the scale has been expanded to 0%–40%. By simply expanding the scale, I made the increase in unemployment rates seem less pronounced.

Many of the polls we see on the Internet represent another potential misuse of statistics. I’m sure you have visited Web sites that encourage you to vote on a question of the day. The results of these informal polls are unreliable simply because those collecting the data have no control over who responds to the polls or how many times they respond. As stated earlier,

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C h a p t e r 1 | 1.4 Ethics and Statistics—It’s a Dangerous World of Data Out There 15

a valid statistical study depends on selecting a sample representative of the population of interest. This is not possible when any person surfing the Internet can participate in the poll. Even though most of these polls state that the results are not scientific, it’s still a natural human tendency to be influenced by the results we see.

Many times statistical mishaps occur simply due to poor decision making. One of the most famous snafus took place during the 1936 presidential race. The Literary Digest, an influential magazine at the time, conducted a presidential election poll, based on question- naires mailed to readers and potential readers, and published the results before the election. Unfortunately, the magazine predicted that Alf Landon would win the election over Franklin D. Roosevelt. Even if history is not your best subject, you should realize that somebody at the Digest had some serious explaining to do after that election day. The problem was that the Digest drew its sample from phone books and automobile registrations. People with phones and cars in 1936 tended to be wealthier Republicans (who voted for Landon) and were not representative of the entire voting population.

A more recent example of poor decision making that was based on statistical analysis occurred in the 1980s when Coca-Cola was feeling the pressure because Pepsi’s market share was rising. To halt the trend, Coca-Cola decided to reformulate its nearly 100-year-old recipe for Coke. The new formula was sweeter than both the old formula and Pepsi’s prod- uct. After scoring well in the taste tests, Coke decided to replace its original product with the new formula. However, the test failed to consider two important factors. First, people might prefer a sweeter taste when they are drinking small quantities of a beverage as they were at time of the test. However, they don’t like to drink larger quantities of it. Second, and more importantly, Coke failed to measure the psychological impact making such a major change to a well-established product would have on consumers. Many Coke drinkers were outraged that the product they loved to drink suddenly no longer existed. After a huge public outcry, Coca-Cola admitted its mistake and brought the original formula back to market as Coke Classic and relabeled the new formulation New Coke. What a mess! The lesson here is that even large, successful corporations can misuse statistics, which can result in poor business decisions.

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■ Statistics is the process of collection, analysis and presentation of data, which can then be used for inference.

• Business statistics helps improve people’s decision making in fields such as marketing, operations, finance, and human resources.

■ Data are the foundation for the field of statistics.

■ Data is defined as the values assigned to specific observations or measurements.

■ Information is derived from data for the purpose of making decisions.

■ Data sources are either primary or secondary.

• Secondary data are data collected by someone else that you are “borrowing.”

• Primary data are data that you have collected for your own use through direct observation, experiments, or surveys.

• Direct observation focuses on gathering data while the subjects of interest are in their natural environment.

• In an experiment, subjects are exposed to certain treatments and data of interest are recorded.

• A survey involves directly asking subjects a series of questions.

■ Data can be classified in different ways.

• Quantitative—characterized by numerical values.

• Qualitative—characterized by terms describing the data.

• By the way they are measured.

• A nominal level of measurement deals strictly with qualitative data.

• Ordinal level data have all the properties of data measured at the nominal level, but we can rank-order the values from highest to lowest.

• At the interval level of measurement, the differences between the data can be measured with actual numbers giving measurable meaning to the differences.

• Ratio data have all the properties of interval data, along with a true zero point.

• A zero data value indicates the absence of the object being measured.

• As time series or cross-sectional data.

• Times series data are values that correspond to a specific measurement over a range of time periods.

• Cross-sectional data are values that correspond to a specific time period.

■ Two main branches of statistics are as follows.

• Descriptive statistics summarize and display sample data.

• Inferential statistics makes a statement about a population based on sample data from that population.

■ Statistics can be misused by the following.

• Employing a biased sample that is not representative of the population of interest.

• Manipulating a graph by adjusting the scale of the data.

• Relying on Internet surveys with no control over who responds or how many times they respond to the survey.

Chapter 1 Summary

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Business statistics. Statistics applied to the business world to help improve people’s decision making in fields such as marketing, operations, finance, and human resources.

Biased sample. A sample that does not represent the intended population and that can lead to distorted findings.

Cross-sectional data. Data values that correspond to a specific time period.

Data. The values assigned to specific observations or measurements. Descriptive statistics. The branch of statistics that focuses on

summarizing, or displaying, data. Direct observation. A method of gathering primary data by

observing subjects of interest in their natural environment. Experiment. A method of gathering primary data by exposing

subjects to certain treatments and recording the data of interest. Focus group. A direct observational technique whereby individu-

als are often paid to discuss their attitudes toward products or services in a group setting controlled by a moderator.

Inferential statistics. The field of statistics that allows us to make claims or conclusions about a population based on a sample of data from that population.

Information. Data that are transformed into useful facts that can be used for a specific purpose, such as making a decision.

Interval level of measurement. Quantitative data that can mea- sure the difference between categories with actual numbers, giving measurable meaning to the differences.

Nominal level of measurement. Qualitative data observations that are assigned to predetermined categories.

Ordinal level of measurement. Data that have all the properties of nominal data but that also permit the rank-ordering of values from highest to lowest.

Parameter. Data that describe characteristics of a population. Population. A population represents all possible outcomes or

measurements of interest. Primary data. Data collected by the person or organization that

eventually uses the data. Qualitative data. Data that use descriptive terms to measure or

classify something of interest. Quantitative data. Data that use numerical values to describe

something of interest. Ratio level of measurement. Data that have all the features of

interval data, with the added benefit of a true zero point. The term true zero point means that a zero data value indicates the absence of the object being measured.

Sample. A subset of a population. Secondary data. Data that somebody else has collected and

made available for others to use. Statistic. Data that describe characteristics about a sample. Statistics. The mathematical science that deals with the collec-

tion, analysis, and presentation of data, which can then be used as a basis for inference and induction.

Survey. A method of collecting primary data by asking subjects a series of questions.

Time series data. Data values that correspond to a specific measurement over a range of time periods.

Identify the type of data (qualitative/quantitative) and the level of measurement for Problems 1.1 through 1.18. Explain your choice. 1.1 The average monthly temperature in degrees Fahrenheit for

the city of Wilmington, Delaware, throughout the year

1.2 The average monthly rainfall in inches for the city of Wilmington throughout the year

1.3 The education level of survey respondents

Level Number of Respondents

High school 168

Bachelor’s degree 784

Master’s degree 212

1.4 The marital status of survey respondents

Status Number of Respondents

Single 28

Married 189

Divorced 62

1.5 The ages of the respondents in a survey

1.6 The genders of the respondents in a survey

1.7 The years in which the respondents to a survey were born

1.8 The voting intentions of the respondents in a survey classi- fied as Republican, Democrat, or Undecided

1.9 The race of the respondents in a survey classified as White, African American, Asian, or Other

1.10 The performance rating of employees classified as Above Expectations, Meets Expectations, or Below Expectations

1.11 The uniform number of each member on a sports team

1.12 A list of the graduating high school seniors by class rank

1.13 Final exam scores for your statistics class on a scale of 0 to 100

1.14 The state in which the respondents in a survey reside

1.15 SAT scores for graduating high school students

1.16 Movie ratings: G, PG, PG-13, R

Chapter 1 Problems

Chapter 1 Key Terms

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1.17 In order to promote the development of electric cars, the government has proposed rating passenger vehicles on a letter scale of A–D based on fuel efficiency and emissions. The following five vehicles were given these ratings:

Vehicle Rating

Nissan Leaf A + Toyota Prius A - Ford Focus B

Jeep Grand Cherokee C + Ferrari 612 Scaglietti D

1.18 S&P provides ratings for the creditworthiness of countries around the world on a letter scale from AAA (highest) to D (lowest). The following table shows the S&P rating for various countries in January 2013.

Country Rating

Germany AAA

United States AA + France AA + Italy BBB + Spain BBB - Portugal BB

Identify the data shown in Problems 1.19 and 1.20 as either time series or cross-sectional.

1.19 The following table shows the closing price of the three major stock markets on February 4, 2013:

Stock Market Closing Price

Dow Jones Industrial 13,880

NASDAQ 3,131

S&P 500 1,496

1.20 The following table shows the closing price of the Dow Jones Industrial stock market during 2012:

Date Closing Price

September 28, 2012 13,437

October 31, 2012 13,096

November 30, 2012 13,026

December 31, 2012 13,104

Problems 1.21 and 1.22 refer to the following data, which show the median weekly earnings of full-time workers by gender over a five-year period:

Year Men Women

2008 $798 $638

2009 $819 $657

2010 $824 $669

2011 $854 $691

2012 $832 $684

1.21 Identify the time series data in the table.

1.22 Identify the cross-sectional data in the table.

Problems 1.23 and 1.24 refer to the following data, which shows the number of photographs sold at Island Art, a retail store in Stone Harbor, New Jersey, by print-size of a four-year period:

Year Print-Size

8 : 10 11 : 14 13 : 19 2010 117 102 160

2011 92 125 146

2012 105 111 133

2013 112 89 138

1.23 Identify the cross-sectional data in the table.

1.24 Identify the time series data in the table.

Identify each of the following as either descriptive or inferential statistics.

1.25 The average salary of a random sample of 50 high school teachers in 2013 was $52,400.

1.26 Based on a random sample of hotels in Chicago and a ran- dom sample of hotels in Atlanta, it was concluded that the average cost of a hotel room in Chicago was greater than one in Atlanta.

1.27 A study has concluded that the average credit card debt of college graduates has increased from 2012 to 2013.

1.28 The average Amazon.com rating of the book The Complete Idiot’s Guide to Statistics by 26 reviewers is 4.6 on a scale of 1 to 5.

1.29 The average American viewer watches 151 hours of TV per month.

1.30 Seventy-eight percent of customers at the Holiday Inn hotel in Dover, Delaware, arrived before 6 pm last week.

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1. Primary data through a survey 2. Primary data through an experiment

3. Secondary data 4. Primary data through direct observation

Y O u r t u r n # 2 a) Quantitative/Interval. The differences between IQ scores are meaningful, but

there is no true zero point because an IQ of 0 does not indicate the absence of intelligence.

b) Quantitative/Ratio. The differences between prices are meaningful, and there is a true zero point because gasoline that is $0 per gallon is free!

c) Qualitative/Ordinal. You can rank letter grades, but the differences between the grades cannot be consistently measured.

d) Quantitative/Ratio. The differences between inventory levels are meaningful, and there is a true zero point because zero boxes on the shelf indicates an absence of the product.

e) Qualitative/Nominal. The types of cars are merely labels with no ranking or meaningful differences.

Y O u r t u r n # 3 1. Cross-sectional 2. Time series

Y O u r t u r n # 4 1. Inferential statistics, because it would not be feasible to survey every

household in the country. These results would be based on a sample of the population and used to make an inference on the entire population.

2. Descriptive statistics, because this data describes a sample, the number of customers last week.

3. Descriptive statistics, because the average exam score describes a sample, which is students in my class.

4. Inferential statistics, because it would not be feasible to survey every American in the country. These results would be based on a sample of the population and used to make an inference about the entire population.

Chapter 1 Solutions to Your Turn

Y O u r t u r n # 1

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I n t h I s c h a p t e r , y o u w I l l l e a r n t o :

■ Use Excel to display data with charts and graphs.

■ Construct a frequency distribution and histogram.

■ Construct and interpret polygons.

■ Display qualitative data using bar charts and pie charts.

■ Display quality control issues with a Pareto chart.

■ Organize data in a contingency table.

■ Present the frequency of data using a stem and leaf display.

■ Show the relationship between two variables using scatter plots.

Displaying Descriptive Statistics

C h a p t e r 2 M a p

2.1 The Role Technology Plays in Statistics | 23

2.2 Displaying Quantitative Data | 25

2.3 Displaying Qualitative Data | 42

2.4 Contingency Tables | 55

2.5 Stem and Leaf Display | 59

2.6 Scatter Plots | 61

2 C h a p t e r

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22 C h a p t e r 2 | Displaying Descriptive Statistics

In Chapter 1, we learned that the purpose of descriptive statistics is to summarize or display data so that we can quickly obtain an overview of the information of interest. In this chapter, we will explore the methods by which data can be displayed in tables and graphs to facilitate better understanding. I find that some students (not you, of course) often underestimate the importance of presenting graphical data and information in a clear and convincing manner to lend support to their ideas. There’s no better way to make your point than with a creative chart your audience can quickly absorb.

By contrast, presenting lots of data in a table format is a surefire way to lose your audi- ence. Our brains do not process information very effectively when faced with reading a list of numbers. We do much better at comprehending data when we are presented with a picture such as a graph or chart. For instance, Table 2.1 shows the selling price, in thousands of dol- lars, of 60 two-bedroom condominiums in Myrtle Beach, South Carolina, during a recent time period. Most people, including me, do not have the attention span to read every single number in this table.

{ table 2.1 | Data Presented as a Table: Condominium Prices in Myrtle Beach

$281.0 $262.1 $291.2 $277.1 $294.5 $298.2

$247.8 $344.9 $322.5 $258.0 $226.3 $320.0

$306.0 $287.5 $228.2 $209.6 $220.0 $218.4

$247.2 $395.0 $291.4 $329.4 $335.5 $325.9

$446.3 $302.3 $257.7 $278.8 $296.8 $343.8

$254.1 $193.0 $174.4 $380.9 $198.6 $270.7

$199.0 $298.8 $268.3 $193.3 $383.6 $348.3

$295.5 $254.1 $301.2 $375.0 $359.9 $260.4

$227.8 $360.3 $299.0 $274.1 $195.9 $337.8

$246.4 $292.6 $237.0 $339.0 $300.2 $271.9

A much better approach would be to construct a chart that provides an effective snap- shot of these 60 data points. You don’t need to work as hard to understand the data in the chart, which is always a good thing, in my book.

By examining the chart in Figure 2.1, we can see the range that contains the highest number of condominiums is $250,000 to under $300,000. This information is not readily observable by merely looking at the data in the table.

{ figure 2.1 Data Presented Graphically:

Condominium Prices in Myrtle Beach

This chapter will provide you with a solid foundation to display descriptive statistics in a way that is sure to impress your future boss and colleagues. Using Microsoft Excel 2013, you will be cranking out fancy graphs like the one shown in Figure 2.1 in no time.

In addition to creating graphs, as a reader and a consumer of information you will also find yourself interpreting data that have been presented to you in a graphical format.

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C H A P t E R 2 | 2.1 The Role Technology Plays in Statistics 23

Publications such as the Wall Street Journal and USA Today present data graphically to cap- ture the attention of their readers and to summarize complex issues efficiently. Being well versed in the art of interpreting graphical data will make you a better decision maker.

2.1 The Role Technology Plays in Statistics

Technology plays a major role in the field of statistics. Throughout this book, we will be using Excel 2013 for Windows to assist you in your efforts to construct various charts and graphs as well as to perform many of the calculations required in statistics.

When I was a youthful engineering undergraduate student in the 1970s, the words personal computer had no meaning. Windows were something you looked through, and no self-respecting consumer electronics company would dream of naming itself after a fruit. I performed calculations on a clever gadget fondly known as a slide rule. For those of you who weren’t even alive during this time period, I’ve included a picture of a slide rule in Figure 2.2.

{ figure 2.2 A Slide Rule Circa 1975.

As you can see, this device looks like a ruler on steroids. It can perform all sorts of mathematical calculations, but is far from being user friendly. During my freshman year in college, I purchased my first handheld calculator, a Texas Instrument model that could per- form only the basic math functions. It was the approximate size of a cash register. I proudly wore this device like a badge of honor on my belt, which I suspect contributed to my current back problems. Oh well, the things we do to look cool!

Fortunately, we have advanced from the Dark Ages and now have awesome, user- friendly computing power at our fingertips. Powerful programs such as SAS, SPSS, Minitab, and Excel are readily accessible to those of us who don’t know a thing about computer pro- gramming. Yet the programs allow us to perform some of the most sophisticated statistical analysis known to humankind. The next section describes how we can prepare Excel 2013 to assist us with various ways to display data and perform calculations required in statistics.

statistical Analysis Using Excel 2013 In addition to the basic statistical functions found in Excel 2013 for Windows, we will be relying on two Excel Add-ins:

• Data Analysis ToolPak—available as a standard feature in Excel (Windows version).

• PHStat—developed by Pearson Education—provides statistical tools not available in Excel’s Data Analysis ToolPak. PHStat is available for download with an access code. Go to www.pearsonhighered.com/phstat to download PHStat with your code. (See page xiv if you need a code.)

If you are a Mac user (like myself), there are some compatibility issues that we have to address for which there is good news and bad news. First the bad news—Excel for Mac (2011 or older) is not compatible with Data Analysis ToolPak. Now the good news—the version of PHStat mentioned in the previous paragraph is compatible with Excel 2011 for Mac. Also (more good news), the basic Excel statistical functions used in this book are

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24 C h a p t e r 2 | Displaying Descriptive Statistics

compatible with Windows and Mac versions. Whenever Excel instructions shown in this textbook are not compatible with the Mac version, I will make a note to direct you to the textbook’s Web site mentioned earlier to download Excel for Mac instructions (for ver- sions 2011 and older) for that particular task. Often, you will be able to rely on PHStat for these situations.

I also need to address those using Excel 2010 for Windows with this textbook. For the most part, the instructions used in this textbook can be easily applied to your software. Whenever there is a discrepancy between the two versions, I will make a note to direct you to the book’s Web site to download Excel 2010 instructions.

We are now ready to activate Excel’s Data Analysis ToolPak (Windows version only) by using the following steps:

1. With Excel 2013 open, click on the File tab found in the upper left corner of your com- puter screen.

2. Click Options shown in the left margin in Figure 2.3. This will open the Excel Options dialog box shown in Figure 2.4.

3. Select Add-Ins in the left margin as shown in Figure 2.4.

4. Click on Go at the bottom of the screen.

{ figure 2.3 Excel’s File Drop-Down Menu

{ figure 2.4 Excel’s Options

5. Select the check boxes for Analysis ToolPak and Analysis ToolPak - VBA in the pop- up menu and click OK as shown in Figure 2.5.

6. Select the Data tab. Click on Data Analysis on the right as shown in Figure 2.6. The Data Analysis pop-up menu should appear in the spreadsheet.

Your Excel software is now ready to perform all sorts of statistical magic as we explore various techniques throughout this textbook. At this point, you can click Cancel and close out Excel. Each time you open Excel in the future, the Data Analysis option will be available under the Data tab.

{ figure 2.5 Excel’s Add-Ins Pop-Up Menu

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 25

2.2 Displaying Quantitative Data In Chapter 1, we made a distinction between quantitative data, which consist of numerical values, and qualitative data, which use descriptive categories. In this section, we will discuss the various methods used to display quantitative data. In a later section, we will cover the qualitative data.

Constructing a Frequency Distribution One of the simplest ways to describe data is through a frequency distribution, which shows the number of data observations that fall into specific intervals. I’ll use the following ex- ample to demonstrate the construction of a frequency distribution using quantitative data. The 50 values in Table 2.2 represent the number of iPads sold at an Apple store in each of the past 50 days.

A frequency distribution is a table that shows the number of data observations that fall into specific intervals.

{ figure 2.6 Excel’s Data Analysis Pop-Up

{ table 2.2 | The Number of iPads Sold in Each of 50 Days

4 2 3 2 5

5 1 3 3 2

3 2 2 3 2

2 2 3 0 1

3 1 1 5 4

1 2 4 3 5

2 0 0 3 2

3 3 3 2 2

0 4 2 4 3

1 1 4 0 1

Table 2.3 on the next page shows these same data as a frequency distribution. The first column in the table lists all the possible daily values that occurred over the 50 days: 0, 1, 2, 3, 4, or 5 iPads sold. The second column counts the number of observations for each of these values during the past 50 days. Notice that the total number of frequencies (50) shown in Table 2.3

Three iPads were sold on this Particular day.

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26 C h a p t e r 2 | Displaying Descriptive Statistics

matches the total number of observations in Table 2.2. Because the number of iPads sold per day is a numerical value rather than a descriptive value, this is an example of a frequency distribution summarizing quantitative data.

Here’s a helpful way to distinguish between continuous and discrete data. Continuous data are often the result of measuring observations, whereas discrete data often involve counting them.

Discrete data are values based on observations that can be counted and are typically represented by whole numbers.

Continuous data are values that can take on any real numbers, including numbers that contain decimal points.

We call each row in a frequency distribution table a class. In our iPad example, the first class represents the number of days that no iPads were sold, the second class shows the num- ber of days one iPad was sold, and so on.

Before we move on, I have one more data type to throw your way. The data in Table 2.2 are known as discrete data, which means the values represent something that has been counted. Discrete data typically take on whole numbers such as 0, 1, 2, 3, and so on. Be- cause selling partial iPads is not a good business decision let alone possible, values such as 2.8 or 4.3 would not be considered discrete data in this example.

Data that can take on any real numbers, including numbers that contain decimal points, are known as continuous data. Continuous data are usually measured rather than counted. Examples of continuous data are weight, time, and distance. I’ll show you an example using continuous data later in this chapter.

The acid test for determining which type of data we are dealing with is to ask ourselves, “How many data values can be found in a specific interval of numbers?” Because discrete data can be counted, they have a finite number of values within an interval, whereas continu- ous data have an infinite number of values available. Let’s take the interval 0–3 as an exam- ple. Remembering that iPad sales were discrete data, only four possible values of iPad sales can occur within this interval (0, 1, 2, or 3). If we were measuring time, which is continuous, there are an infinite number of possible values within this interval (2.0 minutes, 2.04 min- utes, 2.046 minutes, and so on). The only limitation in the number of continuous data values is the level of precision of our measuring instrument.

Discrete vs. continuous data is an important concept throughout this textbook, so let’s take a minute to test your understanding in the following Your Turn box.

A class is a category in a frequency distribution.

{ table 2.3 | Frequency Distribution for the Number of iPads Sold in the Past 50 Days

number sold per day frequency

0 5

1 8

2 14

3 13

4 6

5 4

Total 50

Two iPads were sold in 14 of the 50 days that the data were collected.

Identify the following as either discrete or continuous data:

1. The number of students absent from your statistics class today 2. The number of hours a light bulb burned before going out 3. The number of light bulbs that burned for more than 1,000 hours in a quality control program 4. The number of minutes an airline flight was late 5. The number of red cars a dealer has on its sales lot 6. The number of inches in a person’s height 7. The number of runs scored by the Pittsburgh Pirates tonight

Y o u r t u r n # 1N

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 27

relative Frequency Distributions Another way to display frequency data is by using a relative frequency distribution. Rather than displaying the number of observations in each class, this method calculates the pro- portion of observations in each class by dividing each frequency by the total number of observations. A relative frequency distribution tells us what proportion, or fraction, a class’s observations are relative to the total observations made. Table 2.4 shows the relative fre- quency distribution for our iPad example.

Relative frequency distributions display the proportion of observations of each class relative to the total number of observations.

Notice that the fractions in the relative frequency distribution add up to 1.00. Because of rounding, the total relative frequency will sometimes not add up to exactly 1.00, but it should be very close.

Note that the last cumulative relative frequency should equal 1.00 (or very close to 1.00 due to rounding).

{ table 2.4 | Relative Frequency Distribution for the Number of iPads Sold in the Past 50 Days

number sold per day frequency relative frequency

0 5 5>50 = 0.10 1 8 8>50 = 0.16 2 14 14>50 = 0.28 3 13 13>50 = 0.26 4 6 6>50 = 0.12 5 4 4>50 = 0.08

Total 50 1.00

According to Table 2.4, two iPads were sold on 28% of the days.

Cumulative relative Frequency Distributions A cousin of the relative frequency distribution (a pun I couldn’t resist) simply totals the proportions of each class as you move down the column. It is called the cumulative relative frequency distribution. A cumulative relative frequency distribution provides you with the proportion of observations that are less than or equal to the class you are looking at—the ac- cumulated proportion, in other words. This technique would be useful if the manager of the Apple store wanted to determine the percentage of days that three or fewer iPads were sold. The cumulative relative frequency distribution for our iPad example is shown in Table 2.5.

A cumulative relative frequency distribution totals the proportion of observations that are less than or equal to the class at which you are looking.

{ table 2.5 | Cumulative Relative Frequency Distribution for the Number of iPads Sold in the Past 50 Days

number sold per day

frequency

relative frequency

cumulative relative frequency

0 5 0.10 0.10

1 8 0.16 0.10 + 0.16 = 0.26

2 14 0.28 0.26 + 0.28 = 0.54

3 13 0.26 0.54 + 0.26 = 0.80

4 6 0.12 0.80 + 0.12 = 0.92

5 4 0.08 0.92 + 0.08 = 1.00

Total 50 1.00

Looking at Table 2.5, we can now conclude that three iPads or less were sold on 80% of the business days.

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28 C h a p t e r 2 | Displaying Descriptive Statistics

excel’s FreQUenCY Function Now that I’ve tortured you with constructing frequency distributions by hand, I’ll demon- strate how to use Excel’s FREQUENCY function to construct a distribution and generate a graph known as a histogram. A histogram is simply a graph showing the number of obser- vations in each class of a frequency distribution. Figure 2.1, the chart in which we graphed condominium prices, is an example of a histogram.

Figure 2.7A shows the iPad data from Table 2.2 along with a column labeled “Bins,” which is Excel’s strange term for the classes in our distribution. Use the following steps to construct the frequency distribution.

1. Open the Excel file iPad sales.xlsx or type in the data in Column A and Column C as shown in Figure 2.7A (except Column D).

2. Place your cursor in Cell D2 and highlight cells D2:D7. This is where our frequency distribution will be placed.

3. With cells D2:D7 highlighted, type = FREQUENCY(A2:A51,C2:C7). Do not hit the Enter key just yet!

4. Hit the Control+Shift+Enter keys together. 5. Voila! (pronounced “vwa-la”) The frequency distribution appears before your very eyes

in the highlighted cells in Column D, as shown in Figure 2.7A (circled in red).

A histogram is a graph showing the number of observations in each class of a frequency distribution.

{ figure 2.7a Constructing a Frequency

Distribution with Excel

Using a histogram to Graph a Frequency Distribution The following steps will guide you through the process of constructing a histogram in Excel for Windows. See the textbook’s Web site for Excel for Mac instructions. We’ll start with Figure 2.7B, which shows the iPad data and the bins column.

1. Select the Data tab, and click on Data Analysis in the upper right corner. The Data Analysis pop-up menu will appear as shown in Figure 2.7B.

2. In the pop-up menu, select Histogram and click OK.

3. Figure 2.7C shows the Histogram dialog box. In the Input Range text box, highlight the data in cells A1:A51.

4. In the Bin Range text box, highlight the bin values in cells C1:C7 and check the Labels box.

5. For Output options, select New Worksheet Ply and Chart Output.

6. Click OK.

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 29

7. Figure 2.7D shows the fruits of your labor, but a few extra steps can make it look prettier.

8. Excel tends to produce a histogram that looks like an elephant sat on it. If you click on the histogram, a border will appear around the histogram that will allow you to grab a corner with your cursor and stretch out your chart.

9. By default, Excel adds a “More” bin that doesn’t serve much purpose here. Delete cells A8 and B8 in the frequency distribution to remove this class from your histogram.

10. The labels on the histogram can be modified by highlighting the label and typing over them.

11. The “Frequency” legend to the right of the histogram can be removed by clicking on it and hitting Delete.

{ figure 2.7b Constructing a Histogram with

Excel (Steps 1 and 2)

{ figure 2.7C Constructing a Histogram with

Excel (Steps 3 and 4)

{ figure 2.7D Constructing a Histogram with

Excel (Steps 5–7)

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30 C h a p t e r 2 | Displaying Descriptive Statistics

Figure 2.7E shows our finished histogram, which I think looks much better than Figure 2.7D. By default, Excel inserts gaps between the bars in the histogram, which is the appropriate way to display discrete data. In a later section, I will show you how to display continuous data in a histogram.

{ figure 2.7e Constructing a Histogram with

Excel (Final Result)

the Shape of histograms The shape of the histogram can tell us some useful information about our data. For exam- ple, let’s say I am currently teaching four statistics classes and that I just finished grading exams from each class. Figure 2.8 shows the histogram for each class. You can clearly see that my 8:00 a.m. class performed the best, with many students scoring in the low 90s.

8:00 A.M. Class

60 65 70 75 80 85 90 95 100

9:00 A.M. Class

60 65 70 75 80 85 90 95 100

10:00 A.M. Class

60 65 70 75 80 85 90 95 100

11:00 A.M. Class

Grades

60 65 70 75 80 85 90 95 100

{ figure 2.8 Histograms of My Four

Statistics Classes

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 31

The grade distribution appears symmetrical, with most of the scores in the middle of the range. A symmetrical distribution is one in which the right side of the distribution is the mirror image of the left side of the distribution. The shape of the histogram for the 9:00 a.m. class is similar to the one for the 8:00 a.m. class. However, the 9:00 a.m. class did not do as well—many of the students’ scores were in the mid-80s. The histogram for the 10:00 a.m. class also appears symmetrical and is centered in the mid-80s. However, the scores in this class are more spread out than those of the 9:00 a.m. class. They range from 70 to 95, whereas the 9:00 a.m. class scores range from the upper 70s to 90. Finally, my poor 11:00 a.m. class appears to have really struggled with the exam. Notice that this distribution is more concentrated on the left side, in the low 70s, so it is not symmetrical. Each histogram provides me with a quick overview of how each class performed. We will revisit the shapes of distributions in Chapter 3 when we cover calculating descriptive statistics.

While these concepts are still fresh in your brain, take a crack at a problem in the fol- lowing Your Turn box before moving on to the next topic.

A symmetrical distribution is one in which the right side of the distribution is the mirror image of the left side of the distribution.

Constructing a Frequency Distribution Using Grouped Quantitative Data Each class in the iPad frequency distribution from the last section consisted of a single value. Some data sets, particularly those with continuous data, require several values to be grouped together in a single class. This grouping prevents having too many classes in the frequency distribution, which can make it difficult to detect patterns. Ideally, the number of classes in a frequency distribution should be between 4 and 20.

To demonstrate grouped data in a frequency distribution, I’ll use the following example: In an effort to improve customer service, Dell monitors the time customers wait on hold when calling in for support. Table 2.6 lists the wait times in minutes for the last 50 callers.

{ table 2.6 | Dell’s Customer-Support Hold Times

0.6 3.9 5.4 8.3 10.2

1.2 4.4 5.5 8.7 10.5

1.3 4.4 5.8 9.0 10.9

2.5 4.6 6.1 9.3 12.2

2.8 4.6 6.4 9.3 12.5

3.2 4.6 6.9 9.5 13.1

3.2 4.8 7.0 9.5 13.3

3.5 4.9 8.0 9.7 13.6

3.8 5.1 8.1 9.8 14.4

3.9 5.2 8.1 9.9 17.4

The following table shows the number of stars that 25 reviewers gave Amazon’s Kindle, a device for reading electronic books.

4 4 5 5 5 4 3 5 3 5

5 5 4 5 2 4 5 1 5 4

3 5 5 4 1

Construct the following:

1. A frequency distribution 2. A relative frequency distribution 3. A cumulative relative frequency distribution 4. A histogram

Y o U r t U r n # 2N

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32 C h a p t e r 2 | Displaying Descriptive Statistics

The times are sorted from shortest to longest. These data can also be found in the Excel file Dell hold times.xlsx.

Number of Classes For grouped data, we need to determine the number of classes in our frequency distribution. One method is to use the 2k Ú n rule, where

k = Number of classes n = Number of data points

The trick is to find the lowest value of k that satisfies the rule. For our Dell example, n = 50.

25 = 32 … 50 (k = 5 is too small) 26 = 64 Ú 50 (k = 6 is a good choice)

Class Width Once we decide to set k = 6, we need to determine the width of each class. The width is the breadth, or range, of numbers we plan to put into each class. We can use the following equa- tion to find a good class width:

Estimated class width = Maximum data value - Minimum data value

Estimated class width = 17.4 - 0.6

6 = 2.8

The previous equation provides a rough estimate for the class width. Round this esti- mate to a useful whole number that makes the frequency distribution more readable. For convenience, it would be best to round this class width up to three, but other values such as four could also be considered. There is no one correct answer for the class width. The goal of constructing a frequency distribution is to identify a useful pattern in the data, and often there is more than one acceptable way to accomplish this.

Class Boundaries The class boundaries represent the minimum and maximum values for each class. Choosing the lowest boundary in the first class is an important step. If we choose the minimum time in Table 2.6, which is 0.6 minutes, the first class would be 0.6 to less than 3.6 minutes. This is not a very convenient choice. A more useful class boundary that is easier to read would start at 0 minutes and end at less than 3 minutes. The following list shows the boundaries for the Dell example:

0 to less than 3 minutes

3 to less than 6 minutes

6 to less than 9 minutes

9 to less than 12 minutes

12 to less than 15 minutes

15 to less than 18 minutes

Because the Dell example involves continuous data (minutes), there are no gaps in the boundaries between classes. By establishing the upper boundary of the first class as “less than 3 minutes” and the lower boundary of the next class as “3 minutes,” we eliminate any gap between these classes. In this example, we were able to set up six classes, which matches the value for k that we determined earlier using the 2k 7 n rule. However, it doesn’t always work out this way, which is fine. If we had chosen a class width equal to 4 minutes, we would have ended up with five classes rather than six:

0 to less than 4 minutes

4 to less than 8 minutes

This is one of the few areas in statistics where there is more than one correct answer, so enjoy it!

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 33

8 to less than 12 minutes

12 to less than 16 minutes

16 to less than 20 minutes

Either case would be acceptable. We will continue the Dell example with a class width of three minutes and six classes.

Class Frequencies Figuring out the frequency is the easy part. Simply count up the number of observations in each class and record the total to obtain the frequency. This job is much easier when the data are sorted, as is the case with the Dell data in Table 2.6. Table 2.7 shows our finished frequency distribution along with the relative frequency and cumulative relative frequency.

Sorting the raw data from lowest to highest value makes it easier to count the number of observations contained in each class of a frequency distribution.

Occasionally you will end up with k + 1 or k - 1 classes to cover the entire range of data. No worries!

{ table 2.7 | Frequency Distribution for Dell’s Customer-Support Hold Times

number of minutes

frequency

relative frequency

cumulative relative frequency

0 to less than 3 5 0.10 0.10

3 to less than 6 18 0.36 0.46

6 to less than 9 9 0.18 0.64

9 to less than 12 11 0.22 0.86

12 to less than 15 6 0.12 0.98

15 to less than 18 1 0.02 1.00

Total 50 1.00

Based on this summary, we can conclude that the largest group of customers (18 of them) waited between 3 and 6 minutes and that nearly half the customers (46%) waited less than 6 minutes on the phone for customer support.

When constructing classes for grouped data, be sure to follow these rules:

1. Equal-size classes. All classes in the frequency distribution must be of equal width. Without equal sizes, patterns in the distribution could be misleading.

2. Mutually exclusive classes. This is a fancy term that simply means the class boundaries cannot overlap. For example, the following two classes overlap:

0 to less than 3 minutes

2.5 to less than 5.5 minutes

If a person waited 2.7 minutes for customer support, this observation would appear in both classes and violate the rule. (Violating rules in statistics is a really bad idea!)

3. Include all data values. This rule is pretty self-explanatory. If your data set has 50 observations, make sure that all 50 are accounted for in the total row of the frequency distribution.

4. Avoid empty classes. If our class widths are too narrow, some classes might not have any observations. However, most populations do not have empty intervals; there are at least a few observations within each. Because the observations we collect should be representative of the population, it is undesirable for our histogram to display a class so narrow that there are no observations in it.

5. Avoid open-ended classes (if possible). An example of an open-ended class would be to add 18-over as the last class in the distribution to include any callers who waited more than 18 minutes for service. Open-ended classes violate the first rule of equal class sizes, but sometimes they cannot be avoided.

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34 C h a p t e r 2 | Displaying Descriptive Statistics

Constructing a histogram with Grouped Quantitative Data We will now call on the Histogram tool in the Data Analysis ToolPak in Excel to construct the histogram for the Dell example. For grouped data, the bins in Excel are the upper boundary for each class. Table 2.8 shows the bins that can be used for the Dell data. For example, I need to choose a value to represent “less than 3” which in this case is 2.9. I could have also chosen 2.99.

We can create a histogram by following the steps laid out for the iPad sales example ear- lier in the chapter. Figure 2.9 shows the “raw” histogram that Excel has generated for us, but it needs some work. Because our grouped data are continuous, we need to remove the gaps between the bars in the histogram by taking the following steps:

1. A right-click on any bar in the histogram will cause a pop-up menu to appear.

2. Left-click on Format Data Series.

3. In the dialog box, move the Gap Width slide all the way to the left.

4. Close the Format Data Series dialog box.

{ figure 2.9 Raw Histogram for

Grouped Data

Our final touch is to re-label the bins along the horizontal axis to reflect the entire range of each class. We can do this simply by changing the values in cells A2 to A7 as shown in Figure 2.10. Excel 2010 users should go to the book’s Web site for these instructions. We can also angle the labels along the horizontal axis, as shown in this figure, by doing the following:

1. Right-click the labels for the horizontal axis to open the drop-down menu.

2. Left-click on Format Axis.

3. In the Format Axis dialog box, left-click on the Size & Properties icon, which looks like this:

4. Change the Custom angle to −45 in the text box. 5. Close the Format Axis dialog box.

Now that’s what I call a great-looking histogram for grouped data!

Histograms displaying continuous data do not have gaps between their bars. By contrast, histograms displaying discrete data do have gaps between their bars.

{ table 2.8 | Bins for Dell Data

class bins

0 to less than 3 minutes 2.9

3 to less than 6 minutes 5.9

6 to less than 9 minutes 8.9

9 to less than 12 minutes 11.9

12 to less than 15 minutes 14.9

15 to less than 18 minutes 17.9

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 35

the Consequences of too Few or too Many Classes The purpose of a histogram is to provide some insight into the natural shape of the distribu- tion of interest. Using very wide classes results in a small number of classes, which may ob- scure important patterns. For example, Figure 2.11 shows a histogram for our Dell data with a class width of 9 minutes. A 9-minute width results in only two classes. This looks more like a picture of two toy blocks and tells us little about the distribution of customers waiting on hold for support.

{ figure 2.10 Finished Histogram for

Grouped Data

{ figure 2.11 Too Few Histogram Classes

Creating too many classes in our histogram also has consequences. Figure 2.12 shows the Dell data with a class width of 1 minute, which results in 18 classes. This results in a his- togram with a very jagged shape and three empty classes. I’m sure there are Dell customers in the population who have waited between 11 and 11.9 minutes on hold, but according to our histogram, they do not exist.

{ figure 2.12 Too Many Histogram Classes

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36 C h a p t e r 2 | Displaying Descriptive Statistics

As I mentioned earlier in the chapter, you do have some flexibility in deciding the width and number of classes for your histogram. Just be careful not to wander too far in either di- rection and suffer the consequences discussed here.

are they Discrete or are they Continuous Data? Sometimes the distinction between discrete and continuous data is not obvious. For instance, my golf score next weekend would technically be discrete data because this value is arrived at by counting my total strokes over 18 holes of play. Obviously, this value needs to be a whole num- ber, such as 94, because there is no way to count a partial stroke (even though there are times my golf swing feels like one). However, if our data set contains a wide range of golf scores, such as the final scores for all of the players in the 2012 Masters, a major professional golf tournament (shown in Table 2.9) and very small increments between scores (that is, one stroke), it is common practice to conveniently display this type of information in a grouped-continuous format. These data can also be found in the Excel file titled 2012 Masters Scores.xlsx.

Exercise good judgment when deciding on the number of classes in your histogram. Too few classes can obscure the natural shape of the distribution. Too many classes can result in some classes being empty.

{ table 2.9 | Final Golf Scores of the 2012 Masters

278 286 288 291 293 296

278 286 289 291 294 297

280 286 289 291 294 297

280 286 289 291 294 298

280 286 290 292 295 299

280 287 290 292 295 299

283 287 290 292 295 299

284 288 290 293 296 301

284 288 290 293 296 302

284 288 291 293 296 306

285 288

Using the 2k 7 n rule with n = 62, I’ll set k = 6 because

26 = 64 Ú 62

The estimated class width would be

Estimated class width = 306 - 278

6 = 4.67

Setting the actual class width to either four or five would be reasonable, so I’ll go with five, starting with 278, as shown in Table 2.10. Also in this table I’m showing the frequency, rela- tive frequency, and cumulative relative frequency for each class.

{ table 2.10 | Frequency Distribution for the Final Golf Scores of the 2012 Masters

score

frequency

relative frequency

cumulative relative frequency

278–282 6 0.097 0.097

283–287 12 0.194 0.291

288–292 21 0.339 0.630

293–297 16 0.258 0.888

298–302 6 0.097 0.985

303–307 1 1.016 1.001

Total 60 1.001

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 37

Using bins of 282, 287, 292, 297, 302, and 307 in Excel, I display the histogram of golf scores in a grouped-continuous format in Figure 2.13.

{ figure 2.13 Finished Histogram for the

Final Golf Scores of the 2012 Masters

According to this histogram, the class with the largest percentage of golfers was the one with final scores between 288 and 292.

Income and age are also examples of data that are technically discrete (we count our money and our years, not measure them), but they are normally displayed in a continuous format. To test your skills on frequency distributions with grouped data, take on the follow- ing Your Turn before moving on to the next section.

A survey was performed in the Philadelphia area that recorded the annual salaries of marketing managers. The following table shows the responses from 30 managers. The numbers are in thousands of dollars.

$40.0 $58.7 $63.4 $68.6 $74.8

$46.6 $60.0 $65.7 $68.7 $77.1

$49.0 $60.7 $65.9 $69.2 $78.2

$52.9 $61.2 $66.4 $71.7 $79.2

$54.7 $62.5 $66.8 $72.3 $83.1

$55.5 $63.0 $67.7 $72.6 $87.7

Construct the following:

1. A frequency distribution 2. A relative frequency distribution 3. A cumulative relative frequency distribution 4. A histogram

Y o u r t u r n # 3N

Answers can be found on u page 74

the polygon When you want to compare the shape of two or more distributions on one graph, histograms are not your best choice because the overlapping columns of the two histograms will make the comparison difficult to interpret. A better choice is the percentage polygon, which graphs the midpoints of each class as a line rather than columns. The midpoint is the center position between the upper and lower boundaries. The height of each midpoint represents the rela- tive frequency of the corresponding class. To demonstrate the percentage polygon, consider the data in Table 2.11 on the next page, which lists the number of minutes that call centers in Denver and Phoenix spent with 50 customers. The numbers are shown in ascending order and can be found in the Excel file call center times.xlsx.

The percentage polygon graphs the midpoint of each class as a line rather than a column. The height of each midpoint represents the relative frequency of the corresponding class.

Recall that bins in Excel represent the higher end of each class (something that I sometimes can’t remember).

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38 C h a p t e r 2 | Displaying Descriptive Statistics

PHStat will be used to generate the percentage polygon. First, however, we need to set up the class boundaries for the data. Using the 2k Ú n rule with n = 50, I’ll set k = 6 because

26 = 64 Ú 50

I’ll use the data from Phoenix to estimate the class widths for both distributions as fol- lows. (Using the Denver data would produce similar results.)

Estimated class width = 39 - 12

6 = 4.5

Rounding the class widths up to 5 minutes will result in the class boundaries shown in Table 2.12. I have also indicated the bin and midpoint for each class in the table. I’ll need these values for PHStat.

{ table 2.11 | Number of Minutes Spent with 50 Customers at Each of Two Call Centers

denver

13 13 13 13 14 14 14 14 14 16 16 17 17

18 18 18 18 19 19 19 20 20 20 20 20 21

21 21 21 21 21 21 21 23 23 23 24 24 24

24 25 25 26 26 27 31 32 33 34 36

phoenix

12 18 19 20 21 21 22 22 22 22 23 23 23

24 24 24 24 24 24 24 25 25 25 25 26 26

27 27 28 28 28 28 28 29 29 29 30 30 31

32 33 34 35 36 36 36 37 38 39 39

{ table 2.12 | Classes and Midpoints for the Call Centers

class bin midpoint

10 to less than 15 minutes 14.9 12.5

15 to less than 20 minutes 19.9 17.5

20 to less than 25 minutes 24.9 22.5

25 to less than 30 minutes 29.9 27.5

30 to less than 35 minutes 34.9 32.5

35 to less than 40 minutes 39.9 37.5

I have set up these data in Excel as shown in Figure 2.14A. Only the first nine values from Table 2.12 are shown in Columns A and B. Your actual Excel sheet would contain all 50 values. PHStat requires the first bin of a distribution to be open-ended toward negative infinity. Consequently, I added a class of “less than 10,” which has a bin equal to 9.9 in the spreadsheet. Because this class is open-ended (the lowest value could be any number less than 10), it does not have a midpoint. The first midpoint in Column D (12.5) corresponds to the second bin in Column C (14.9).

Take the following steps to use PHStat to construct the polygon for these data (Windows and Mac):

1. Go to Add@Ins + PHStat + Descriptive Statistics + Histogram & Polygons, as shown in Figure 2.14B.

2. Fill in the Histogram & Polygons dialog box as shown in Figure 2.14C. Click OK.

{ figure 2.14a Constructing a Polygon with

PHStat (Data)

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 39

Figure 2.14D on the next page shows our percentage polygon for the two call centers. We can conclude from the distributions that the call times for Denver tend to be shorter than the call times for Phoenix. At the lower end of the distribution (shorter times) the Denver percentages are higher than the Phoenix percentages. For instance, nearly 20% of Denver’s call times fall into the 12.5-minute interval compared with less than 5% of the calls to Phoenix. On the higher end of the distribution, roughly 15% of Phoenix’s calls are in the 37.5-minute interval, whereas less than 5% of Denver’s are.

As Figure 2.14E shows on the next page, using PHStat you can also create a cumula- tive percentage polygon, also known as an ogive. A cumulative percentage polygon is a line graph that plots the cumulative relative frequency distribution. For example, in Figure 2.14E, 20% of the calls in Denver fall in the … 14.9-minute interval, 40% fall in the … 19.9-minute interval, 80% fall in the … 24.9-minute interval, and so forth. A full 100% of the calls fall in the … 39.9-minute interval. There are no calls that last 40 minutes or longer. Now let’s look at the calls for Phoenix: Roughly 40% of them fall into the … 24.9-minute interval, compared with 80% of the Denver times. Notice that the Phoenix line is to the right of the Denver line. In other words, the Phoenix line is falling at the higher end of the distribution, where the times are longer. This is consistent with our findings in Figure 2.14D.

{ figure 2.14b Constructing a Polygon with

PHStat (Step 1)

{ figure 2.14C Constructing a Polygon with

PHStat (Step 2)

The cumulative percentage polygon, or ogive, is a line graph that plots the cumulative relative frequency distribution.

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40 C h a p t e r 2 | Displaying Descriptive Statistics

While these steps are fresh in your mind, take some time to construct polygons to ob- serve just how badly my Pittsburgh Pirates have been performing lately compared with a team that has a winning tradition, the New York Yankees.

{ figure 2.14D The Percentage Polygon

{ figure 2.14e The Cumulative Percentage

Polygon (Ogive)

The following table shows the number of wins per season for the New York Yankees and Pittsburgh Pirates from 1974 until 2012, not including the strike-shortened seasons of 1981, 1994, and 1995. These data can also be found in the Excel file Pirates Yankees wins.xlsx.

new york yankees

95 97 95 103 89 94 97 95 101 101 103 95

87 98 114 96 92 88 76 71 67 74 85 89

90 97 87 91 79 103 89 100 100 97 83 89

pittsburgh pirates

79 72 57 62 67 68 67 67 72 75 72 62

69 78 69 79 73 75 96 98 95 74 85 80

64 57 75 84 84 83 98 88 96 92 92 88

Use seven classes, each with a class width of 10. Start with 51–60, 61–70, and so on, and then con- struct a percentage and cumulative percentage polygon using PHStat. What conclusions can you draw when you compare the distributions of the two teams?

Y o U r t U r n # 4N

Answers can be found on u page 74

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C h a p t e r 2 | 2.2 Displaying Quantitative Data 41

2.2 Section Problems Basic Skills 2.1 Using the 2k Ú n rule, determine the number of classes

needed for the following data set sizes.

a. n = 100 b. n = 300 c. n = 1,000 d. n = 2,000

2.2 A data set that consists of 50 numbers has a minimum value of 16 and a maximum value of 74. Determine the class boundaries using the 2k Ú n rule if the data are a. discrete b. continuous

2.3 Consider the following discrete data:

2 4 3 3 1 1 5 2 2 1 1 3 5 2 1 4 3 4 2 2 1 3 4 5

a. Construct a frequency distribution. b. Construct a relative frequency distribution. c. Construct a cumulative relative frequency distribution.

2.4 Consider the following continuous data.

17.5 19.8 29.9 28.1 15.7 21.7 33.6 21.1 19.8 32.7 15.1 13.9 31.8 39.8 33.1 19.6 23.5 42.8 21.7 36.8 35.6 14.5 24.2 14.7 19.2 21.8 20.7 26.9 26.7 35.2

a. Construct a frequency distribution using the 2k Ú n rule.

b. Construct a relative frequency distribution. c. Construct a cumulative relative frequency distribution.

applications 2.5 The results of a survey that collected the current credit

card balances for 36 undergraduate college students are shown in the following table. These data can also be found in the Excel file college credit card.xlsx.

$2,467 $3,373 $434 $1,426 $4,628 $167 $5,927 $4,064 $4,503 $2,600 $3,673 $4,231 $1,788 $5,805 $5,395 $733 $4,890 $1,455 $5,817 $1,145 $846 $739 $4,370 $986 $5,729 $1,159 $1,021 $1,563 $967 $499 $353 $2,297 $289 $375 $162 $1,740

a. Using the 2k Ú n rule, construct a frequency distribu- tion for these data.

b. Using the results from part a, calculate the relative fre- quencies for each class.

c. Using the results from part a, calculate the cumulative relative frequencies for each class.

d. Construct a histogram for these data.

2.6 The following table shows the number of students in the 25 classes being taught this semester at Goldey-Beacom College.

32 42 42 24 33 25 37 23 35 18 46 37 34 25 45 38 28 42 32 43 43 36 28 27 40

a. Using the 2k Ú n rule, construct a frequency distribu- tion for these data.

b. Using the results from part a, calculate the relative fre- quencies for each class.

c. Using the results from part a, calculate the cumulative relative frequencies for each class.

d. Construct a histogram for these data.

2.7 A major U.S. airline records the number of no-shows on a flight that operates each day from Philadelphia to Paris. A no-show is a passenger who purchases a ticket but fails to arrive at the gate at time of departure. The following table indicates the number of no-shows during the last 70 flights. These data can also be found in the Excel file no-shows.xlsx.

1 2 1 2 3 2 1 2 2 2 1 3 3 3 3 4 2 4 3 3 1 2 4 2 4 2 4 1 0 1 1 3 2 1 2 3 1 1 3 1 2 3 1 1 2 2 0 0 3 2 4 2 1 2 1 2 3 1 4 2 3 1 2 1 1 2 1 3 4 2

a. Construct a frequency distribution for these data. b. Using the results from part a, calculate the relative fre-

quencies for each class. c. Using the results from part a, calculate the cumulative

relative frequencies for each class. d. Construct a histogram for these data.

2.8 A major electronics retailer was recently sued by 40 employees who were laid off by the company. The employees claimed they were victims of age discrimina- tion, which is illegal in the United States. The first of the two following tables shows the ages of the 40 employees who were laid off. The second of the following tables shows the age of a random sample of 40 employees currently working for the company. These data can be found in the Excel file employee ages.xlsx.

Ages of Laid-Off Employees

38 35 39 44 54 59 39 43 60 61 56 57 56 53 51 56 53 43 19 42 42 56 75 64 38 62 39 61 43 74 65 44 40 51 46 51 44 45 67 51

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42 C h a p t e r 2 | Displaying Descriptive Statistics

Ages of a Sample of Current Employees

34 46 41 33 39 38 35 44 34 19 39 36 37 36 35 28 37 28 35 30 76 31 29 32 42 39 32 30 47 21 39 26 27 42 27 30 33 32 60 38

a. Construct a percentage polygon for these data. b. Construct a cumulative percentage polygon for these

data. c. Does it appear that the employees suing this company

have a case?

2.9 The Excel file labeled Lowes.xlsx lists the receipt total for 350 randomly selected customers for Lowes, a home improvement store.

a. Using Excel and the 2k Ú n rule, construct a frequency distribution for these data.

b. Using the results from part a, calculate the relative fre- quencies for each class.

c. Using the results from part a, calculate the cumulative relative frequencies for each class.

d. Construct a histogram for these data.

2.10 The Hawaii Island Chamber of Commerce collects ocean temperature data to help promote tourism for local busi- nesses. The Excel file titled Hawaii ocean temps.xlsx lists daily ocean temperatures for the past 125 days.

a. Using Excel, construct a frequency distribution for these data.

b. Using the results from part a, calculate the relative fre- quencies for each class.

c. Using the results from part a, calculate the cumulative relative frequencies for each class.

d. Construct a histogram for these data. e. For what percentage of days was the ocean temperature

less than 77 degrees Fahrenheit?

2.3 Displaying Qualitative Data Qualitative data are values that are categorical (nominal or ordinal measurement levels) and that describe a characteristic, such as gender or level of education. This type of data is handled differently when displayed.

Constructing a Frequency Distribution Using Qualitative Data Frequency distributions help display qualitative data by indicating the number of occur- rences of various categories. To demonstrate this procedure, we’ll use Table 2.13, which shows the final letter grade of 30 students from my statistics class. These data can also be found in the Excel file final grades.xlsx. I would like to determine the percentage of stu- dents who passed the course with a grade of C or better.

{ table 2.13 | Final Grades in a Statistics Class

B C A A B F B C B D

B C A B B C F A C B

B B C B A A A B B B

In an earlier section, we used Excel’s FREQUENCY function to summarize quantita- tive data. However, this function does not work with qualitative data, so I will use Excel’s COUNTIF function as shown in Figure 2.15A. This neat function will count the number of A’s, B’s, C’s, D’s, and F’s in our data set using the following characteristics:

= COUNTIF(range, criteria)

where

range = the range of cells from which you are counting criteria = the value that you are counting

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C h a p t e r 2 | 2.3 Displaying Qualitative Data 43

1. Open the Excel file final grades.xlsx or enter the grade data in Column A and the Bins in Column C as seen in Figure 2.15B.

2. To count the number of A’s in the class, type = COUNTIF($A$2:$A$31,C2) in Cell D2. 3. Copy and paste this formula in Cells D3 to D6 to count the number of B’s, C’s, D’s, and F’s.

4. The formulas for the relative and cumulative frequencies are also shown in Columns E and F. These are the same calculations we did manually in the previous sections.

Figure 2.15B shows the frequency distribution for our final grades example.

{ figure 2.15a Excel’s COUNTIF Function

{ figure 2.15b The Frequency Distribution for

the Final Grades Example

Based on the value calculated in Cell F4 in Figure 2.15B, 90% of the students passed the course with a grade of C or better. I’m sure you’re itching to try your hand at generating frequency distributions with Excel, so take a shot at the following Your Turn.

Y o U r t U r n # 5N

Answers can be found on u page 75

In 2013, the companies that dominated the U.S. cell phone carrier market were AT&T, Sprint, Verizon, and T-Mobile. The following table shows the results of a survey that asked 20 people to name the com- pany that provides their current cell phone service. These data can be found in the Excel file cell phone carriers.xlsx.

Sprint AT&T Sprint AT&T Sprint Sprint Verizon AT&T AT&T Sprint Verizon Verizon AT&T Sprint AT&T Verizon T-Mobile AT&T Sprint AT&T

Construct the following:

1. Frequency distribution 2. Relative frequency distribution

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44 C h a p t e r 2 | Displaying Descriptive Statistics

Bar Charts Bar charts are a good tool for displaying qualitative data that have been organized in cat- egories. The bars in a bar chart can be arranged either horizontally or vertically. We’ll start with a vertical bar chart and use Excel 2013 for Windows to construct the chart. See the textbook’s Web site for Excel 2011 for Mac or Excel 2010 for Windows instructions. Let’s explore the world of bar charts with the data in Figure 2.16A, which shows the grade distri- bution from my statistics class.

Bar charts are a good tool for displaying qualitative data that have been organized in categories and can be arranged in a vertical or horizontal orientation.

{ figure 2.16a Constructing a Vertical Bar

Chart with Excel (Steps 1–3)

1. Using the frequency distribution from Figure 2.15, highlight Cells C2:D6 as shown in Figure 2.16A.

2. Select the Insert tab and the Insert Column Chart icon which are both circled in red in Figure 2.16A.

3. Click on the first 2-D Column icon in the pop-down menu.

4. To add labels to your bar chart, click on the bar chart Excel created to activate the Chart Tools and choose the Design tab as shown in Figure 2.16B.

{ figure 2.16b Constructing a Vertical Bar

Chart with Excel (Steps 4–7)

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C h a p t e r 2 | 2.3 Displaying Qualitative Data 45

5. Select Add Chart Element + Axis Titles + Primary Horizontal as shown in Figure 2.16B to insert a title along the horizontal axis.

6. Select Add Chart Element + Axis Titles + Primary Vertical to insert a title along the vertical axis.

7. Select Add Chart Element + Chart Title + Above Chart to insert a title at the top of the histogram.

Figure 2.16C shows our finished vertical bar chart displaying the qualitative grade data.

{ figure 2.16C Constructing a Vertical Bar

Chart with Excel (Final Result)

If you’ve just realized Figure 2.16C looks very much like the histograms that we’ve already discussed, you are right. However, there is a subtle difference. Histograms display quantitative data using vertical bars that are separated for discrete data and joined together for continuous data. Bar charts, on the other hand, display qualitative (categorical) data. The bars can be vertical or horizontal.

Bar charts can also be displayed in the horizontal direction and are called—you guessed it—horizontal bar charts. The following steps show how these charts are constructed in Excel.

1. Highlight the Grade data from Figure 2.16A in Cells C2:D6.

2. Click on the Insert tab and select the Insert Bar Chart icon which is to the right of the Insert Column Chart icon that was just used.

3. Click on the first 2-D Bar icon.

4. Add titles as we did with the vertical bar chart (note the axes are reversed here).

The resulting horizontal bar chart is shown in Figure 2.16D.

Horizontal bar charts display the bars in a horizontal direction.

{ figure 2.16D Horizontal Bar Chart with Excel

Now that you have mastered basic bar charts, let’s take it up a notch with the next ex- ample, shown in Table 2.14 on the next page, which provides the total points scored by the football teams in the NFC (National Football Conference) North Division in 2011 and 2012. These data can be found in the Excel file NFC.xlsx.

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46 C h a p t e r 2 | Displaying Descriptive Statistics

{ table 2.14 | Points Scored by the Football Teams in the NFC North Division, 2011 and 2012

team points scored 2011 2012

Green Bay Packers 560 433

Minnesota Vikings 340 379

Chicago Bears 353 375

Detroit Lions 474 372

Because each category (in this case, teams) has two data points, we can display the data in a clustered bar chart. The clustered bar chart will display the two data points together, or “clustered” with one another, as Figure 2.17B shows. To get the data to look like this, per- form the following steps, as shown in Figure 2.17A.

1. Open the Excel file NFC.xlsx or insert the data from Table 2.14 into an Excel file, as shown in Figure 2.17A.

2. Highlight the data in Columns A, B, and C.

3. Click on the Insert tab and the Insert Column Chart icon which are both circled in red in Figure 2.17A.

4. Click on the Clustered Column icon, which is the first 2-D Column icon in the row.

{ figure 2.17a Constructing a Clustered Bar

Chart in Excel

Do not insert a title in cell A1 in Figure 2.17A. Doing so will interfere with the proper construction of the clustered bar chart in Excel. If you don’t believe me, go on and try it!

Figure 2.17B shows each team’s pair of data points in a bar graph format. According to this chart, we can see that Green Bay and Detroit scored fewer points in 2012, whereas Chi- cago and Minnesota both scored more when compared with the 2011 season.

{ figure 2.17b Finished Clustered Bar Chart

Clustered bar charts group several values side by side within the same category in a vertical direction.

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C h a p t e r 2 | 2.3 Displaying Qualitative Data 47

We can also arrange our data in a stacked bar chart, as shown in Figure 2.17C. To con- struct this chart in Excel, follow the steps in the clustered bar chart example but choose the second 2-D Column icon rather than the first.

Stacked bar charts group several values in a single column within the same category in a vertical direction.

{ figure 2.17C Stacked Bar Chart

Looking at Figure 2.17C, we can easily see that Green Bay scored the most points over the two-year span, with Detroit scoring the second-most points.

Choosing between clustered and stacked bar graphs depends on the purpose of the chart. Clustered charts work well when you are comparing data within categories, such as which team scored more points in 2011 when compared with 2012. Stacked bar charts are more ef- fective at displaying totals in each category, such as which team scored the most points over the two-year period (which, in case you forgot, was my favorite team since 1967, the Green Bay Packers).

Take a stab at bar charts with some Facebook data in the next Your Turn problem.

Clustered charts work well when you are comparing data within categories, whereas stacked bar charts are more effective at displaying totals in each category.

The data in the following table show the unemployment rate (as a percentage of the workforce) for five states at the end of 2011 and 2012. These data can be found in the Excel file unemployment.xlsx.

State 2011 (%) 2012 (%)

Arizona 9.0 7.9

Florida 9.9 8.0

Nebraska 3.7 4.2

North Carolina 10.4 9.2

Virginia 6.1 5.5

Construct the most appropriate bar chart (stacked or clustered) if the goal is to display changes in un- employment rate from 2011 to 2012 for each state.

Y o U r t U r n # 6N

Answers can be found on u page 75

pareto Charts Pareto charts are a specific type of bar chart used in quality control programs by businesses to display the causes of problems graphically. The charts are essentially histograms that show, in decreasing order, the frequency of the categories that cause problems. In other words, the most problematic categories are listed first. Pareto charts are named after Vilfredo Pareto (1848–1923), an Italian industrialist who helped develop the field of microeconomics (which has tortured more students than the field of statistics, your author included). Pareto observed that 80% of the land in Italy was owned by 20% of the population. In the 1940s, the quality control consultant Joseph Juran came across Pareto’s work and began calling it Pareto’s 80/20 rule. Pareto’s rule states that 20% of defects cause 80% of quality control problems.

Pareto charts are essentially bar charts that show the frequency of the categories that cause quality control problems. The charts show the categories in a decreasing order.

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48 C h a p t e r 2 | Displaying Descriptive Statistics

I’ll use the following example to demonstrate the use of Pareto charts. QVC, a major home shopping retailer (and my wife’s very favorite company in the world) maintains re- cords of the reasons customers return products they purchased. The data in Table 2.15 show the reasons for the returns and the number of incidences for each category during the past few weeks. These data can be found in the Excel file QVC.xlsx.

{ figure 2.18a Spreadsheet for Constructing a

Pareto Chart

Use the following steps to construct our Pareto chart with Excel 2013. See the text- book’s Web site for Excel 2011 for Mac or Excel 2010 for Windows instructions.

1. Highlight Cells A1:A6.

2. Holding down the CTRL key, highlight Cells C1:C6.

3. Holding down the CTRL key, highlight Cells D1:D6.

4. Click on the Insert tab and select Column.

5. Click on Clustered Column, the first 2-D Column icon shown in the list.

Your chart will look like the one in Figure 2.18B. But we still have a little work to do here.

1. Right-click on one of the Cumulative Relative Frequency bars (the red ones).

2. In the drop-down menu, select Change Series Chart Type.

{ table 2.15 | Reasons for and Number of Returns to QVC

reason for return frequency

Product defective 46

Disappointed with product 22

Product no longer wanted 14

Late delivery of product 5

Other 3

Pareto charts also plot the cumulative relative frequency as a line on the chart. This line is known as an ogive, which was discussed earlier in the chapter. To construct a Pareto chart in Excel, we first need to calculate the relative and cumulative relative frequencies for our data, as shown in Figure 2.18A. See Figure 2.15 if you need a refresher on how to calculate the values in Columns C and D.

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C h a p t e r 2 | 2.3 Displaying Qualitative Data 49

The Change Chart Type dialog box will appear as seen in Figure 2.18C.

1. Click on the drop-down box to the right of Cumulative Relative Frequency in the Series Name box which is circled in Figure 2.18C.

2. Select the first Line graph in the drop-down menu as seen in Figure 2.18C.

3. Click OK.

{ figure 2.18b Preliminary Pareto Chart

{ figure 2.18C Excel’s Change Chart Type

Dialog Box

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50 C h a p t e r 2 | Displaying Descriptive Statistics

All that is left is to change the scale of the vertical axis and add data points to the line graph with the following steps.

1. Right-click on one of the numbers on the vertical axis.

2. In the drop-down menu, left-click Format Axis.

3. The Format Axis dialog box will appear on the right side of your spreadsheet.

4. Under Axis Options, change the Maximum value from 1.2 to 1.0.

5. Under Axis Options, change the Major value from 0.1 to 0.2.

6. Click on Number at the bottom on the Format Axis menu to expand the options.

7. In the Category box, select Percentage.

8. In the Decimal places box, change to 2 to 0.

9. Close the Format Axis dialog box by clicking the X in the right-hand corner.

10. Right-click on the red ogive line on the chart and left-click on Add Data Labels.

Figure 2.18D shows our completed Pareto chart for QVC. As you can see, Pareto charts are an efficient way to summarize quality control issues. From this example, you can see by looking at the ogive, or cumulative relative frequency line, that 75.6% of the returns are due to the first two categories—defective merchandise and customers being disappointed with the merchandise.

Apply your Pareto chart skills to the airline industry in this next Your Turn section.

{ figure 2.18D Finished Pareto Chart

This red line represents the ogive in the Pareto chart.

The following table shows an airline company’s flight-delay data. The table shows the reasons for the delays and the relative frequency of each type of delay.

Reason for Delay Relative Frequency (%)

Bad weather 33

Air traffic congestion 22

Mechanical issues 18

Crew shortages 13

Too long to refuel 9

Other 5

Construct a Pareto chart for these data.

Y o U r t U r n # 7N

Answer can be found on u page 75

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C h a p t e r 2 | 2.3 Displaying Qualitative Data 51

pie Charts Pie charts are another excellent tool for comparing proportions for qualitative (categorical) data. To demonstrate, we’ll use Table 2.16, which shows the number of computers shipped by leading computer manufacturers during the third quarter of 2012 in the United States. These data can be found in the Excel file computer shipments.xlsx.

Pie charts are another excellent tool for comparing proportions for categorical data. Each category occupies a segment of the pie that represents the relative frequency of that category.

{ table 2.16 | Estimated U.S. Computer Shipments, Third Quarter, 2012

company units shipped (000s)

HP 4,142

Dell 3,271

Apple 2,079

Lenovo 1,358

Acer 990

Others 2,548

Based on: http://www.gartner.com.

Perform the following steps, as shown in Figure 2.19A, to construct a pie chart with Excel 2013. See the textbook’s Web site for Excel 2011 for Mac or Excel 2010 for Windows instructions.

1. Highlight Cells A1:B7.

2. Click on the Insert tab and select the Insert Pie icon circled in Figure 2.19A.

3. Click on the first 2-D Pie icon.

{ figure 2.19a Constructing a Pie Chart in

Excel (Steps 1–3)

The pie chart will appear in your spreadsheet on the next page. Click on the chart and take the following steps to populate your pie chart with percentages:

4. Select the Design tab as seen in Figure 2.19B.

5. Select the Chart Style circled in Figure 2.19B.

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52 C h a p t e r 2 | Displaying Descriptive Statistics

Figure 2.19C shows our finished pie chart, complete with U.S. market share percentages for the computer industry during the third quarter of 2012. Looking at this chart, it is easy to conclude that HP and Dell together controlled over half the U.S. market share in computers shipped during this time period.

{ figure 2.19b Constructing a Pie Chart in

Excel (Steps 4–5)

{ figure 2.19C Constructing a Pie Chart in

Excel (Final Result)

One word of warning about constructing pie charts: All categories in the data set must be included in the pie. For instance, in the last example, it may be tempting to omit the “Other” category. However, doing so would misrepresent the market share for the remaining categories. The last thing any respectable statistician wants to do is misrepresent anything.

As you can see, bar charts and pie charts work with the same type of data—qualitative data. I’m sure your inquisitive mind is now screaming with the question, “How do I choose between a pie chart and a bar chart?” If you want to compare the relative sizes of the classes to one another and together they comprise all possible categories, use a pie chart. Bar charts are more useful when you want to highlight the actual data values and when the classes com- bined don’t form a whole. For example, Figure 2.20 is a bar chart showing the number of in- dividuals 16 years or older in various employment categories in the United States from 2004 to 2013. The chart allows us to observe how the trends in the categories have changed over

{ figure 2.20 Employment Status for

Individuals 16 Years or Older, 2004–2013

Use a bar chart when you want to highlight the actual data values and when the classes combined don’t form a whole.

Use a pie chart if you want to compare the relative sizes of the classes to one another and together they comprise all possible categories.

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C h a p t e r 2 | 2.3 Displaying Qualitative Data 53

time. One pie chart would not be able to display all this information. It could not display the trends in these categories over the 10-year span. The best we could do with a pie chart is to show the three categories for a specific year.

Before we move on, try your hand at making some Excel pie charts.

A recent Gallup Poll surveyed 1,145 people who had visited emergency rooms within the past year. The people surveyed were asked to rate the quality of care they received. The results are shown in the fol- lowing table.

Rating Frequency

Very good 527

Good 492

Poor 80

Very poor 46 Based on: USA Today/Gallup Poll.

Construct a pie chart summarizing these data.

Y o U r t U r n # 8N

Answer can be found on u page 75

2.3 Section Problems applications 2.11 The following table shows the results of a survey that

asked users of the Internet to identify their favorite search engine. These data can be found in the Excel file search engine.xlsx. (For your information, Baidu is a Chinese- language search engine.)

Google Google Yahoo Google Google Bing Google Yahoo Google Google Baidu Google Google Other Yahoo Google Google Google Yahoo Bing Baidu Google Google Google Google Google Google Yahoo Google Google

a. Using Excel, construct a frequency distribution for these data.

b. Using the results from part a, calculate the relative fre- quencies for each search engine.

c. Construct a vertical bar chart for these data.

2.12 A restaurant manager surveys her customers after their dining experience. Customers rate their experience as Excellent (E), Good (G), Fair (F), or Poor (P). The following table records the experience of 60 customers and can be found in the Excel file dining experience. xlsx.

G G E G F G G E G E F G P G E E G F G G E E F E G E P G G F E G E G G G G F E G E E G P G G E G P F E F G G P G G G G G

a. Construct a frequency distribution for these data. b. Using the results from part a, calculate the relative fre-

quencies for each class. c. Using the results from part a, calculate the cumulative

relative frequencies for each class. d. Construct a horizontal bar chart for these data. e. What percentage of customers rated their dining expe-

rience as either Excellent or Good?

2.13 Recently, a salary survey was conducted to measure the impact of having an MBA degree in various industries with four years of work experience. The following table summarizes these results:

Four Years’ Experience

Occupation No MBA MBA

Consulting $66,271 $101,137

Energy $66,461 $84,701

Finance $80,236 $89,169

Manufacturing $71,013 $90,435

Media $63,745 $97,295

Telecom $74,378 $106,142 Based on: http://www.topmba.com.

Construct a clustered bar chart that summarizes these data.

2.14 A recent business trend is for companies to purchase on- line software applications (applications that can be run on the Internet) rather than to own and install the applications

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54 C h a p t e r 2 | Displaying Descriptive Statistics

on their own computer systems. A recent survey asked 210 small business owners if they would pay a monthly subscription fee to use the various online software ap- plications shown in the following table. The percentages of owners who answered yes were determined for January and May and are shown in the following table:

Use of Software January (%) May (%)

Business applications 22 44

Support for information technology

14 42

Computer security 15 49

Electronic storage 15 37

Based on: AMI Partners data.

Construct a clustered bar chart that summarizes these data.

2.15 The following table shows the number of graduating students in 2012 and 2013 with various majors at Sussex College.

Major 2012 2013

Management 38 30

Marketing 24 39

Finance 31 40

Economics 13 16

Accounting 25 20

With the goal of comparing the number of graduating students from 2012 and 2013 for each major, construct a stacked bar chart that summarizes these data.

2.16 The following table lists the reasons customers provided when they rated their dining experience unsatisfactory at a particular restaurant:

Reason Relative Frequency (%)

Service too slow 37

Long wait for table 16

Mistake with order 14

Poor food quality 13

Poor atmosphere 11

Other 9

Construct a Pareto chart for these data.

2.17 A manager at a local restaurant records the number of customers who ordered each of four types of entrées. The following table summarizes the results:

Entrée Type Frequency

Beef 25

Fish 17

Chicken 11

Vegetarian 7

Construct a pie chart summarizing these data as percentages.

2.18 The following table shows the breakdown of gasoline prices in terms of its components.

Component Percentage

Crude oil 60.9

Taxes 15.1

Refining 13.7

Distribution and marketing 10.3

Based on: Energy Information Administration data.

Construct a pie chart summarizing these data.

2.19 Opinion Research Corporation asked adults to describe the category that best describes their health care costs. The following table shows the number of adults in each category:

Category Number of Adults

Fairly priced 426

Expensive 274

Low cost 256

Don’t know 47

Construct a display that best describes these data.

2.20 The following table shows the economic output, or gross domestic product (GDP), for various countries in 2002 and 2012:

GDP ($ trillion)

Country 2002 2012

United States $10.5 $15.7

Japan $4.0 $6.0

China $1.3 $8.3

Germany $2.0 $3.4

France $1.4 $2.6

United Kingdom $1.6 $2.4

Italy $1.2 $2.0

Construct a display that best describes these data.

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C h a p t e r 2 | 2.4 Contingency Tables 55

2.4 Contingency Tables All of the examples so far in this chapter have consisted of observations with a single value. For instance, each student had one grade and each manager had one salary. Contingency tables provide a format to display the frequencies of two qualitative variables. The following example will be used to demonstrate this procedure.

The manager of a 7-Eleven convenience store recorded the genders of the last 20 cus- tomers and their types of payment (credit, debit, or cash), which are shown in Table 2.17. These data can also be found in the Excel file payment types.xlsx.

Contingency tables provide a format to display the frequencies of two qualitative variables.

{ table 2.17 | Genders and Payment Types for 7-Eleven’s Customers

customer gender payment customer gender payment

1 Female Cash 11 Male Cash

2 Female Credit 12 Male Credit

3 Female Credit 13 Female Cash

4 Male Cash 14 Male Credit

5 Male Debit 15 Male Cash

6 Female Credit 16 Female Debit

7 Male Cash 17 Female Credit

8 Female Credit 18 Female Debit

9 Female Debit 19 Male Cash

10 Female Credit 20 Female Credit

The raw data in Table 2.17 are summarized more efficiently as the contingency table shown in Table 2.18.

{ table 2.18 | Contingency Table for 7-Eleven’s Customers

gender

payment female male total

Cash 2 5 7

Credit 7 2 9

Debit 3 1 4

Total 12 8 20

The “Female” column in Table 2.18 shows the number of female customers who paid with cash (2), credit cards (7), and debit cards (3), which accounts for all 12 female custom- ers in Table 2.17. The “Male” column shows the number of male customers who paid with cash (5), credit cards (2), and debit cards (1), which accounts for all eight male customers in Table 2.17. The last column in Table 2.18 shows the total number of customers, both male and female, who paid with cash (7), credit cards (9), and debit cards (4). This column ac- counts for all 20 customers.

We can also convert the frequencies in Table 2.18 to relative frequencies, as we did with the frequency distributions earlier in the chapter, by dividing all the values in Table 2.18 by 20, the total number of observations. These results are shown in Table 2.19 on the next page.

Five customers from Table 2.17 were male and paid in cash.

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56 C h a p t e r 2 | Displaying Descriptive Statistics

Contingency tables help us identify relationships between two or more variables. Based on the 20 customers from Table 2.18, we can conclude that over half the female customers paid with credit (7 out of 12) and over half the male customers paid with cash (5 out of 8). From Table 2.19 we can conclude that 10% of the customers were both female and paid with cash. These conclusions certainly were not obvious looking at the raw data shown in Table 2.17 (at least not to me).

Constructing a Contingency table in excel (Windows) Contingency tables can be rather tedious to construct by hand (especially when there are lots of data), so I’m going to show you a cool feature in Excel known as a pivot table, which is really just a contingency table in disguise. First, arrange the data from Table 2.17 as shown in Figure 2.21A. Because we are counting the number of customers who are female or male and have paid by cash, credit, or debit, we need to include a Customer column in our spread- sheet. This new column allows Excel to count the number of rows (customers) associated with each category.

1. Click on any cell within your data.

2. Choose the Insert tab.

3. Click on the Pivot Table icon as shown in Figure 2.21A.

4. A Create Pivot Table dialog box will appear. Simply click OK.

Five males out of 20 paid with cash, so 5>20 = 0.25

{ table 2.19 | Relative Contingency Table for 7-Eleven’s Customers

gender

payment female male total

Cash 0.10 0.25 0.35

Credit 0.35 0.10 0.45

Debit 0.15 0.05 0.20

Total 0.60 0.40 1.00

{ figure 2.21a Creating a Pivot Table in

Excel (Steps 1–4)

Excel for Mac users can find instructions for pivot tables on the textbook’s Web site.

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C h a p t e r 2 | 2.4 Contingency Tables 57

5. A new worksheet will be created for your pivot table. From the Pivot Table Fields found to the right of this worksheet,

a. Drag the Gender field down into the Column Labels box below, as shown in Figure 2.21B.

b. Drag the Payment field down into the Row Labels box below. c. Drag the Customer field down into the Values box below.

6. Your spreadsheet should look like Figure 2.21B.

{ figure 2.21b Creating a Pivot Table in Excel

(Steps 5–6)

Don’t panic when you see that the values in our pivot table are not correct. By default, Excel sums up all the values in the Customer column in our pivot table, which doesn’t help us since these values are arbitrary customer numbers from 1 to 20. We want Excel to count the number of customers in each category. To do this,

7. Left-click anywhere within the pivot table.

8. Right-click on the cell you selected.

9. In the drop-down menu, select Summarize Data By + Count.

You can now let out a big sigh of relief as our pivot table converts the observations to a frequency distribution and counts the number of them in each category, as shown in Figure 2.21C.

{ figure 2.21C Creating a Pivot Table in Excel

(Final Result)

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58 C h a p t e r 2 | Displaying Descriptive Statistics

After a little practice with this, you will find pivot tables very useful tools for organizing complex data sets. Pivot tables work with both quantitative and qualitative data, as well as grouped data. These tables can also work with more than two variables in a data set.

Test your newly discovered pivot table skills with this next Your Turn problem.

The digital camera manager at Best Buy was investigating the relationship between the camera brand a customer purchased and whether the person also purchased an extended warranty. The following table shows the purchasing behavior of the last 20 customers who bought a digital camera.

Customer Brand Warranty Customer Brand Warranty

1 Canon No 11 Nikon No

2 Nikon No 12 Sony Yes

3 Canon No 13 Canon No

4 Sony Yes 14 Sony Yes

5 Nikon No 15 Sony No

6 Canon No 16 Canon No

7 Canon No 17 Canon No

8 Sony No 18 Nikon No

9 Nikon Yes 19 Canon No

10 Canon Yes 20 Nikon No

Construct a contingency table and a relative contingency table for these data. What conclusions might the manger draw?

Y o U r t U r n # 9N

Answers can be found on u page 75

2.4 Section Problems applications 2.21 The following table shows the final grades and genders of

24 students in a statistics class. These data can be found in the Excel file grade distribution.xlsx.

Grade Gender Grade Gender

B Male C Female

A Female A Male

B Female A Female

C Male B Male

B Male B Female

B Female B Male

A Female C Male

C Female A Female

B Male B Male

C Male A Male

A Female B Female

B Male B Female

Develop a contingency table for these data. What conclu- sions can be drawn about the grade distribution?

2.22 A regional manager at Sears compares customer satisfac- tion ratings (1, 2, 3, or 4 stars) at the company’s Media, Pennsylvania, store (M); Exton, Pennsylvania, store (E); and Darby, Pennsylvania, store (D). The following table shows these data from 50 customers, which can also be found in the Excel file Sears ratings.xlsx.

Store Rating Store Rating Store Rating

M 3 M 2 E 1

D 4 M 3 E 2

E 3 D 3 M 3

M 2 E 3 D 3

E 3 E 2 M 2

M 1 D 3 M 4

D 4 D 2 E 3

E 2 D 4 E 3

D 3 E 3 E 1

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C h a p t e r 2 | 2.5 Stem and Leaf Display 59

2.5 Stem and Leaf Display The stem and leaf display is another graphical technique you can use to display quantitative data. A statistician named John Tukey originated the concept during the 1970s. The major benefit of this approach is that all the original data points are visible on the display and that they are very easy to construct by hand.

To demonstrate the method, I will use test scores from a recent statistics exam, which are shown in Table 2.20.

The stem and leaf display splits the data values into stems (the larger place values) and leaves (the smaller place value). By listing all of the leaves to the right of each stem, we can graphically describe how the data are distributed.

Store Rating Store Rating Store Rating

M 4 M 1 E 3

E 4 M 2 D 2

M 2 M 3 M 1

E 4 M 3 D 4

D 3 D 4 M 2

M 2 D 3 E 4

Store Rating Store Rating Store Rating

D 4 M 2 M 3

M 3 D 4

Develop a contingency table for these data. What conclu- sions can be drawn about the store location and customer satisfaction?

{ table 2.20 | Exam Scores

81 86 78 80 81 82 92 90

79 83 84 95 85 88 80 78

84 79 80 83 79 87 84 80

The first step in constructing a stem and leaf display is to sort the data from lowest to highest, as shown in Table 2.21.

{ table 2.21 | Sorted Exam Scores

78 78 79 79 79 80 80 80

80 81 81 82 83 83 84 84

84 85 86 87 88 90 92 95

Figure 2.22 shows the stem and leaf display for the scores.

The “stem” in this display is the first column of numbers, which represents the first digit of the exam scores. The “leaf” in the display is the second digit of the exam scores, with one digit for each score. The values 0, 2, and 5 on the 9 stem represent the exam scores 90, 92, and 95.

If we choose to, we can break this display down further by adding more stems. Figure 2.23 on the next page shows this approach of splitting the stems in half. Here, the stem labeled 7 (5) stores all the scores between 75 and 79. The stem 8 (0) stores all the scores between 80 and 84. After examining this display, I can see a pattern that’s not as obvious when looking at Figure 2.22: The majority of the exam scores are in the low 80s.

7 | 8 8 9 9 9

8 | 0 0 0 0 1 1 2 3 3 4 4 4 5 6 7 8

9 | 0 2 5

{ figure 2.22 A Stem and Leaf Display for the

Exam Scores

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60 C h a p t e r 2 | Displaying Descriptive Statistics

A common mistake that my students make is to omit stems that have no leaves. Figure 2.24 shows a stem and leaf display for the following data, which represent the weekly demand for milk at a local Costco warehouse store.

105 111 111 113 116 119 132 138 140 143 143 147

7(5) | 8 8 9 9 9

8(0) | 0 0 0 0 1 1 2 3 3 4 4 4

8(5) | 5 6 7 8

9(0) | 0 2

9(5) | 5

{ figure 2.23 Splitting the Stems in Half

10 | 5

11 | 1 1 3 6 9

12 |

13 | 2 8

14 | 0 3 3 7

{ figure 2.24 A Stem and Leaf Display with

an Empty Stem

Because each data point has three digits, we’ll use the first two digits for the stems and the third digit for the leaves. Even though the 12 stem has no leaves, including this stem in the display provides a more accurate picture of this distribution. The lesson here is “don’t cut your leafless stems.”

The beauty of the stem and leaf display is its simplicity. It does not require computer software to construct, shows each of the data values, and provides a histogram-like picture of the distribution. Before moving on to the next section, use the following Your Turn example to test your stem and leaf skills on this next data set.

Sorting the data from lowest to highest makes constructing the stem and leaf display much easier and less prone to mistakes.

Be sure to include all the stems in the distribution, even the ones with no leaves, as shown in Figure 2.24.

The space between the numbers in the leaves for a stem and leaf display should be equal so that the shape of the frequency distribution is properly displayed. In other words, a string of six numbers should appear twice as long as a string of three numbers.

2.5 Section Problems Basic Skills 2.23 Construct a stem and leaf display for the following data,

which also can be found in the Excel file Prob 2-23 data .xlsx.

72 120 83 79 137 99 75 86 71 122 87 125 130

111 134 118 110 97 116 139 112 115 80 100 94 73

86 90 78 78 107 101 87 90 126 134 120 107 111

2.24 Construct a stem and leaf display for the following data, which also can be found in the Excel file Prob 2-24 data .xlsx.

130 123 122 121 113 114 112 100 125 150 113 127 139

132 111 110 105 121 123 102 140 126 137 136 137 140

108 127 145 129 146 115 121 108 137 142 109 114 132

The following table shows the ages of employees who work in an AT&T Wireless retail facility. Construct a stem and leaf display for these data by splitting the stems in half.

Y o U r t U r n # 1 0N

Answer can be found on u page 75

48 40 38 37 37 37 36 34 33 33

33 33 32 32 29 29 29 28 28 27

27 27 26 26 25 25 25 25 25 25

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C h a p t e r 2 | 2.6 Scatter Plots 61

2.6 Scatter Plots Up to this point, most of the data we have summarized involved only a single value, such as the number of computers shipped by HP or the number of defective returns to QVC. Scatter plots provide a picture of the relationship between two quantitative variables that are paired together. To demonstrate scatter plots, we’ll use the data in Table 2.22, which shows the selling price of various sizes of Samsung LCD televisions that were listed recently on Best Buy’s Web site. These data can also be found in the Excel file Samsung TV.xlsx.

applications 2.25 The following table shows the number of cars that arrived

at a local Burger King drive-through during lunch hour (12:00–1:00 p.m.) over the past 39 days. These data can also be found in the Excel file Burger King.xlsx.

50 27 52 22 29 54 44 37 35 40 55 40 24

51 42 49 48 44 38 67 23 43 45 59 41 50

21 43 37 37 13 45 52 64 51 58 16 48 47

a. Construct a stem and leaf display for these data. b. Construct a stem and leaf display, splitting the stems

from part a.

2.26 The following table shows the ages of the top 25 golfers on the 2012 PGA money list which represents the top earners that year. These data can also be found in the Excel file PGA money list.xlsx.

23 37 32 35 34 36 32 42 30 26 34 42 35

35 30 43 27 45 28 35 24 31 37 39 32 Based on: http://www.espn.com

a. Construct a stem and leaf display for these data. b. Construct a stem and leaf display, splitting the stems

from part a.

{ table 2.22 | Size and Price Data for Samsung LCD Televisions

size (in.) price ($) size (in.) price ($)

43 500 60 1,100

55 900 46 1,600

51 900 19 200

32 400 55 2,200

51 1,200 60 1,700

37 500 55 2,000

60 2,800 22 300

60 1,100 40 600

46 1,600 40 900

Based on: http://www.bestbuy.com.

There are a total of 18 observations (televisions) in Table 2.22, each consisting of a size and the corresponding price of a television. There is an important relationship between the size and price of a television. Common sense tells us that as the size of the television in- creases, the price will also increase. In this example, we call the television price the dependent variable because we believe that the price will depend on the television’s size. The television size is known as the independent variable.

When constructing a scatter plot, the independent variable is placed on the horizontal axis, or x-axis. The dependent variable is placed on the vertical axis, or y-axis. Perform the following steps, as shown in Figure 2.25A on the next page, to construct the scatter plot in Excel 2013. See the textbook’s Web site for Excel 2011 for Mac or Excel 2010 for Windows instructions.

The dependent variable, which is placed on the vertical axis, of the scatter plot is influenced by changes in the independent variable, which is placed on the horizontal axis.

Scatter plots provide a picture of the relationship between two quantitative variables that are paired together.

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62 C h a p t e r 2 | Displaying Descriptive Statistics

1. Open the Excel file Samsung TV.xlsx or type in the data as seen on Figure 2.25A. If you type in the data, be sure to place the independent variable (size) in the first column and the dependent variable (price) in the second column.

2. Highlight Cells A1:B17.

3. Select the Insert tab and select the Insert Scatter Chart icon circled in Figure 2.25A.

4. Click on the first Scatter icon.

{ figure 2.25a Constructing a Scatter Plot in

Excel (Steps 1–4)

{ figure 2.25b Constructing a Scatter Plot in

Excel (Final Result)

There are 16 data points on the scatter plot in Figure 2.25B. Each point represents one television from Table 2.22. For instance, the far-left data point corresponds to the 19-in.

We now have a preliminary scatter plot that could use a little work. Perform the follow- ing steps to construct the scatter plot shown in Figure 2.25B:

5. Add axis titles by clicking on the scatter plot and selecting the Design tab under Chart Tools.

6. Insert the chart title and axis titles by clicking on Add Chart Element (like we did in Figure 2.16B).

7. Format the television prices on the vertical axis to currency by right-clicking on one of the prices and use the Format Axis menu (like we did in Figure 2.18D).

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C h a p t e r 2 | 2.6 Scatter Plots 63

television that sells for $200. Our scatter plot shows an upward trend in price as the size of the television increases. This makes sense, as we would expect larger televisions to cost more.

Use caution when deciding which variable in your scatter plot is independent and which is dependent. The relationship between independent and dependent variables exists only in one direction, as shown in the following equation:

Independent variable (x) S Dependent variable (y) Size (x) S Price (y)

The relationship does not make sense in the opposite direction. For instance, if Best Buy reduces the price of a television, doing so does not reduce the size of the television.

Line Charts A line chart is a special type of scatter plot in which the data points in the scatter plot are connected with a line. To demonstrate this type of chart, I’ll use the data in Table 2.23, which shows the annual revenues for The Walt Disney Company from 1993 to 2012. These data can also be found in the Excel file Disney revenue.xlsx.

A line chart is a special type of scatter plot in which the data points in the scatter plot are connected with a line.

{ table 2.23 | Annual Revenue for The Walt Disney Company

year revenue ($ billions)

1993 8.5 2003 27.1

1994 10.4 2004 30.8

1995 12.5 2005 31.9

1996 18.7 2006 33.7

1997 22.5 2007 35.5

1998 23.0 2008 37.8

1999 23.4 2009 36.1

2000 25.4 2010 38.1

2001 25.8 2011 40.9

2002 25.4 2012 42.3

Based on: Disney Annual Reports.

The data set shown in Table 2.23 is known as a time series because each data point is associated with a specific point in time. (We first discussed time series data and charts in Chapter 1, if you recall.) We can construct a line chart using these data by performing the following steps:

1. List the data in Columns A and B as shown in Figure 2.26A on the next page. Because Year is the independent variable, place it in Column A.

2. Highlight Cells A1:B19.

3. Select the Insert tab and select the Insert Scatter Chart icon circled in Figure 2.26A.

4. Click on the second Scatter icon.

A data set is known as a time series when each data point is associated with a specific point in time.

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64 C h a p t e r 2 | Displaying Descriptive Statistics

When graphing a time series such as the Disney example, the convention is to place the time data on the horizontal axis of the graph. This is accomplished by placing the time data in Column A in the Excel spreadsheet and the other data in Column B, as shown in Figure 2.26A.

Let’s wrap this chapter up with one final Your Turn to test your scatter plot skills by en- gaging in one of my favorite activities—car shopping.

{ figure 2.26a Constructing a Line Chart in

Excel

{ figure 2.26b Line Chart Showing Disney’s

Revenues, 1993–2012

A line chart is a special type of scatter plot in which the data points in the scatter plot are connected with a line.

Figure 2.26B shows our line chart with titles added. This chart provides much better in- formation than Table 2.23. We can see that Disney experienced the most significant revenue growth in the mid-1990s, followed by a few years of flat revenues and then significant in- creases from 2004 to 2008. Connecting the data points with a line helps readers of the chart visualize the movement in the data over time. Now, who wants to go see Mickey?

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C h a p t e r 2 | 2.6 Scatter Plots 65

2.6 Section Problems 2.27 Netflix is a company that provides online video streaming

and a DVD-by-mail service for members since 1997. The following table shows the number of U.S. subscribers, in millions, each quarter from 2007 to 2012. These data can also be found in the Excel file Netflix subscribers.xlsx.

Quarter 2007 2008 2009 2010 2011 2012

1 6.8 8.2 10.3 14.0 22.8 23.4

2 6.7 8.4 10.6 15.0 24.6 23.9

3 7.0 8.7 11.1 16.8 23.8 25.1

4 7.5 9.4 12.3 19.5 24.4 27.1

Construct a line chart using these data. Describe the rela- tionship between time and number of subscribers.

2.28 Suppose the Internet retailer Buy.com would like to investi- gate the relationship between the amount of time in minutes a purchaser spends on its Web site and the amount of money he or she spends on an order. The following table shows the data from a random sample of 12 customers. These data can also be found in the Excel file titled Buy.com.xlsx.

Time Order Size ($) Time Order Size ($)

18 69 5 58

13 26 37 365

26 94 6 127

Time Order Size ($) Time Order Size ($)

23 199 36 160

2 49 24 75

14 38 9 250

Construct a scatter plot using these data. Describe the relationship between the amount of time in minutes a pur- chaser spends on its Web site and the amount of money he or she spends on an order.

2.29 Take-Two Interactive Software, Inc., produces video games such as Grand Auto Theft and NBA 2K9. The fol- lowing table lists the company’s quarterly profits and losses from 2007 to 2012. These data can also be found in the Excel file Take Two.xlsx.

Q 2007 ($) 2008 ($) 2009 ($) 2010 ($) 2011 ($) 2012 ($)

1 -21.5 -38.0 -53.8 -33.8 -22.0 -66.8 2 -51.2 98.2 -10.1 16.9 6.4 -93.8 3 -58.5 51.8 -55.5 12.4 -47.4 -12.5 4 -7.1 -15.0 -22.0 40.9 14.1 71.3 http:/ir.take2games.com

Construct a line chart using these data.

A few years ago, I had the opportunity to “bond” with my son Brian as we shopped for his first car when he turned 16. Brian had visions of Mercedes and BMWs dancing in his head, whereas I was thinking more along the line of Chevys and Toyotas. After many “discussions” on the matter, we compromised by looking for 1999 Volkswagen Jettas. However, Brian had two requirements:

• The car had to be black.

• It had to be the “new body” style.

Apparently, somebody at Volkswagen had the brilliant idea back in 1999 to subtly change the design of the Jetta halfway through the production year. Personally, I would never have noticed the difference. Brian, on the other hand, wouldn’t be caught dead driving in the original body style, which essentially eliminated half the used 1999 Volkswagen Jettas on the market. Undeterred, I searched the world over, asking each seller, “Is it the new body style?” The things we do for our children. What follows is a table showing the mileage of eight Jettas with the new body style and the asking price of each:

Mileage Price ($) Mileage Price ($)

21,900 16,000 65,900 10,300

34,300 11,400 72,200 12,400

41,000 13,500 76,700 8,000

53,500 14,700 84,800 9,600

Construct a scatter plot for these data. Describe the relationship between the mileage and the price of Jettas with these data.

Y o U r t U r n # 1 1N

Answers can be found on u page 75

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■ Technology, such as Excel, provides an excellent tool to display descriptive data in a graphical format.

■ A frequency distribution shows the number of data observations that fall into specific intervals.

• Relative frequency distributions display the percentage of observations of each class relative to the total number of observations.

• Cumulative relative frequency distributions indicate the percentage of observations that are less than or equal to the current class.

• A histogram graphically displays the frequency distribution as a bar graph showing the number of observations in each class as the height of each bar.

• Ideally, the number of classes in a frequency distribution should be between 4 and 20.

• Some data sets require several values to be grouped together in a single class using the follow- ing rules:

• Create equal class sizes.

• Use mutually exclusive classes.

• Include all data values.

• Avoid open-ended classes.

• Histograms displaying

• discrete data have gaps between the bars indicating the frequency of each class.

• continuous data do not have gaps between the bars.

• A percentage polygon graphs the midpoints of each class as a line rather than columns.

• The height of each midpoint represents the relative frequency of the corresponding class.

• The cumulative percentage polygon, or ogive, is a line graph that plots the cumulative fre- quency distribution.

■ Contingency tables provide a format to display the frequencies of two qualitative variables.

• Contingency tables are constructed with Excel as pivot tables.

■ The stem and leaf display is a graphical technique used to present quantitative data by splitting the data values into stems (the first digits in the value) and leaves (the remaining digit in the value).

■ Bar charts are used to display data that has been organized in categories and can be arranged in a vertical or horizontal orientation.

■ Pareto charts show in a decreasing order the frequency of the classes that cause quality-control problems.

• A Pareto chart also plots the cumulative relative frequency as a line on the chart known as an ogive.

■ Pie charts are a tool for comparing proportions for categorical data.

• Each category occupies a segment of the pie that represents the relative frequency of that category.

■ Scatter plots display the relationship between two quantitative variables that are paired together on an x-y graph.

• A line chart is a type of scatter plot in which the data points in the scatter plot are connected with a line.

Chapter 2 Summary

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Bar chart. A chart that displays qualitative data that have been organized in categories and can be arranged vertically or horizontally.

Class. A category in a frequency distribution. Clustered bar chart. A bar chart that groups several values side

by side within the same category in a vertical direction. Contingency table. A table that provides a format to display the

frequencies of two qualitative variables. Continuous data. Values that can take on any value within a

given range; continuous data are often the result of measuring observations rather than counting them.

Cumulative percentage polygon (ogive). A line graph that plots the cumulative relative frequency distribution.

Cumulative relative frequency distribution. A frequency distri- bution that totals the proportion of observations that are less than or equal to the class at which you are looking.

Dependent variable. The dependent variable is the variable in the scatter plot that is influenced by changes in the other variable, the independent variable; the dependent variable is placed on the vertical axis.

Discrete data. Values based on observations that can be counted; they are restricted to integer values (whole numbers).

Frequency distribution. A table that shows the number of data observations that fall into specific intervals.

Histogram. A bar graph showing the number of observations in each class as the height of each bar.

Horizontal bar chart. A bar chart that displays the bars in a horizontal direction.

Independent variable. The independent variable is the variable in the scatter plot that is not affected by changes in the other vari- able; the independent variable is placed on the horizontal axis.

Line chart. A special type of scatter plot in which the data points in the scatter plot are connected with a line.

Pareto chart. A bar chart that shows in a decreasing order the frequency of the categories that cause quality control problems.

Percentage polygon. Graphs the midpoint of each class as a line rather than a column. The height of each midpoint represents the relative frequency of the corresponding class.

Pie chart. A chart that displays categories in the shape of a pie; each segment of the pie represents the relative frequency of that category.

Relative frequency distribution. A frequency distribution that displays the proportion of observations of each class relative to the total number of observations.

Scatter plot. A chart that provides a picture of the relationship between two quantitative variables that are paired together on a graph with a horizontal and vertical axis.

Stacked bar chart. A bar chart that groups several values in a single column within the same category in a vertical direction.

Stem and leaf display. A chart that splits data values into stems (the larger place values) and leaves (the smaller place values).

Symmetrical distribution. The shape of a distribution when its two halves mirror one another.

Time series. A data set where each data point is associated with a specific point in time.

Chapter 2 Key Terms

Chapter 2 Problems 2.30 The following table shows the number of complaints re-

ceived each day by a customer service department over the past 40 days. These data can also be found in the Excel file complaints.xlsx.

8 10 8 2 4 13 23 10 6 3

18 16 11 3 3 17 12 8 10 6

3 19 9 10 21 11 8 10 0 7

20 13 18 3 5 19 2 8 10 9

a. Using the 2k Ú n rule, construct a frequency distribution for these data.

b. Using the results from part a, calculate the relative fre- quencies for each class.

c. Using the results from part a, calculate the cumulative relative frequencies for each class.

d. Construct a histogram for these data.

2.31 The Arizona Diamondbacks and the city of Phoenix are considering building a new baseball stadium for the Ma- jor League Baseball team. A decision needs to be made whether to enclose the new facility with a roof. To help make this decision, data were gathered on the number of days per month it rained during the baseball season, which

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are shown in the following table. These data can also be found in the Excel file Diamondbacks.xlsx.

0 3 5 0 3 0 1 0 5 0 2 1 0 4 1 1 0 2 3 2 0 1 1 3 3 3 3 4 0 3 2 4 0 4 0 1 3 0 2 0 1 0 0 4 2 2 1 0 3 3

a. Construct a frequency distribution for these data. b. Using the results from part a, calculate the relative fre-

quencies for each class. c. Using the results from part a, calculate the cumulative

relative frequencies for each class. d. Construct a histogram for these data. e. What is the likelihood that a month during the baseball

season will experience one day or less of rain?

2.32 The following table shows the number of minutes that 48 Verizon Wireless customers used on their cell phone accounts last month. These data can also be found in the excel file wireless.xlsx.

235 406 339 544 78 327 922 58 76 860 549 179 492 63 1,187 188 288 90 235 353 964 577 524 43 731 45 343 170 178 338 169 136 728 678 74 483 522 279 863 373 255 692 87 486 238 218 550 402

a. Using the 2k Ú n rule, construct a frequency distribution for these data.

b. Using the results from part a, calculate the relative fre- quencies for each class.

c. Using the results from part a, calculate the cumulative relative frequencies for each class.

d. Construct a histogram for these data.

2.33 The following table lists the math SAT scores for 72 stu- dents. These data can also be found in the excel file math SAT scores.xlsx.

586 621 613 493 552 449 559 672 428 409 571 449 498 457 578 369 511 726 566 412 795 382 607 701 739 359 354 517 384 196 496 515 439 493 508 535 507 509 428 612 530 556 660 576 626 522 515 374 190 601 499 593 538 509 450 690 571 252 599 362 505 555 323 519 524 281 580 412 553 394 555 541

a. Using the 2k Ú n rule, construct a frequency distribution for these data.

b. Using the results from part a, calculate the relative fre- quencies for each class.

c. Using the results from part a, calculate the cumulative relative frequencies for each class.

d. Construct a histogram for these data.

2.34 The following data show the exam grades for two differ- ent sections of a statistics class that I teach. The first group of data is from a day section of traditional students. The second group of data is from my evening section, which is comprised mostly of older students who work full-time dur- ing the day. These data can also be found in the Excel file statistics grades.xlsx. Note: Consider these data as discrete data because fractional grades will not occur.

Day

71 69 68 90 87 76 90 66 90 71

66 75 78 80 75 90 76 77 80 100

75 75 66 86 85 86 77 75 66 77

Evening

96 98 88 94 88 86 100 99 95 89

96 98 92 85 93 83 90 88 91 89

87 82 83 87 88 88 81 93 77 90

a. Construct a percentage polygon for these data. b. Construct a cumulative percentage polygon for these data. c. What conclusions can be drawn about the grade distribu-

tion of these two sections?

2.35 On-time arrival is a critical measurement in the airline in- dustry. The Excel file titled airline arrivals.xlsx lists the number of minutes that a particular flight either arrived early (negative value) or late (positive value) over the past 300 days of operation.

a. Using Excel and the 2k Ú n rule, construct a frequency distribution for these data.

b. Using the results from part a, calculate the relative fre- quencies for each class.

c. Using the results from part a, calculate the cumulative relative frequencies for each class.

d. Construct a histogram for these data. e. What approximate percentage of flights did not arrive

late during this time period?

2.36 The following data show the average monthly revenue a store earned per customer by selling six different types of cell phones:

Phone Average Revenue Per Customer ($) iPhone 95

Blackberry 85

LG 60

Samsung 55

Motorola 55

Prepaid 20

a. Construct a vertical bar chart for these data. b. Construct a horizontal bar chart for these data.

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2.37 The recent downturn in the housing market can be mea- sured by the number of days homes are on the market before they are sold. To compare the health of the housing market between the communities of Wayne, Pennsylvania, and Dover, Delaware, these data were collected for 100 homes from each city and can be found in the Excel file days on the market.xlsx.

a. Construct a percentage polygon for these data. b. Construct a cumulative percentage polygon for these

data. c. What conclusions can be drawn about the housing mar-

ket in these two communities?

2.38 Charles Schwab conducted a survey that asked 1,252 young adults between the ages of 22 and 28 the following ques- tion: “What most surprised you as you began life on your own?” The following table summarizes the responses:

Response Percentage

High cost of living 26

Difficulty saving money 22

Number of financial decisions 13

Difficulty paying all the bills 12

Challenges finding a job 10

Based on: Charles Schwab.

a. Construct a vertical bar chart for these data. b. Construct a horizontal bar chart for these data.

2.39 The following table shows the number of wins for baseball teams in the National League Central Division for the 2011 and 2012 seasons:

Team 2011 2012

St. Louis Cardinals 90 88

Chicago Cubs 71 61

Milwaukee Brewers 96 83

Cincinnati Reds 79 97

Houston Astros 56 55

Pittsburgh Pirates 72 79

Based on: http://www.espn.com.

Construct a horizontal stacked bar chart that summarizes these data.

2.40 The following table lists the reasons customers provided over the phone when they rated their customer service ex- perience with a particular cable company unsatisfactory.

Reason Frequency

Too long on hold 47

Not knowledgeable 22

Not courteous 18

Hard to understand 15

Reason Frequency

Too many transfers 10

Other 8

Construct a Pareto chart for these data.

2.41 AutoBeef is a Web site to which customers post complaints about their cars. The following table lists the number of complaints and the type of complaints posted about used Ford Explorers:

Complaint Frequency

Transmission 721

Body 437

Wheels 164

Drivetrain 139

Windows 89

Engine 55

Interior 45

Electrical 44

Steering 42

Suspension 41

Air conditioner/heater 26

Brakes 22

Other 47 Based on: http://www.carcomplaints.com/Ford/Explorer/2002.

Construct a Pareto chart for these data.

2.42 A survey conducted by NetApplications in January 2013 on the use of Internet browsers reported the following market shares:

Browser Market Share (%)

Internet Explorer 55.2

Firefox 20.0

Chrome 17.5

Safari 5.2

Opera 1.8

Other 0.3

Construct a pie chart to display these data.

2.43 SnagAJob.com surveyed 551 hiring managers to ask them what recruiters looked for when hiring seasonal employees. The following table shows the results:

Response Percentage

Positive attitude 36

Ability to work the daily schedule 27

Previous experience in the industry 23

Commitment to work for the entire summer 14

Construct a pie chart to display these data.

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2.44 The following table shows the sources of our electricity, according to the U.S. Department of Energy:

Source Percentage

Coal 48

Natural gas 21

Nuclear 20

Hydro 6

Other 4

Oil 1

Construct a pie chart to display these data.

2.45 Data were collected about the time per day children 8 to 18 years old spent with various types of media. The following table reports the average time the children spent with each media category:

Media Time

Television content 4 hours 29 minutes

Music/audio 2 hours 31 minutes

Computer 1 hour 29 minutes

Video games 1 hour 13 minutes

Print 38 minutes

Movies 25 minutes

Construct a display that best describes these data.

2.46 The following table shows the average number of view- ers aged 18–49 who watch the ABC, CBS, FOX, or NBC networks at 10:00 p.m. The data were collected by Nielsen during the last week of January 2012 and 2013.

Network Average Number of Viewers (Thousands)

2012 2013

CBS 4,071 3,542

NBC 3,163 3,669

ABC 3,174 2,910

FOX 4,162 3,155

Construct a display that best describes these data.

2.47 The following table indicates the market share for LCD televisions in the fourth quarter of 2011.

Brand Market Share %

Samsung 23.6

Vizio 15.4

LG 12.4

Sony 8.0

Toshiba 7.8

Construct a display that best describes these data.

2.48 A Scottrade survey asked adults how much money they have saved for retirement. The data are as follows:

How Much Percentage

Less than $25,000 39

$25,001 to $100,000 14

$100,001 to $250,000 9

$250,001 to $500,000 7

More than $500,000 7

Not sure 24

Construct a display that best describes these data.

2.49 The number of television coverage hours for six Winter Olympics is shown in the following table:

Winter Olympics Television Coverage Hours

Albertville (1990) 116

Lillehammer (1994) 120

Nagano (1998) 124

Salt Lake City (2002) 376

Torino (2006) 398

Vancouver (2010) 835

Construct a display that best describes these data.

2.50 The following table shows the average 401(k) balances, in thousands of dollars, by age group according to the AARP.

Age Group Balance ($)

50–54 224

55–59 250

60–64 234

65–69 228

70 and older 230

Construct a display that best describes these data.

2.51 The following table displays the average mathematics scores from the Trends in International Mathematics and Science Study (TIMSS) for fourth graders in 1995 and 2011 for various countries.

Average Math Score

Country 1995 2011

Singapore 590 606

Japan 567 585

Great Britain 484 542

United States 518 541

Australia 531 508

Construct a display that best presents these data with the goal of comparing scores between 1995 and 2012 for each country.

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2.52 A survey conducted to determine how students pay for col- lege reported the following results.

Source Percentage

Grants/scholarships 29

Parent income and savings 28

Student borrowing 18

Student income and savings 12

Parent borrowing 9

Friends/relatives 4

Construct a display that best presents these data.

2.53 The following table provides the new housing starts, in thousands, in different regions in the United States over a three-year period.

Region 2010 2011 2012

Northeast 71.6 67.7 77.8

Midwest 97.9 100.9 128.8

South 297.5 307.8 398.7

West 119.9 132.5 174.7

Based on: U.S. Department of Commerce.

Construct a display that best presents these data with the goal of comparing total housing starts over this three-year period for each region in the country.

2.54 The following data shows the ages of 55 current BMW owners. These data can also be found in the Excel file BMW owners.xlsx.

28 50 37 51 42 27 40 38 46 48 56 29 44 40 42 46 50 30 60 30 43 66 32 36 35 48 47 43 39 26 44 32 35 47 42 47 34 39 34 60 43 26 50 43 61 42 32 45 46 49 54 51 39 26 49

a. Using the 2k Ú n rule, construct a frequency distribution for these data.

b. Using the results from part a, calculate the relative fre- quency distributions for each class.

c. Using the results from part a, calculate the cumulative relative frequency distributions for each class.

d. Construct a histogram for these data.

2.55 A survey asked 614 small-business owners if they found their company’s Facebook presence valuable. The results are shown in the following table.

Response Percentage

Yes 42

No 30

Don’t have a presence 23

No longer have a presence 5

Construct a display that best presents these data.

2.56 The following data represent the number of student ab- sences for a large, two-semester statistic class. These data can also be found in the Excel file absence.xlsx.

3 5 2 5 5 3 2 6 2 2 0 2 4 4 0 5 5 2 4 2 3 5 2 3 6 5 4 3 2 7 3 7 3 4 5 3 3 5 6 4 4 3 2 3 3 3 2 6 2 4 4 5 2 1 3 4 7 3 4 3

a. Construct a frequency distribution for these data. b. Using the results from part a, calculate the relative fre-

quency distributions for each class. c. Using the results from part a, calculate the cumulative

relative frequency distributions for each class. d. Construct a histogram for these data. e. What is the likelihood that more than four students will

be absent from this statistics class?

2.57 The following data represent the number of cars per day that used the drive-through lane for a fast food restaurant.

117 104 102 98 97 96 95 91 90 90 89 88 87 87 86 85 83 82 81 78 77 77 76 75 74 71 71 71 70 70 70 69 69 67 67 64

a. Construct a stem and leaf display for these data. b. Construct a stem and leaf display for these data, splitting

the stems from part a.

2.58 A marketing research firm would like to display the re- lationship between a family’s weekly food costs and the number of family members living in the household. The following table shows the monthly food costs and the num- ber of family members for 16 families. These data can also be found in Excel file food costs.xlsx.

People Costs ($) People Costs ($)

4 926 6 1,263

5 1,155 3 1,023

3 962 8 1,265

4 909 3 1,114

4 1,084 7 1,322

2 755 2 937

4 949 5 962

4 1,000 3 829

Construct a scatter plot using these data. Describe the re- lationship between the number of household members and the family’s monthly food costs.

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2.59 The following table represents the average new-home size, in square feet, in the United States over various years.

Year Average Size

1985 1,785

1990 2,080

1995 2,100

2000 2,266

2005 2,434

2010 2,392

Construct a display that best describes these data.

2.60 The term never events in the health care industry refers to mistakes that should never happen in the operating room. Researchers identified the following “never events” and their relative frequency of occurring.

Event Relative Frequency (%)

Foreign objects left behind 49.8

Wrong procedure 25.1

Wrong site 24.8

Wrong patient 0.3

Construct a Pareto chart for these data.

2.61 As a long-time lover of Cheez-It crackers, I find myself of- fended when somebody claims that Cheese Nips, which is a competing product, is just as good. To put an end to this silly argument, I performed a taste test with 24 unbiased individuals, asking them to state their preference and listed the results in the following table. Those with no preference were recorded as “No Prefer.” These data can also be found in the Excel file titled Cheezits.xlsx.

Cheez- Its

Cheese Nips

Cheez- Its

Cheese Nips

Cheez- Its

No Prefer Cheez- Its

Cheese Nips

Cheez- Its

No Prefer Cheese Nips

Cheez- Its

No Prefer Cheez- Its

Cheese Nips

Cheese- Its

Cheez- Its

a. Construct a frequency distribution for these data. b. Construct a relative frequency distribution for these

data.

2.62 The following data represents the total annual snowfall (in inches) for 30 cities.

12 13 15 18 21 21 23 26 26 27 27 28 30 33 33 35 36 36 39 40 40 42 42 44 46 47 49 50 51 57

Construct a stem and leaf display using these data.

2.63 The following table shows the percentage of children at various age groups who use smartphones and laptop com- puters according to a survey conducted by the NPD Group.

Age Smartphones (%) Laptops (%)

4–5 37 22

6–8 35 28

9–11 38 37

12–14 47 43

Construct a display that best presents these data with the goal of comparing percentages between devices for each age group.

2.64 A grocery store manager records the purchases of soda by the brands Coke (C), Pepsi (P), and Mt. Dew (M). The type of soda is also noted (regular or diet). The following table shows the brand and type of soda for the last 33 customers. These data can also be found in the Excel file soda.xlsx.

Brand Type Brand Type Brand Type

P Reg C Diet M Reg

C Diet M Reg P Diet

P Diet P Diet C Reg

M Reg C Diet P Reg

C Reg P Reg M Reg

P Reg C Reg P Reg

C Diet M Reg C Reg

C Diet M Diet M Reg

M Diet C Reg P Reg

C Reg M Reg C Diet

P Diet P Reg M Reg

Develop a contingency table for these data. What conclusions can be drawn about the brands and types of soda sales?

2.65 Golf Galaxy, a major golf equipment retailer, would like to investigate the relationship between the brand of golf clubs (Taylor Made, Nike, and Callaway) and the age of the golfer who purchased them. The Excel file titled golf clubs. xlsx lists the golf club brand purchased by the last 100 cus- tomers along with the customers’ ages. Using Excel, con- struct a pivot table for these data. What conclusions can be drawn about the golf club brand and age of the customer? Hint: Group the ages of the customers with a class width of 10 years using the following steps:

a. After the initial pivot table is constructed, right-click on one of the ages.

b. In the drop-down menu, left-click on Group. c. In the Group dialog box, input the following values:

1. Starting at = 20 2. Ending at = 59 3. By = 10

d. Click OK.

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2.66 The class sizes for 40 courses currently being taught at Neumann University this semester are shown in the fol- lowing table. These data can also be found in the Excel file class size.xlsx.

28 26 35 28 20 43 31 22 25 18

35 39 19 29 23 31 45 35 22 28

28 20 41 36 32 32 39 33 25 30

19 43 23 31 31 51 46 20 36 25

a. Construct a stem and leaf display for these data. b. Construct a stem and leaf display, splitting the stems

from part a.

2.67 The following table shows the number of customers who have visited a T-Mobile retail store during the past 40 days. These data can also be found in the Excel file T-Mobile.xlsx.

72 99 99 100 119 91 101 74 102 130

81 76 70 97 77 70 130 85 82 131

94 125 97 112 118 72 93 93 75 95

93 104 92 105 77 138 88 77 90 92

a. Construct a stem and leaf display for these data. b. Construct a stem and leaf display, splitting the stems

from part a.

2.68 The Excel file MLB payroll.xlsx provides the payroll and number of wins for all the Major League Baseball teams

during the 2012 season. Construct a scatter plot using these data. Describe the relationship between payroll and number of wins.

2.69 The Excel file titled gasoline prices.xlsx lists the average monthly price per gallon of regular gasoline in the United States from August 1990 to January 2013 from the U.S. Energy Information Administration. Construct a line chart using these data and describe the trend.

2.70 The Excel file titled airline performance.xlsx lists the status of 100 flights as either “on time” or “late” along with the airline for that flight.

a. Construct a contingency table for these data. b. Construct a relative contingency table for these data. c. What conclusions could be drawn from these data?

2.71 The amount of time required to navigate through airport se- curity can vary from city to city. The Excel file titled TSA times.xlxs shows the number of minutes that 60 passengers spent going through security at Chicago’s O’Hare Interna- tional Airport and Seattle International Airport. Construct a display that will allow a direct comparison of the security times for both airports.

2.72 The Excel file titled US Housing Starts.xlsx lists the num- ber of housing starts in the United States each year from 1970 to 2012 provided by the U.S. Department of Com- merce. Construct a display that best describes these data.

Chapter 2 Solutions to Your Turn

Y o u r T u r n # 1 1. Discrete 2. Continuous 3. Discrete 4. Continuous

5. Discrete 6. Continuous 7. Discrete

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Y o U r t U r n # 3

25 = 32 Ú 30, therefore use five classes.

Estimated class width = $87.7 - $40.0

5 = $9.54 ≈ $10

Salary Frequency Relative Frequency

Cumulative Relative Frequency

40–49.9 3 0.100 0.100

50–59.9 4 0.133 0.233

60–69.9 14 0.467 0.700

70–79.9 7 0.233 0.933

80–89.9 2 0.067 1.000

Total 30 1.000

Y o U r t U r n # 4

Number of Stars

Frequency

Relative Frequency

Cumulative Relative Frequency

1 2 0.08 0.08

2 1 0.04 0.12

3 3 0.12 0.24

4 7 0.28 0.52

5 12 0.48 1.00

Total 25 1.00

Y o U r t U r n # 2

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Y o u r T u r n # 5

Y o u r T u r n # 6

A clustered bar chart would be more appropriate because it provides a convenient way to compare unemployment rates for all five states between 2011 and 2012.

Y o u r T u r n # 7

Y o u r T u r n # 8

Y o u r T u r n # 9 Warranty

Brand Yes No Total

Canon 1 8 9

Nikon 1 5 6

Sony 3 2 5

Total 5 15 20

Warranty

Brand Yes No Total

Canon 0.05 0.40 0.45

Nikon 0.05 0.25 0.30

Sony 0.15 0.10 0.25

Total 0.25 0.75 1.00

Based on these data, it appears that Sony customers are more likely to purchase an extended warranty when compared with Canon and Nikon customers.

As expected, as mileage increases, the selling price decreases.

Y o u r T u r n # 1 1

2(5) � 5 5 5 5 5 5 6 6 7 7 7 8 8 9 9 9

3(0) � 2 2 3 3 3 3 4

3(5) � 6 7 7 7 8

4(0) � 0

4(5) � 8

Y o u r T u r n # 1 0

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C h a p t e r

I n t h I s c h a p t e r , y o u w I l l l e a r n t o :

■ Calculate the mean, weighted mean, median, and mode.

■ Identify the most appropriate measure of central tendency to describe data.

■ Calculate and interpret the range, variance, and standard deviation.

■ Understand the coefficient of varia- tion, z-score, the empirical rule, and Chebyshev’s Theorem.

■ Calculate the mean and standard deviation of grouped data.

■ Calculate percentiles, percentile rank, and quartiles for a data set.

■ Construct and interpret box-and- whisker plots.

■ Calculate and interpret the covari- ance and the correlation coefficient for two data sets.

Calculating Descriptive Statistics

C h a p t e r 3 M a p

3.1 Measures of Central Tendency | 78

3.2 Measures of Variability | 91

3.3 Using the Mean and Standard Deviation Together | 101

3.4 Working with Grouped Data | 112

3.5 Measures of Relative Position | 116

3.6 Measures of Association Between Two Variables | 126

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78 C h a p t e r 3 | Calculating Descriptive Statistics

The emphasis in Chapter 2 was on demonstrating ways to display our data graphically so that our brain cells could quickly absorb the big picture. With that task behind us, we can proceed to the next step—summarizing our data numerically. Chapter 3 focuses on how to calculate descriptive statistics manually and, if you so choose, how to verify these results with our good friend Excel. The Stats in Practice at the start of this chapter relied on the median to present income data over a five-year period. We will explore this measurement, along with many others, throughout Chapter 3. Also, I hope the opening example sends you another message loud and clear: Stay in school and get your degree!

3.1 Measures of Central Tendency Most data sets have a tendency to concentrate around a specific value. For example, the his- togram in Figure 3.1 shows that U.S. adult males from a sample have a tendency to cluster around a height of 70 in. This value is known as the central tendency for our data set. As we learned in Chapter 2, a histogram is a great tool for providing a picture to show where the central tendency can be found. In this chapter, we will explore how to precisely calculate the

Stats in practice: Using the Median to Compare Salaries

The Bureau of Labor Statistics (BLS) is an agency of the fed- eral government that collects and reports labor data for the economy. Their Web site, www.bls.gov, provides a wealth of information on earnings and employment status for various characteristics such as age, gender, and education. Accord- ing to BLS, women’s wages have increased at a higher rate than men’s since the start of the recent economic recession. Between 2007 and 2012, women’s wages have risen 12.5% compared to an 11.5% increase for men.

The median weekly wage for men in 2012 was $854, compared with a median wage of $691 per week for women. However, men had a higher unemployment rate during 2012 when com- pared to women: 8.2% vs. 7.9%.

According to BLS, education also played a role in wages. An in- dividual without a high school diploma earned a median of $451 a week in 2011, compared to $1,150 for an individual with at least a bachelor’s degree.

Based on: http://www.bls.gov.

Central tendency is a single value used to describe the center point of a data set.

{ figure 3.1 Heights of U.S. Adult Males

60 to

< 62

62 to

< 64

64 to

< 66

66 to

< 68

68 to

< 70

70 to

< 72

72 to

< 74

74 to

< 76

76 to

< 78

78 to

< 80

80 to

< 82

82 to

< 84

Height (in.)

Heights of U.S. Adult Males

100

200

300N u

m b

er o

f A

d u

lt s

400

900

800

700

600

500 The central tendency is the value where data points tend to cluster.

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C h a p t e r 3 | 3.1 Measures of Central Tendency 79

where

x = The sample mean xi = The values in the sample (x1 = the first data value, x2 = the second data value,

and so on)

a n

i = 1 xi = The sum of all the data values in the sample

n = The number of data values in the sample To see how the sample mean is calculated, look at the following data set, which shows the an-

nual household incomes (in thousands of dollars) of five randomly selected families in Delaware.

$87.2 $118.9 $76.2 $107.7 $61.5

These data values are a sample because they represent a subset of all of the households in Delaware.

The sample mean is found as follows:

x = a

i = 1 xi

n =

87.2 + 118.9 + 76.2 + 107.7 + 61.5 5

= 451.5

5 = $90.3

The average income from our sample of five Delaware households is $90,300. The population mean is calculated in the same way as the sample mean; the only difference

is in the notation. The formula for determining the population mean is shown in Equation 3.2.

Formula 3.1 for the sample Mean

x = a

i = 1 xi

Formula 3.2 for the population Mean

m = a N

i = 1 xi

The mean, or average, which is the most common measure of central tendency, is calculated by adding all the values in a data set and then dividing the result by the number of observations.

value that represents the central tendency of our data. As we will see in the following sec- tions, our choices of central tendency measures are the mean, median, and mode.

the Mean The most common measure of central tendency is the mean, or average. We calculate the mean by adding all the values in our data set and then dividing the result by the number of observations. The mathematical formula for the mean differs slightly depending on whether you’re referring to the mean of the sample or the mean of the population. The formula for the sample mean is shown in Equation 3.1.

Don’t panic when you see the

symbol a n

i = 1 xi. If our data

sample contains the values 5, 8, and 2, then n = 3, x1 = 5, x2 = 8, and x3 = 2, resulting in the following expression:

a 3

i = 1 xi = x1 + x2 + x3

= 5 + 8 + 2 = 15

where

m = The population mean N = The number of data values in the population Let’s say you have just completed a statistics course that had four exams, and your

scores are as follows:

89 95 87 97

This data set is a population because all of your exam scores for the course are included.

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80 C h a p t e r 3 | Calculating Descriptive Statistics

The population mean is shown as follows:

m = a N

i = 1 xi

N =

89 + 95 + 87 + 97 4

= 368

4 = 92

Your exam average was a 92. Very impressive!

the Weighted Mean When we calculated the average exam score in the previous example, we gave each data value the same weight in the calculation. A weighted mean refers to a mean that needs to go on a diet. Just kidding; I was checking to see whether you were paying attention. Let’s try this again. A weighted mean allows you to assign more weight to certain values and less weight to others. For example, let’s say your statistics grade this semester will be based on a combination of your final exam score, a homework score, and a final project, each weighted according to Table 3.1.

Formula 3.3 for the weighted Mean

x = a

i = 1 1wixi2

a n

i = 1 wi

The symbol a n

i = 1 1wi xi2 means

“the sum of w multiplied by x.” Each pair of w and x is first multiplied together, and these results are then summed.

Note that in this equation, we are dividing by the sum of the weights rather than by the number of data values.

{ table 3.1 | Scores from a Statistics Class

type score weight (%)

Exam 94 50

Project 92 35

Homework 100 15

We can calculate your final grade using the following formula for a weighted average (Equation 3.3).

where

wi = The weight for each data value xi

a n

i = 1 wi = The sum of the weights

Table 3.2 sets up how to use Equation 3.3 for our statistics class example.

{ table 3.2 | Data Used to Calculate a Weighted Mean

type score weight weight : score i xi wi 1wixi2

Exam 1 94 0.50 47.0

Project 2 92 0.35 32.2

Homework 3 100 0.15 15.0

A weighted mean allows you to assign more weight to certain values and less weight to others.

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C h a p t e r 3 | 3.1 Measures of Central Tendency 81

a 3

i = 1 wi = 0.50 + 0.35 + 0.15 = 1.0

a 3

i = 1 1wixi2 = 47.0 + 32.2 + 15.0 = 94.2

We can obtain the same result by plugging the numbers directly into Equation 3.3, the formula for a weighted average:

x = 10.5021942 + 10.3521922 + 10.15211002

0.50 + 0.35 + 0.15 =

47.0 + 32.2 + 15.0 1.0

= 94.2

Congratulations—you earned an A! The weights in a weighted average do not need to add up to 1.0 like they do in our example.

For instance, the mean of a frequency distribution is simply a weighted mean calculation, as shown in the next example.

An airline company recorded the number of no-shows (people who fail to arrive at the gate on time to board a plane) for the past 120 flights. The frequencies are shown in Table 3.3. For 37 of the 120 flights, all of the passengers made it on board in time. There were zero no-shows, in other words. Thirty-one flights each had one passenger missing, 20 flights each had two passengers missing, and so on down the table. We would like to calculate the aver- age number of no-shows per flight.

Even though we can’t have 1.56 passengers not show up for a particular flight, the average number of no-shows can be a fraction.

{ table 3.3 | Frequency of No-Shows for Airline Flights

number of no-shows number of flights

0 37

1 31

2 20

3 16

4 12

5 4

To calculate the average number of no-shows per flight, use Equation 3.3. The weights would be the number of flights for each number of no-shows.

x = a wi xia wi = 1372102 + 1312112 + 1202122 + 1162132 + 1122142 + 142152

37 + 31 + 20 + 16 + 12 + 4

x = 187

120 = 1.56 passengers

The average number of no-shows per flight is 1.56 passengers.

the advantages and Disadvantages of Using the Mean to Summarize Data The mean is a convenient measurement for summarizing a data set with a single value. Most students will agree with me when I say it’s a pretty easy calculation to perform and understand. However, you need to beware of the mean’s potential pitfalls.

I’ll use the following example to demonstrate the first drawback of the mean. Suppose you have just been hired as a quality control manager for Kellogg’s, and you are responsible for ensuring that each box of Frosted Flakes is properly filled with 18 ounces of cereal.

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82 C h a p t e r 3 | Calculating Descriptive Statistics

On Friday afternoon, your staff informs you that a sample of filled cereal boxes has an aver- age weight of 18.1 ounces. Satisfied that the process is working well, you head home for a relaxing weekend, only to return Monday to customer complaints that their Frosted Flakes boxes were missing some Frosted Flakes! How can this be? When you ask to see the data from Friday’s sample, you are shocked. The weights of the boxes sampled ranged widely. As the following equation shows, one box contained only 13.1 ounces of Frosted Flakes, whereas another contained a whopping 22.7 ounces!

x = a

i = 1 xi

n =

14.2 + 19.6 + 22.7 + 13.1 + 20.9 5

= 90.5

5 = 18.1 ounces

Obviously, the process was over-filling and under-filling boxes, resulting in a sample mean that led you to believe all was well. This example highlights one drawback of using the mean: When you summarize data with a single value, you lose information about the origi- nal data, which in this case was critical. On Tuesday, you find yourself updating your resume and searching the Internet for a new job.

Another concern about relying on the mean involves how it is affected by outliers in the data. Outliers are extreme values above or below the mean that require special consideration. They could be caused by data entry errors or measurement errors. In this case, you should consider deleting these values from the data set to avoid distorting the analysis. However, the outlier could be a legitimate value that just happens to be very large or small and should remain in the analysis. Consequently, you will have to judge carefully whether to include the outlier. For example, the following data represent the appraised value of five homes in a particular cul-de-sac:

$300,000 $320,000 $270,000 $210,000 $8,000,000

The average value of the homes, in thousands of dollars, is

m = a N

i = 1 xi

N =

300 + 320 + 270 + 210 + 8,000 5

= 9,100

5 = $1,820

As you can see, the average home value of $1,820,000 is much higher than most of the houses in the cul-de-sac. The one extreme home value of $8,000,000 distorts the population mean, making the rest of the homes in the cul-de-sac seem much more expensive than they actually are. We will discuss more about outliers and how to deal with them later in the chapter.

When you rely on the mean to summarize many data values, you lose important information about the original data.

This outlier is causing the mean of this data set to be much higher than most of the remaining values, which can be misleading.

I don’t mean to be mean, but I’m going to have you test your mean skills on this Your Turn.

1. The following data indicate the number of minutes eight randomly selected airline flights were either early (negative values) or late (positive values) arriving at their destinations. Calculate the sample mean.

14 - 17 31 - 10 0 11 5 18

2. A company has four locations at which customers were surveyed about their satisfaction levels. The following table shows the average customer rating for each of the four locations along with the number of customers at each location who responded. Calculate the average customer rating for the entire company.

Location Average Rating Number of Customers

1 7.2 116

2 8.8 80

3 6.0 67

4 7.5 91

Y o U r t U r n # 1N

Answers can be found on u page 141

Outliers are values that are much higher or lower than most of the data. The mean can be heavily influenced by outliers.

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C h a p t e r 3 | 3.1 Measures of Central Tendency 83

the Median The next measure of central tendency that we will explore is the median. The median is the value in the data set for which half the observations are higher and half the observations are lower. To demonstrate the technique for finding the median, consider the following data set, which represents nine quiz scores for a particular statistics student:

86 95 74 83 92 70 88 81 91

We find the median by first arranging the data values in ascending order:

70 74 81 83 86 88 91 92 95

To determine the position of the median in our data set, we calculate the index point. The index point, i, marks the middle of the data values. We can calculate it using Equation 3.4.

Formula 3.4 for the Index point for the Median

i = 0.51n2

The median is the value in the data set for which half the observations are higher and half the observations are lower.

where n equals the number of data points, which in this case is nine.

i = 0.51n2 = 0.5192 = 4.5 Whenever the index point is not a whole number, we round the value up to the next

highest whole number, which, in this case, is five. The median exam score is therefore in the fifth position of our sorted data set and is shown in red:

70 74 81 83 86 88 91 92 95

The median of our quiz scores is 86. Notice that our index point using Equation 3.4 found the middle value, which is the definition of the median. There are four quiz scores below and four quiz scores above the median.

In the last section, we discussed the fact that outliers in our data set can heavily affect the mean. This is not so with the median, as I will now demonstrate. Let’s replace the quiz score of 70 with an abysmal score of 20, as shown as follows. (Aren’t you glad it wasn’t you that scored the 20?)

20 74 81 83 86 88 91 92 95

Notice that the change has no impact on the median calculation. With nine data points, the median will always be in the fifth position, regardless of the values above and below it. One of the nice features of the median is that it is not influenced by outliers as the mean was in the previous section. Unfortunately, most instructors don’t calculate grades based on the median but on the mean. The mean quiz score from this last example is 78.9 with the outlier data point, much lower than the median score of 86!

When we have an even number of data values, the procedure for finding the median dif- fers slightly. Table 3.4 shows the closing stock prices for Dick’s Sporting Goods at the end of each month between July and December 2012.

{ table 3.4 | Closing Stock Prices for Dick’s Sporting Goods

date stock price ($)

July 31 49.12

August 31 49.76

September 28 51.85

October 31 50.00

November 30 52.51

December 31 45.49

The index point is used to determine the position of the median in the data set.

The outlier in this data set does not influence the median, which remains at 86.

When there are an odd number of data values, the median is always the middle value in the data set.

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84 C h a p t e r 3 | Calculating Descriptive Statistics

First, sort the stock prices from low to high:

$45.49 $49.12 $49.76 $50.00 $51.85 $52.51

Next, calculate the index point using Equation 3.4:

i = 0.51n2 = 0.5162 = 3 Here is where the difference lies when there is an even number of data points: When the

index point is an even whole number, the position of the median is halfway between the in- dex point (i) and the next highest data point (the i + 1 position). Because i = 3, the median is the midpoint between the 3rd value ($49.76) and 4th value ($50.00).

Median

$45.49 $49.12 $49.76 $50.00 $51.85 $52.51

In other words, the median is halfway between the two middle values whenever we have an even number of data points. The following equation shows you how to find it:

Median = $49.76 + $50.00

2 = $49.88

1. The following data represent the ages of 10 employees in a marketing department at a specific company. Identify the median age.

26 45 39 28 57 31 43 45 50 62

2. The following data are the math SAT scores of nine students from St. Mark’s High School. Identify the median SAT score.

560 610 490 670 700 460 510 590 650

Y o U r t U r n # 2N

Answers can be found on u page 141

the Mode The last measure of central tendency we’ll discuss in this chapter is the mode, which gets the least amount of attention when compared with the mean and median. Nonetheless, the mode has some useful characteristics. The mode is simply the value that appears most often in a data set, as seen in the following example.

Table 3.5 shows last week’s sales for a particular style of a woman’s dress at my wife’s favorite clothing store. The sales are shown by dress size.

{ table 3.5 | Dress Sales by Size

size number sold

4 7

6 7

8 14

10 15

12 5

14 9

A concept that some of my students struggle with is correctly assigning the median to the index point rather than the point between the middle two values in the data set. You can practice right now with this Your Turn section.

The mode is the value that appears most often in a data set.

When you have an even number of data points, and the index point is a whole number, the median lies halfway between the index point and next highest data point.

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C h a p t e r 3 | 3.1 Measures of Central Tendency 85

A word of warning: Some of my students mistakenly think that the mode equals 15, which is the number of dresses sold. Other students make the mistake of telling me the mode in this problem is 7 because it is the frequency that appears most often.

Note that the mode is not 21. Nor is the mode equal to 11, the frequency that appears most often.

Be careful not to claim that the mode is zero in this example. That is an incorrect statement because the value zero does not occur most frequently.

The mode for this data set is Size 10 because this is the dress size that had the most sales last week (15).

For this example, the mean or median would not be very useful. For instance, to report that the average dress size sold last week was 9.1 isn’t very helpful to the manager of this department because there is no such dress size in the store.

The mode is a particularly useful way to describe categorical data such as the dress sizes in Table 3.5. Table 3.6 shows another distribution of categorical data—television sales categorized by brand. Suppose the data in Table 3.6 show the number of televisions by brand sold by a particular Walmart store last month.

{ table 3.6 | Televisions Sold by Walmart

brand number sold

Sony 21

Panasonic 9

Samsung 18

Sharp 11

Toshiba 15

Vizio 11

I’m hoping that you recognize the data in Table 3.6 as categorical data from what you learned in Chapters 1 and 2. The mode is the only choice for describing this type of data because it is qualitative rather than quantitative. The Sony television brand would be the mode here because it had the highest frequency (21). When dealing with categorical data, the mode is the category with the highest frequency, rather than the actual frequency for that category.

If no data value or category repeats more than once, then we say that the mode does not exist. Consider the following data, which represent the number of years of service for the faculty members in my department:

16 10 3 7 9 2 6

Because no value in this data set repeats itself, there is no mode.

Also, more than one mode can exist, as the following data set shows. The data set indi- cates the number of students absent from my last 10 statistics classes.

2 0 1 3 1 4 2 0 1 2 We can sort the data to make identifying the mode easier.

0 0 1 1 1 2 2 2 3 4

The modes for the data, which are shown in red, are one student and two students be- cause both values appear three times. Because the distribution has two modes, it is called a bimodal distribution.

Even if you’re not in the “mode,” I encourage you to take a minute to be sure you under- stand how to find the mode in this Your Turn section.

Categorical data are data that are described as a category or a label. Examples are gender (male or female), marital status (married, single, widowed), or yes/no responses.

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86 C h a p t e r 3 | Calculating Descriptive Statistics

1. The following data represent the average hourly earnings in the United States for each month during 2012 according to www.data360.org. Determine the mode for these data.

Month Earnings ($) Month Earnings ($)

January 2012 16.90 July 2012 16.85

February 2012 16.85 August 2012 16.75

March 2012 16.75 September 2012 16.71

April 2012 16.74 October 2012 16.73

May 2012 16.74 November 2012 16.86

June 2012 16.81 December 2012 16.94

2. The following data show the number of touchdown passes thrown by the NFL’s top quarterbacks during the 2012 football season. Determine the mode for these data.

43 39 37 34 32 28 27 27 26 26 26 26 24 23 22 22 22 21 20 20 19 19 18 14

3. The following data represent the number of customers per day at a hardware store during the past 14 days. Determine the mode for these data.

67 62 57 31 69 32 83 49 80 53 38 36 46 41

4. The demand for various flavors of Rita’s Water Ice yesterday is shown in the following table. Identify the mode.

Flavor Number Sold

Lemon 37

Grape 46

Cherry 60

Orange 46

Lime 19 Answers can be found on u page 141

Y o U r t U r n # 3N

the Shapes of Frequency Distributions The positions of the mean and median will influence the shape of the frequency distribution for the data. Figure 3.2 shows a case in which the mean and median are approximately equal. This results in a symmetrical frequency distribution. Recall from Chapter 2 that a symmetri- cal distribution is one in which the right side of the distribution looks like the mirror image of the left side of the distribution.

{ figure 3.2 A Symmetrical Frequency

Distribution

In a symmetrical frequency distribution, the mean and median are roughly equal to each other.

Mean Median

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C h a p t e r 3 | 3.1 Measures of Central Tendency 87

Using excel to Calculate the Mean, Median, and Mode Excel provides a convenient method for determining the mean, median, and mode for your data set. We will demonstrate this with the following example. The following data represent the number of magazine subscriptions from 12 households:

4 1 0 8 2 5 4 6 1 5 2 1

Figure 3.5 shows the Excel equations used to calculate the mean, median, and mode for this data set. The Excel functions for these calculations are as follows:

= AVERAGE1data values2 = MEDIAN1data values2 = MODE.SNGL1data values2

As you can see from Figure 3.5, the mean = 3.25, the median = 3, and the mode = 1.

In a left-skewed distribution, the median is larger than the mean.

Figure 3.3 shows a distribution that’s not symmetrical. Such a distribution is said to be skewed. Notice that higher data values dominate the data set. As you can see in this figure, the median is larger than the mean. As a result, the shape of the distribution is left-skewed, even though the data are concentrated on the right side. An easy way to remember this distri- bution is left-skewed is to observe that if you were skiing down the long slope of the graph, you would be moving in the left direction. For example, if grade inflation exists at your institution, which occurs when the grade point average (GPA) of the student body steadily increases over time, then a histogram of student GPAs will resemble a left-skewed distribu- tion, as shown in Figure 3.3.

Figure 3.4 shows the opposite scenario, one in which the data are concentrated on the lower end of the distribution. Here the median is less than the mean. The shape of the distribution is right-skewed because we ski toward the right as we move down the slope. For example, annual income data tend to be right-skewed because most of us hardworking individuals tend to be in either the middle or somewhat to the left side of the income scale, whereas a few ultra-rich Warren Buffet–types of people are spread far to the right side of the distribution.

{ figure 3.3 A Left-Skewed Frequency

Distribution

Mean Median

This distribution is left-skewed because if we were skiing down this slope, we would be moving toward the left.

{ figure 3.4 A Right-Skewed Frequency

Distribution

MeanMedian

This distribution is right-skewed because if we were skiing down this slope, we would be moving toward the right.

In a right-skewed distribution, the me- dian is less than the mean.

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88 C h a p t e r 3 | Calculating Descriptive Statistics

If the data set does not have a mode, Excel displays “#N/A” in the cell. If there is more than one mode, Excel will incorrectly display only one mode from the data set, so use cau- tion when relying on this feature. Also, Excel will not identify the mode for categorical data.

The mean, median, and mode can also be obtained using the Data Analysis tool in Excel for Windows. See the textbook’s Web site (www.pearsonhighered.com/donnelly) for Excel for Mac instructions. If you haven’t activated the tool yet, refer to Chapter 2 for instructions. Then determine the mean, median, and mode for our data by taking the following steps:

1. Place the data in Cells A1 through A12 as shown in Figure 3.6.

2. Select Data + Data Analysis. 3. In the Data Analysis dialog box, select Descriptive Statistics and click OK.

{ figure 3.5 Excel Equations Used to

Calculate the Mean, Median, and Mode

4. In the Descriptive Statistics dialog box, click on the textbox for Input Range: and select Cells A1:A12 as shown in Figure 3.7.

5. Click on the textbox for Output Range: and select Cell C1.

6. Check Summary statistics, and click OK.

The results are shown highlighted in red in Figure 3.8, which shows the mean = 3.25, the median = 3, and the mode = 1. We will discuss other values in this figure later on in the chapter.

Be sure you’re aware of some of Excel’s deficiencies when handling the mode.

{ figure 3.6 Using Excel’s Data Analysis Tool

to Calculate the Mean, Median, and Mode (Steps 1–3)

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C h a p t e r 3 | 3.1 Measures of Central Tendency 89

Which Measure of Central tendency Should You Use: the Mean, Median, or Mode? Each of these measures of central tendency has its advantages and disadvantages. The mean is relatively easy to determine and most widely understood by people with little statistical training. As a result, it is by far the most popular choice for measuring the central tendency. However, as you learned, the mean can be heavily affected by outliers, which at times can be misleading.

When outliers are present in the data, the median is often the best choice because it is unaffected by them. However, the median requires a little more effort to determine if you are not using computer software such as Excel, especially when several data points are in- volved. The data need to be sorted from low to high values, and the index point needs to be calculated. In the many statistics exams I have given over the years (please don’t ask how many), I have found that students struggle more with calculating the median when com- pared with calculating the mean. So give the median some extra practice when preparing for your next exam.

When describing the central tendency for categorical data, the mode is your only choice. However, there are two main drawbacks to using the mode. First, in many data sets, a mode does not exist. Second, there may be more than one mode in the data set, a situation that Excel does not properly identify. Table 3.7 summarizes our discussion of central tendency.

{ figure 3.7 Using Excel’s Data Analysis Tool

to Calculate the Mean, Median, and Mode (Steps 4–6)

{ figure 3.8 Using Excel’s Data Analysis Tool

to Calculate the Mean, Median, and Mode (Final Result)

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90 C h a p t e r 3 | Calculating Descriptive Statistics

3.1 Section Problems Basic Skills 3.1 Consider the following data values:

18 12 7 10 15 12 6 12 14

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Describe the shape of this distribution.

3.2 Consider the following data values:

6 14 17 1 16 17 6 19

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Describe the shape of this distribution.

3.3 Consider the following data values:

14 62 0 20 26 38 8

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Describe the shape of this distribution.

3.4 Consider the following data values:

25 18 - 7 10 34 - 12 a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Describe the shape of this distribution.

applications 3.5 The following table shows the monthly unemployment

rate in the United States from January 2011 until January 2013. These data can also be found in the Excel file titled US unemployment.xlsx.

9.1 9.0 8.9 9.0 9.0 9.1 9.0 9.0 9.0 8.9 8.6 8.5 8.3

8.3 8.2 8.1 8.2 8.2 8.2 8.1 7.8 7.9 7.8 7.8 7.9

Based on: http://www.bls.gov.

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Describe the shape of this distribution.

3.6 Bank of America stock prices at the end of each month from February 2012 until January 2013 are shown as below. These data can also be found in the Excel file titled Bank of America.xlsx.

$7.97 $9.57 $8.11 $7.35 $8.18 $7.34

$7.99 $8.83 $9.32 $9.86 $11.61 $11.32

Based on: http://www.bigcharts.com

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Describe the shape of this distribution.

{ table 3.7 | A Summary of the Advantages and Disadvantages of Using the Mean, Median, and Mode

Mean

Advantages

• Relatively easy to calculate • Easy to understand

Disadvantages

• May be heavily affected by extreme values

Median

Advantages

• Not affected by extreme values

Disadvantages

• Requires more effort to determine when compared with the mean

Mode

Advantages

• Can be used with categorical data

Disadvantages

• Might not exist in some data sets • Might not be unique (more than one mode may exist)

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C h a p t e r 3 | 3.2 Measures of Variability 91

3.7 The following frequency distribution shows the various levels of demand for a particular laptop computer sold by Costco during the last 50 business days. Determine the average number of laptops sold per day.

Number of Laptops Frequency

0 14

1 8

2 8

3 12

4 5

5 3

3.8 The following table shows the 2009 carbon emissions, in tons per capita, for various countries around the world. These data can also be found in the Excel file titled emissions.xlsx.

Country Emissions Country Emissions

United States 17.7 Canada 16.2

Mexico 4.0 France 6.3

Italy 7.0 Russia 11.2

China 5.8 United Kingdom

8.4

Australia 19.6 Qatar 79.8

Brazil 2.1 Peru 1.3

India 1.4 Japan 8.6

Based on: International Energy Agency data.

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Which of these three measurements would best

describe the central tendency of the data? e. Describe the shape of this distribution.

3.9 The following data represent the exchange rates for the euro in terms of U.S. dollars from March 2011 until February 2013. These data can also be found in the Excel file titled Euro.xlsx.

1.40 1.44 1.43 1.44 1.43 1.43

1.38 1.37 1.36 1.32 1.29 1.32

1.32 1.32 1.28 1.25 1.23 1.24

1.29 1.30 1.28 1.31 1.33 1.35

Based on: http://www.oanda.com.

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Which of these three measurements would best

describe the central tendency of the data? e. Describe the shape of this distribution.

3.10 Mark McGwire, a former Major League Baseball player who broke Roger Maris’s single-season record for home runs in 1998, recently admitted to using performance-enhancing steroids, primarily during the second half of his 16-year career. McGwire claimed he used the steroids strictly for rehabilitation purposes and that they did not give him more power to hit home runs. The following data show the number of home runs McGwire hit during his first eight years of his career followed by the number hit during his last eight years.

1986–1993

3 49 32 33 39 22 42 9

1994–2001

9 39 52 58 70 65 32 29

For both data sets,

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Describe the shape of this distribution. e. What conclusions can be drawn concerning Mark

McGwire’s claim?

Measures of variability show how much spread is present in the data.

3.2 Measures of Variability Another way to describe a data set is through measures of variability, which determine how much spread is present in the data. For example, a student who gets A’s in half of his classes and F’s in the other half has the same GPA as a student who gets straight C’s. However, the first student—the one with more variability in his grades—has to retake many courses. The second student—the one with less variability—does not have to retake any courses and can graduate on schedule. In business, less variability is an indication of more consistency, which is usually more desirable, especially in the area of quality control. Recall my example earlier in the chapter regarding the filling process for Frosted Flakes and how troublesome high variability was.

The most common measures of variability are the range, variance, and standard devia- tion. Although these measures are not as widely used by the typical person as the mean and median, they are extremely important to those in the field of statistics.

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92 C h a p t e r 3 | Calculating Descriptive Statistics

The following data set shows five online donations recently received by the American Red Cross:

$35 $10 $50 $20 $60

The range for these data is shown as follows:

Range = $60 - $10 = $50

Formula 3.5 for the range

Range = Highest value - Lowest value

The range, which is one measure of variability, can found by subtracting the lowest value from the highest value in the data set.

The range is the easiest measure of variability to calculate, but it has the most limitations, one being that it is highly affected by outliers.

Stats in practice: Using the range to Measure Volatility

{ figure 3.9 Number of Volatile Days in the Stock Market

The variance measures the variability, or spread, of the data points in a set around the set’s mean.

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C h a p t e r 3 | 3.2 Measures of Variability 93

where

s2 = The sample variance x = The sample mean n = The sample size (number of data values)

1xi - x2 = The difference between each data value and the sample mean Using our Red Cross donation data, we start the sample variance calculation by determining the sample mean:

x = 35 + 10 + 50 + 20 + 60

5 =

175

5 = 35

Table 3.8 summarizes the following steps used to calculate the variance:

Step 1. Calculate the mean of the data set.

Step 2. Subtract the mean 1x2 from each donation 1xi2 in order to find the distance, or difference, between the mean and each donation. The differences are shown in the third column of Table 3.8.

Step 3. Square the differences that we calculated in Step 2. The squared differences are shown in the fourth column of Table 3.8.

Step 4. Add up the squared differences in the fourth column in Table 3.8, which in our Red Cross example equals 1,700.

The Variance and Standard Deviation for a Sample If our data set is a sample from a larger population, the sample variance, s2, is found using Equation 3.6.

{ table 3.8 | Sample Variance Calculations for the Red Cross Donations

xi x 1xi − x2 1xi − x2 2 35 35 0 0

10 35 - 25 625 50 35 15 225

20 35 - 15 225 60 35 25 625

Total = a 5

i = 1 1xi - x22 = 1,700

We are now ready to calculate the sample variance by plugging our values into Equation 3.6:

s2 = a

i = 1 1xi - x22

n - 1 =

1,700

5 - 1 = 425

Your curious mind may be asking why we square the differences in Table 3.8. Well, I commend your curiosity. If we simply sum the differences between the mean and each data value (the third column in Table 3.8), we get the following:

a 5

i = 1 1xi - x2 = 0 + 1 - 252 + 15 + 1 - 152 + 25 = 0

Formula 3.6 for the sample Variance

s2 = a

i = 1 1xi - x22

n - 1

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94 C h a p t e r 3 | Calculating Descriptive Statistics

This result will happen with any data set—the sum of these differences will always be zero. The negative differences cancel out the positive differences. By squaring the differ- ences, we remove the negative signs and get a true measure of the variation in our data set. This total squared difference is known as the sum of squares, which you will see more of in future chapters.

Now, I have a question for you: What is the unit of measure for the variance? Because the original data set is in dollars, and the variance calculation has squared the difference between each donation and the average donation, the unit of measure is dollars2 (dollars squared). I know this unit of measure sounds strange, which leads me to the next measure of variability, the standard deviation.

The standard deviation also measures the average distance between all of a set’s data points and the mean. However, a nice feature of the standard deviation is that unlike the variance, the unit is not squared. Consequently, it matches those of the original data.

Next, let’s look at how the standard deviation is calculated for a sample and then for a population.

The sample standard deviation, s, is the square root of the sample variance. Equation 3.7 shows the formula for finding the sample standard deviation.

The standard deviation is the square root of the variance. The unit of measure for the standard deviation is the same as the unit for the original data.

When all the data values in the sample are the same, the sample standard deviation equals zero.

Formula 3.7 for the sample standard Deviation

s = 2s2 = H ani = 11xi - x22n - 1 In our Red Cross donation example, the sample standard deviation is determined as follows:

s = 2s2 = 2425 = 20.62 Therefore, the standard deviation from our donation sample is $20.62. Feel better? As you can see from the variance and standard deviation equations, we are simply

measuring how far on average each data value is from the mean of the sample. The farther each value is from the mean, the larger a distribution’s variance and standard deviation are. To help you grasp this important concept, Table 3.9 calculates the standard deviation for a sample in which all five donations are $35. Because all five donations are $35, the mean (average) donation is also $35.

s = H ani = 11xi - x22n - 1 = A 05 - 1 = $0 Because there is no variability in the sample (all numbers are the same), the standard

deviation equals $0. Note that the variance and the standard deviation should never be negative values. If you obtain a negative result in your calculations, stop and go back to find your mistake.

The standard deviation is the most common and most important measure of variability throughout this text- book. Keep this section in mind for future reference!

{ table 3.9 | Sample Variance Calculations for the Red Cross Donations

xi x 1xi − x2 1xi − x2 2

35 35 0 0

a 5

i = 1 1xi - x22 = 0

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C h a p t e r 3 | 3.2 Measures of Variability 95

where

a n

i = 1 x i

2 = The sum of each data value after it has been squared

a a n

i = 1 xib

= The square of the sum of all data values

n = The sample size

Short-Cut Formulas for the Sample Variance and Standard Deviation Now that I’ve tortured you with the nasty standard deviation equation, I’m going to show you a short-cut equation that gives you the same results with a little less work. I recommend using this method when you are calculating the variance and standard deviation manually.

Formula 3.9 for the sample standard Deviation (short-cut)

s = a ani = 1x i2 - a a n

i = 1 xib

n n - 1

Formula 3.8 for the sample Variance (short-cut)

s2 = a

i = 1 x i

2 - a a

i = 1 xib

n - 1

where

a n

i = 1 x i

2 = The sum of each data value after it has been squared

a a n

i = 1 xib

= The square of the sum of all data values

n = The sample size

I can imagine the look on your face right now. It’s the same expression that I get from my students when I show them this so-called short-cut method. I admit that these equations appear nastier than the first ones we used, but before you call me crazy, let me demonstrate why they really aren’t, using our Red Cross data.

The following steps, summarized in Table 3.10 for our Red Cross data, are used to cal- culate the variance using the short-cut formula:

Step 1. Sum up the total of the data values 1xi2, shown in the first column of Table 3.10. For our Red Cross example, the sum of the donations is a

i = 1 xi = 175.

Step 2. Square the result of Step 1. For our Red Cross example, the squared sum of the

donations is a a n

i = 1 xib

= 117522 = 30,625. Step 3. Square each value from the first column 1x i22. The squared values are shown in the

second column of Table 3.10.

Step 4. Sum up the values in the second column of Table 3.10. For our Red Cross example,

the sum of the squared donations is a 5

i = 1 x i

2 = 7,825.

I recommend using this short-cut equation when calculating the standard deviation by hand, especially when the sample size is relatively large.

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96 C h a p t e r 3 | Calculating Descriptive Statistics

Now carefully plug in the values from Table 3.10 into Equation 3.8. Note that n = 5 because we have five donations in our sample.

s2 = a

i = 1 x i

2 - a a

i = 1 xib

n - 1 =

7,825 - 30,625

5 - 1 =

7,825 - 6,125 4

= 425

This result matches the sample variance we found earlier with less work. The short-cut method for calculating the variance by hand is less work than using Equation 3.6 because you don’t have to (1) calculate the sample mean and (2) subtract the sample mean from each and every observation. These benefits become more substantial as the number of data points in the sample increases.

The Variance and Standard Deviation for a Population If our data set represents an entire population rather than a sample from a population, the variance is determined using Equation 3.10.

Formula 3.10 for the population Variance

s2 = a N

i = 1 1xi - m22

Formula 3.11 for the population standard Deviation

s = R aNi = 11xi - m22N

{ table 3.10 | Sample Variance Calculations for the Red Cross Donations (Short-Cut Method)

data value 1xi2 data value squared 1xi22 35 1,225

10 100

50 2,500

20 400

60 3,600

Sum of the data values: a 5

i = 1 xi = 175 Sum of the squared data values: a

i = 1 x i

2 = 7,825

Squared sum of the data values: a a n

i = 1 xib

= 117522 = 30,625

To calculate a n

i = 1 xi

2 (the sum of

the squared values), square each data value and then add up the squared terms. To calculate

a a n

i = 1 xi b

(the square of the sum),

add up all the data values first and then square that sum.

The variance and the standard deviation should never be negative values. If you obtain a negative result in your calculations, stop and go back to find your mistake.

where

s2 = The population variance m = The population mean N = The population size

1xi - m2 = The difference between each data value and the population mean The population standard deviation would simply be the square root of Equation 3.10.

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C h a p t e r 3 | 3.2 Measures of Variability 97

where

a N

i = 1 x i

2 = The sum of each data value after it has been squared

a a N

i = 1 xib

= The square of the sum of all data values

For example, suppose that at your school there are six statistics classes with the follow- ing student sizes:

34 25 41 32 25 29

These data would be considered a population because all current classes are included. Table 3.11 summarizes the population variance calculations.

Formula 3.12 for the population Variance (short-cut)

s2 = a N

i = 1 x i

2 - a a

i = 1 xib

{ table 3.11 | Population Variance Calculations for the Class Size

data value 1xi2 data value squared 1xi22 34 1,156

25 625

41 1,681

32 1,024

25 625

29 841

Sum of the data values: a 6

i = 1 xi = 186 Sum of the squared data values: a

i = 1 x i

2 = 5,952

Squared sum of the data values: a a 6

i = 1 xib

= 118622 = 34,596

Rather than use these equations to calculate the population variance and standard devia- tion, I present the population short-cut versions to make your life easier.

As with our previous example, we plug the results from Table 3.11 into Equation 3.12 to obtain the population variance for our class sizes.

s2 = a N

i = 1 x i

2 - a a

i = 1 xib

N =

5,952 - 34,596

6 =

5,952 - 5,766 6

= 31

Formula 3.13 for the population standard Deviation (short-cut)

s = a aNi = 1x i2 - a a N

i = 1 xib

N N

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98 C h a p t e r 3 | Calculating Descriptive Statistics

Measures of sample variability can also be determined using the Data Analysis Add-In. The following steps describe this procedure using our Red Cross donation data:

1. Place the Red Cross donation data in Cells A1 through A5.

2. Select Data + Data Analysis. 3. In the Data Analysis dialog box, select Descriptive Statistics and click OK.

4. In the Descriptive Statistics dialog box, click on the text box for Input Range: and select Cells A1:A5.

5. Click on the text box for Output Range: and select Cell C1.

6. Check Summary statistics, and click OK.

Figure 3.11 shows the Data Analysis output for our Red Cross example. The spreadsheet shows the sample standard deviation, variance, and range values in red that we calculated earlier in this section.

The population standard deviation is simply the square root of the population variance.

s = 2s2 = 231 = 5.57 students Using Excel to Calculate the Variance and Standard Deviation We can also rely on Excel to calculate the sample variance and standard deviation from our Red Cross example, as shown in Figure 3.10. The Excel functions for the sample variance and standard deviation are shown here. Notice the letter S in VAR.S and STDEV.S, which indicates “sample.”

= VAR.S1data values2 = STDEV.S1data values2

{ figure 3.10 Calculating the Sample

Variance and Standard Deviation Using Excel Formulas

{ figure 3.11 Calculating the Sample

Variance and Standard Deviation Using Excel Data Analysis

Excel also provides the following functions to calculate the population variance and standard deviation, for our class size example as shown in Figure 3.12. Notice the letter P in VAR.P and STDEV.P, which indicates “population.”

= VAR.P1data values2 = STDEV.P1data values2

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C h a p t e r 3 | 3.2 Measures of Variability 99

Why Divide by n − 1 for the Sample Variance? A common question that my students often ask is why the denominator in the population variance equation is N and the denominator for the sample variance equation is n - 1? As you will see in later chapters, we use the sample variance to provide a good estimate for the population variance. This is because we rarely know the population variance but we always have enough information to calculate the sample variance. As it turns out, using n - 1 in the denominator for the sample variance in Equation 3.6 is a much better estimate of the true population variance when compared to dividing by n. There are two ways to convince you of this phenomenon. The first way is to use a mathematical proof which will cause your eyes to glaze over and send your mind off to some faraway place. Or I can use a numerical example that might possibly result in an “ahha moment.” I choose the latter.

Suppose I play the following game with my two-year old grandson, Caleb. I have three plastic cups turned upside down on the table in which I have hidden one, three, and five M&Ms, respectively. After moving the cups around on the table like a shell game, Caleb gets to pick one cup and eat the contents of his choice. (I’m assuming his mother will not be reading this. After all, that’s what grandparents are for!) What I have described is a popula- tion with values 1, 3, and 5 1N = 32 with the following mean and variance:

m = a N

i = 1 xi

N =

1 + 3 + 5 3

= 3

s2 = a N

i = 1 1xi - m22

N =

11 - 322 + 13 - 322 + 15 - 322 3

= 8

3 = 2.67

After replacing the M&M’s that Caleb has eaten from his first choice, I move the cups around and we repeat the game one more time. At this point, I have drawn a sample size of two from this population 1n = 22 by allowing Caleb to pick a cup twice. The real challenge is to now convince Caleb that the game is over. Table 3.12 shows the following information:

Column 1: All of the possible combinations of M&M’s that Caleb will eat after twice choosing a cup.

Column 2: The sample mean for each of the combinations in Column 1 using Equation 3.2.

{ figure 3.12 Calculating the Population

Variance and Standard Deviation Using Excel Formulas

The following data represent the number of books that seven adults have read during the past 12 months. Determine the variance and standard deviation of these data.

10 10 4 8 13 6 11

A local PNC Bank branch employs eight employees whose ages are shown here. Determine the variance and standard deviation of these data.

32 40 27 29 34 38 46 24

Y o U r t U r n # 4N

Answers can be found on u page 141

Before you go any further in this chapter, try your hand at calculating the variances and standard deviations in the next Your Turn box.

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100 C h a p t e r 3 | Calculating Descriptive Statistics

Column 3: The sample variance for each of the combinations in Column 1 using Equation 3.6.

Column 4: The result of replacing n - 1 with n in the denominator of Equation 3.6.

The variance calculations for the first two samples in this table are shown for you.

{ table 3.12 | Possible Combinations of Choosing a Sample of Size Two from a Population of Size Three

(1) (2) (3) (4)

possible combinations sample mean sample variance

x = a

i = 1 xi

n s2 =

a n

i = 1 1xi - x22

n - 1

a n

i = 1 1xi - x22

11,12 1 11 - 122 + 11 - 122 2 - 1

= 0 11 - 122 + 11 - 122

2 = 0

11,32 2 11 - 222 + 13 - 222 2 - 1

= 2 11 - 222 + 13 - 222

2 = 1

11,52 3 8 4 13,12 2 2 1 13,32 3 0 0 13,52 4 2 1 15,12 3 8 4 15,32 4 2 1 15,52 5 0 0

If the sample variance is a good estimate of the true population variance, then I would expect that the average sample variance from all nine samples (Column 3) in Table 3.12 would be equal to the population variance. Let’s check it out:

Average Sample Variance = 0 + 2 + 8 + 2 + 0 + 2 + 8 + 2 + 0

9 =

9 = 2.67

Even though none of the sample variances in Column 3 of Table 3.12 equals the popula- tion variance (2.67), their average does equal 2.67. Because if this phenomenon, we call the sample variances found in Column 3 unbiased estimators of the population variance.

The same can’t be said for the average of Column 4 where we used n rather than n - 1 in the denominator:

Average of Column 4 = 0 + 1 + 4 + 1 + 0 + 1 + 4 + 1 + 0

9 =

9 = 1.33

I hope this little demonstration convinced you that using n - 1 in the denominator of the sample variance calculation provides a better estimate of the true population variance in comparison to using n. (Special thanks to Caleb Nelson for his invaluable participation in this highly important experiment.)

The sample variance is an unbiased estimator of the population variance because we would expect the average of many sample variances taken from a population would be equal to the true population variance.

Caleb might not understand statistics (yet) but he sure loves finding five M&Ms!

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C h a p t e r 3 | 3.3 Using the Mean and Standard Deviation Together 101

3.2 Section Problems Basic Skills 3.11 Consider the following sample data values:

18 12 7 10 15 12 6 12 14

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.12 Consider the following sample data values:

5 11 17 23 14 16 11 17

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.13 Consider the following population data values:

14 41 0 20 26 38 8

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.14 Consider the following population data values:

25 18 - 7 10 34 - 12 a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

applications 3.15 A local Holiday Inn asked customers to rate their stays on

a scale of 1–10. The following list shows the ratings given by the last 10 customers:

8 8 7 10 6 9 5 7 6 8

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.16 The following list represents the number of absent students from my last nine statistics classes:

2 4 0 1 6 1 3 2 3

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.17 Every 10 years, the U.S. Census Bureau asks people about the number of people living within their households. The following list shows how eight households responded to the question.

5 1 2 6 4 4 3 5

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.18 The following values represent the number of iPhones sold per day at a local AT&T retail store during the last 10 business days:

14 5 6 11 18 10 10 12 3 15

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.19 The following list shows the starting salaries, in thousands of dollars, for 11 recent college graduates with business degrees:

36 32 40 35 32 41 35 44 38 37 40

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.3 Using the Mean and Standard Deviation Together

The standard deviation is a common measure of consistency in business applications, particularly in the area of quality control. Let’s go back to our Frosted Flakes example from earlier in the chapter, where we sampled five 18-ounce boxes and weighed the contents. Sample 1 in Table 3.13 represents the sample that caused customers to complain about missing Frosted Flakes. The mean of the sample appears to be satisfactory at 18.1 ounces. Consequently, you figured the pop- ulation of boxes was OK, which cost you your job. The problem lies with too much variability in the sample, which is reflected by a standard deviation of 4.23 ounces. Sample 2 in Table 3.13 is a more acceptable sample from a quality control perspective. The sample mean is still 18.1 ounces, but the variability is much lower: The standard deviation is only 0.12 ounces. The standard devia- tion calculations are also shown in Table 3.13. This example demonstrates the fact that the sam- ple mean alone does not provide sufficient information in a quality control setting. The sample standard deviation must also be investigated to confirm the filling process is working properly.

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102 C h a p t e r 3 | Calculating Descriptive Statistics

{ table 3.13 | Frosted Flakes Samples: The Mean vs. the Standard Deviation

sample 1 (OZ) sample 2 (OZ)

Box 1 14.2 18.2

Box 2 19.6 17.9

Box 3 22.7 18.1

Box 4 13.1 18.1

Box 5 20.9 18.2

Mean 18.1 18.1

Standard deviation 4.23 0.12

In quality control settings, businesses prefer a smaller standard deviation, which is an indication of more consistency in the process.

In this example, we were able to compare two different standard deviations because the means from both samples were identical. When sample means are very different, com- paring standard deviations can be misleading. This is because the size of the standard deviation is affected by the scale of the data. Larger data values will tend to generate larger standard deviations. This scenario is covered in the following section.

the Coefficient of Variation The coefficient of variation, CV, measures a distribution’s standard deviation in terms of its percentage of the mean. A high CV is an indication of high variability, or less consistency, in the data. In other words, the data points vary widely among themselves.

Comparing the consistency between two data sets when their means are very differ- ent is best done with the coefficient of variation. It is a useful measure because two different data sets are likely to have different means. The coefficient of variation gives us a common

The coefficient of variation, CV, measures the standard deviation in terms of its percentage of the mean.

sample 1 sample 2

x 1oz2 x2 1oz2 2 x 1oz2 x2 1oz2 2 14.2 201.64 18.2 331.24

19.6 384.16 17.9 320.41

22.7 515.29 18.1 327.61

13.1 171.61 18.1 327.61

20.9 436.81 18.2 331.24

Total 90.5 1,709.51 90.5 1,638.11

sample 1 sample 2

s1 = a ani = 1x i2 - a a n

i = 1 xib

n - 1 s2 = a ani = 1x i2 - a a

i = 1 xib

n - 1

s1 = H 1,709.51 - 190.52255 - 1 s1 = H 1,638.11 - 190.52255 - 1 s1 = 4.23 oz s2 = 0.12 oz

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C h a p t e r 3 | 3.3 Using the Mean and Standard Deviation Together 103

Formula 3.14 for the Sample Coefficient of Variation

CV = s

x 11002

where

s = The sample standard deviation x = The sample mean

Formula 3.15 for the Population Coefficient of Variation

CV = s

m 11002

where

s = The population standard deviation m = The population mean As you can see, the coefficient of variation measures the standard deviation in terms of

its percentage of the mean. When comparing two samples, the smaller CV indicates more consistency (or less variability) within the sample or population.

To demonstrate this approach, let’s compare the stock price of Nike to that of Google during the same six-month period to determine which company’s stock price was more con- sistent. Table 3.14 shows the stock prices along with the mean and standard deviation for each of the two stocks.

A smaller coefficient of variation indicates more consistency within a set of data values.

{ table 3.14 | Stock Prices for Nike and Google

date nike ($) gOOgle ($)

September 14, 2012 48.32 709.68

October 15, 2012 47.81 740.98

November 15, 2012 45.42 647.26

December 14, 2012 48.46 701.96

January 15, 2013 53.64 724.93

February 15, 2013 54.95 792.89

Mean 49.77 719.62

Standard deviation 3.70 47.96

Based on: http://www.bigcharts.com.

measure—a percentage—we can use to compare the two sets. That way, instead of compar- ing apples to oranges, we can compare apples to apples (or oranges to oranges, depending on the fruit you prefer).

Equation 3.14 shows the formula for the coefficient of variation for a sample. Equation 3.15 shows the formula for the coefficient of variation for a population.

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104 C h a p t e r 3 | Calculating Descriptive Statistics

The coefficient of variation for each sample is computed as follows:

Nike: CV = s

x 11002 = $3.70

$49.77 11002 = 7.4%

Google: CV = s

x 11002 = $47.96

$719.62 11002 = 6.7%

Even though Google’s stock price has a higher standard deviation than Nike’s does, it is more consistent because its coefficient of variation is lower. Take this opportunity to check out the following Your Turn box. Here you will have the chance to compare the home-run consistency of two Major League Baseball players. Sounds like fun to me!

The following table shows the number of home runs hit by Jimmy Rollins and Albert Pujols over eight consecutive Major League Baseball seasons. These data can also be found in the Excel file titled homerun cv.xlsx. Who was the more consistent home-run hitter?

Year Rollins Pujols

2005 12 41

2006 25 49

2007 30 32

2008 11 37

2009 21 47

2010 8 42

2011 16 37

2012 23 30

Based on: http://www.espn.com.

Answer can be found on u page 142

Y o u r t u r n # 5N

Stats in practice: Variability Is not always a Bad thing

Earlier in this section, we discussed the problems that a high variability causes in a quality control setting. However, there are situations in the business world in which variability is viewed positively. For instance, Herr Foods, a producer of a variety of salty snacks (including my favorite potato chips), performs taste tests to compare how consumers react to its products. A prod- uct such as horseradish cheddar chips is considered a polariz- ing product because it has a very strong taste. Consumers tend to rate the product very high or very low, which results in highly variable taste-test ratings for the product. A more standard product, such as regular tortilla chips, typically receives more consistent ratings. Suppose the following table shows how seven customers rated each product on a scale of 1–10. The mean and standard deviation for each sample are also shown in the table.

Horseradish Cheddar Chips

Tortilla Chips

4 8

10 8

10 7

3 8

10 8

9 10

10 7

Mean 8.0 8.0

Standard deviation 3.1 1.0

(continued)

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C h a p t e r 3 | 3.3 Using the Mean and Standard Deviation Together 105

Formula 3.16 for the Population z-score

z = x - m s

the z-Score One of the most common measurements in statistics is the z-score, which identifies the num- ber of standard deviations a particular value is from the mean of its population or sample. Equation 3.16 shows how the z-score for a population is computed.

The z-score identifies the number of standard deviations a particular value is from the mean of its distribution.

I assure you that this equation will be used over and over again throughout this textbook, so be sure to become familiar with it.

Even though both products have the same average (mean) rat- ing, the horseradish cheddar results are viewed more positively from a marketing perspective because of the higher standard deviation. Herr Foods does not consider high variability in the testing scores a problem if the product has a strong flavor.1

where

x = The data value of interest m = The population mean s = The population standard deviation If our data are from a sample, we would use Equation 3.17 to compute the z-score.

where

x = The data value of interest x = The sample mean s = The sample standard deviation To illustrate z-scores, we’ll use Table 3.15, which shows the number of calories contained

in a sample of fast-food hamburgers. Also shown are the mean and standard deviation for the sample, which you can verify using Excel. (Try not to drool lest you get your textbook wet.)

Formula 3.17 for the Sample z-score

z = x - x

{ table 3.15 | Calories for Various Fast-Food Hamburgers

hamburger type restaurant calOries

Cheeseburger McDonald’s 300

Single with everything Wendy’s 430

Big Mac McDonald’s 540

Whopper Burger King 670

Bacon cheeseburger Sonic 780

Baconator Wendy’s 840

Triple whopper with cheese Burger King 1,230

2>3 lb. Monster thickburger Hardee’s 1,420 Sample Mean 776.3

Sample Standard Deviation 385.1

1 Special thanks to Don Kirkley from Herr Foods for providing this material.

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106 C h a p t e r 3 | Calculating Descriptive Statistics

Let’s calculate the z-score for the Triple Whopper with Cheese using Equation 3.17:

z = x - x

s =

1,230 - 776.3 385.1

= 453.7

385.1 = 1.18

What does the number 1.18 tell us? Well, our Triple Whopper w/Cheese is 453.7 calories above the sample mean. If one standard deviation is 385.1 calories, then our burger is slightly more than one standard deviation (1.18) above the sample mean.

The z-score for the Big Mac demonstrates this concept for values below the mean:

z = x - x

s =

540 - 776.3 385.1

= - 236.3 385.1

= - 0.61

The Big Mac is 236.3 calories below our sample mean, which translates to - 0.61 standard deviations. Our z-score is negative, indicating our data value is lower than the sample mean.

Had one of our hamburgers in our sample been exactly 776.3 calories, the z-score for this selection would be as follows:

z = x - x

s =

776.3 - 776.3 385.1

= 0

385.1 = 0.0

In general, z-scores will always have the following attributes:

• Positive for values above the mean

• Negative for values below the mean

• Zero for values equal to the mean

The z-score is analogous to expressing the original data in other units, such as convert- ing 2 feet to 24 in. In our example, the 540 calories for the Big Mac is being converted to a z-score of - 0.61. However, the z-score itself has no units, even though the original values will normally be expressed in units such as dollars, years, pounds, and calories.

The following Stats in Practice box highlights another use of the z-score.

Think of the z-score calculation as simply converting the value of x to a different set of units, like converting feet to inches.

Stats in practice: Comparing olympic performances

How does one compare great performances of athletes com- peting in different sports? Take, for example, the 2010 Winter Olympic Games in Vancouver. How does the performance of Switzerland’s Simon Ammann, who won the Large Hill Ski Jumping event by a margin of 14.2 points, compare to the United State’s Shani Davis, who won the 1,000-m Speedskat- ing event by 0.18 seconds? Which performance was more dominant?

This question was answered in the Wall Street Journal article titled “The Best Olympic Performances So Far” by Matthew Futterman. In order to compare performances across sporting events, we need to convert the winning scores to a common measurement scale. The z-scores that we have been discussing in this chapter provide the perfect solution.

Mr. Futterman calculated the z-scores for the winning time or score for Olympic events. These values represent the number of standard deviations the winning performance was from the average performance in that particular event. Congratulations to France’s Jay Vincent, who is credited with the most dominant performance of the 2010 Winter Olympics. His 12.2-second margin of victory in the 10K Sprint Biathlon event is 2.13 standard deviations below the average time in this event.

Now, back to our first question of whose performance was more dominant: Simon Ammann’s Large Hill Ski Jumping Event or Shani Davis’ 1,000-m Speedskating event. Ammann’s 14.2-point margin was 2.01 standard deviations above the average score whereas Shani Davis’ 0.18-second margin was less—1.81 standard deviations below the average time.

Based on: Matthew Futterman, “The Best Olympic Performances So Far,” Wall Street Journal, February 24, 2010, http://online.wsj.com.

Finally, we can use z-scores to identify outliers in the data set. Data values that have z-scores above +3 or below −3 are categorized as outliers. Table 3.16 shows the z-scores for the calories in each of our fast-food hamburgers.

Outliers are extreme values in the data set that require special consideration.

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C h a p t e r 3 | 3.3 Using the Mean and Standard Deviation Together 107

None of the hamburgers listed in Table 3.16 would be classified as an outlier because all of the z-scores are within {3 standard deviations from the sample mean. Even the gut- busting, stroke-inducing Monster Thickburger didn’t qualify as an outlier because its z-score was only 1.67 standard deviations above the mean. (Please don’t use this as an endorsement to consume one!)

The STANDARDIZE function in Excel will calculate z-scores for us using Equation 3.17 and is executed as follows:

= STANDARDIZE1x, mean, standard deviation2 Using our Big Mac example, we have the following:

= STANDARDIZE1540, 776.3, 385.12 = - 0.61 In Chapter 8 we will use z-scores to explore how statistics can be used to make infer-

ences about a population based on a sample taken from that population. As I mentioned earlier, z-scores will be used extensively in this textbook. So file this information somewhere safe in your brain for easy access later.

the empirical rule The values of many large data sets sometimes cluster around the mean or median so that the data distribution in the histogram resembles a bell-shape, symmetrical curve. When this is the case, according to the empirical rule (sounds like a decree from the emperor) approxi- mately 68%, 95%, and 99.7% of the values will fall within { one, two, and three standard deviations of the mean, respectively. In other words, almost all of the values—99.7%—will fall within { three standard deviations of the mean. (The empirical rule is also sometimes called the 68-95-99.7 rule.)

For example, suppose that the average exam score for my large statistics class is 80 points, the standard deviation is 5.0 points, and the distribution of grades is bell-shaped around the mean, as shown in the shaded region in Figure 3.13. Because one standard deviation above the mean would be 85 180 + 52 and one standard deviation below the mean would be 75 180 - 52, the empirical rule tells us that approximately 68% of the exam scores will fall between 75 and 85 points.

We can also calculate these exam scores by rearranging Equation 3.16 in terms of x (the population) rather than z (the z-score), as shown in Equation 3.18 on the next page.

According to the empirical rule, if a distribution follows a bell-shaped, symmetrical curve centered around the mean, we would expect approximately 68%, 95%, and 99.7% of the values to fall within one, two, and three standard deviations above and below the mean, respectively.

{ table 3.16 | z-Scores for Fast-Food Hamburgers

hamburger type calOries

xi 1xi - x2 z Cheeseburger 300 - 476.3 - 1.24

Single with everything 430 - 346.3 - 0.90

Big Mac 540 - 236.3 - 0.61

Whopper 670 - 106.3 - 0.28

Bacon cheeseburger 780 3.7 0.01

Baconator 840 63.7 0.17

Triple whopper with cheese 1,230 453.7 1.18

2>3 lb. Monster thickburger 1,420 643.7 1.67

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108 C h a p t e r 3 | Calculating Descriptive Statistics

We can find the exam scores that are one standard deviation above and below the mean by setting z = 1 and z = - 1.

x = m + zs x = 80 + 112152 = 85 x = 80 + 1 - 12152 = 75

The empirical rule applies only to bell-shaped curves that are relatively symmetrical.

{ figure 3.13 Percentage of Exam Scores That

Are One Standard Deviation Around the Mean

75 85

68%

80 907065 95 Exam Scores

The empirical rule also states that approximately 95% of the data values will fall within two standard deviations above and below the mean. In our example, the exam scores that are two standard deviations from the mean are as follows:

x = 80 + 122152 = 90 x = 80 + 1 - 22152 = 70

According to Figure 3.14, approximately 95% of the exam scores will be between 70 and 90 points.

If the data are bell-shaped and somewhat symmetrical, we expect practically all the data values to lie with three standard deviations from the center of the data.

{ figure 3.14 Percentage of Exam Scores That

Are Two Standard Deviations Around the Mean

75 85

95%

80 907065 95 Exam Scores

Taking this one final step, the empirical rule states that, under these conditions, approxi- mately 99.7% of the data values will fall within three standard deviations above and below the mean. In our example, the exam scores that are three standard deviations from the mean are as follows:

x = 80 + 132152 = 95 x = 80 + 1 - 32152 = 65

According to Figure 3.15, I would expect nearly all the exam scores to be between 65 and 95.

Formula 3.18 for Expressing the z-Score in Terms of x

x = m + zs

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C h a p t e r 3 | 3.3 Using the Mean and Standard Deviation Together 109

This is further evidence that values outside three standard deviations around the mean are rare and are considered outliers.

Chebyshev’s theorem Chebyshev’s Theorem is a mathematical rule similar to the empirical rule except that it applies to any distribution rather than just bell-shaped, symmetrical distributions. The theorem states that for any z score greater than 1, the percent of the values that fall within z standard devia- tions above and below the mean will be at least the result obtained using Equation 3.19.

Chebyshev’s Theorem can be applied to any distribution regardless of its shape.

{ figure 3.15 Percentage of Exam Scores That

Are Three Standard Deviations Around the Mean

75 85

99.7%

80 907065 95 Exam Scores

Formula 3.19 for Chebyshev’s Theorem

a1 - 1 z 2 b * 100

So, according to the theorem, using Equation 3.19 we can state the following:

• When z = 2, at least a1 - 1 22

b * 100% = 75% of the data values will fall within two standard deviations above and below the mean.

• When z = 3, at least a1 - 1 32

b * 100% = 89% of the data values will fall within three standard deviations above and below the mean.

• When z = 4, at least a1 - 1 42

b * 100% = 94% of the data values will fall within four standard deviations above and below the mean.

In other words, regardless of whether a distribution is bell-shaped, at least 94% of the data values will fall within four standard deviations of the mean; at least 89% of the data values will fall within three standard deviations of the mean; and at least 75% of the data values will fall within two standard deviations of the mean. Chebyshev’s Theorem can be applied to any distribution of data but can only be stated for z scores that are greater than 1.

Let’s check out Chebyshev’s Theorem to see whether it really works. Airlines define turnaround time as the number of minutes required to prepare a plane for its next flight. Reducing this time is a key factor in the airline’s efforts to improve profitability in a very competitive market. Table 3.17 shows the turnaround times, in minutes, for a sample of 30 domestic flights.

Pafnuty Chebyshev (also known as Tchebysheff) was a Russian mathematician who lived from 1821–1894.

Chebyshev’s Theorem states that for any number z greater than 1, the percent of the values that fall within z standard deviations from the mean will be at least

a1 - 1 z 2 b * 100.

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110 C h a p t e r 3 | Calculating Descriptive Statistics

Using Excel, we can determine the following:

x = 34.3 1sample mean2 s = 11.4 1sample standard deviation2

To verify that at least 75% of the observations fall within two standard deviations from the mean, we’ll rearrange Equation 3.17 in terms of x rather than z, as seen in Equation 3.20. We are using x and s here because our data represent a sample rather than a population.

{ table 3.17 | Turnaround Time (minutes)

73 64 57 49 49 45 41 39 38 38

37 37 37 36 36 34 34 34 34 34

33 31 31 30 30 29 27 27 27 25

25 25 25 25 25 23 23 22 22 21

Formula 3.20 for Expressing the z-Score (z) in Terms of x (Sample)

x = x + zs

We can find the turnaround times that are two standard deviations above and below the mean by setting z = 2 and z = - 2.

x = x + zs x = 34.3 + 122111.42 = 57.1 x = 34.3 + 1 - 22111.42 = 11.5

All but the top two turnaround times fall within this interval. Therefore, the percent of observations that falls within two standard deviations of the mean is as follows:

40 11002 = 95%

Because 95% is greater than 75%, Chebyshev’s Theorem holds true. Table 3.18 verifies that Chebyshev’s Theorem also holds true for three and four standard

deviations from the mean of our sample.

It’s important to remember that Chebyshev’s Theorem states that at least 75% of the data will be two standard deviations from the mean. Therefore, 95% satisfies the theorem.

{ table 3.18 | Verifying Chebyshev’s Theorem

z x s x + zs x - zs chebyshev’s percent actual percent 2 34.3 11.4 57.1 11.5 75 95

3 34.3 11.4 68.5 0.1 89 97.5

4 34.3 11.4 79.9 -11.3 94 100

As you can see in each case, the actual percent of values for 2, 3, and 4 standard deviations above and below the mean exceeds Chebyshev’s percent, which confirms the theorem works. One downside of using Chebyshev’s Theorem is that when it is applied to a bell-shaped, symmetrical distribution, it will tend to understate the percentage of observations that will fall within an interval. Now, to see if you have been paying attention, let’s see if you can handle the following Your Turn section before moving on in the chapter.

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C h a p t e r 3 | 3.3 Using the Mean and Standard Deviation Together 111

1. Assume that the average tuition for a four-year, in-state public college is $7,000 per year with a standard deviation of $850 per year. Calculate the z-score for the following tuitions:

a. $6,000 b. $8,400

2. Assume that the average weekly rental charge for a compact car from Avis is $322 with a standard deviation of $25 per week and that the distribution for these data is bell-shaped. Find the interval of weekly charges that includes the following:

a. 68% of the car rentals around the mean b. 95% of the car rentals around the mean c. 99.7% of the car rentals around the mean

3. Assume that the average annual premium for health care insurance for an employee is $1,920 per year with a standard deviation of $160 per year and that the distribution of these data is not bell- shaped. Find the interval of annual premiums that will include at least the following:

a. 75% of the employees around the mean b. 85% of the employees around the mean (Hint: You’ll need to calculate z using Chebyshev’s

Theorem.) c. 94% of the employees around the mean

Y o u r t u r n # 6N

Answers can be found on u pages 142–143

3.3 Section Problems Basic Skills 3.20 Consider the following two sample data sets.

Set 1:

4 7 6 9 5

Set 2:

2 11 14 4 3

a. Calculate the coefficient of variation for each data set. b. Which data set has more variability?

3.21 Consider the following two sample data sets.

Set 1:

11 21 15 24 20

Set 2:

3 8 0 5 6

a. Calculate the coefficient of variation for each data set. b. Which data set has more variability?

3.22 Consider a sample with a mean equal to 40 and a standard deviation equal to 12. Calculate the z-scores for the fol- lowing values:

a. 50 b. 65 c. 35 d. 10

3.23 Consider the following sample data:

9 7 15 10 4 12

Calculate the z-scores for the following values:

a. 13 b. 16 c. 8 d. 4

applications 3.24 Assume the average price of a laptop computer is $800

with a standard deviation of $75. The following data represent the prices of a sample of laptops at Best Buy. Calculate the z-score for each of the following prices:

a. $699 b. $949 c. $625 d. $849 e. $999

3.25 The following data represent the number of minutes cus- tomers at Wendy’s had to wait in the drive-through line before their order was taken.

1 4 6 0 4 3 10 0 4

Calculate the z-scores for the following times:

a. 1 minute b. 3 minutes c. 7 minutes d. 12 minutes

3.26 Assume the average selling price for houses in a certain county is $325,000 with a standard deviation of $40,000.

a. Determine the coefficient of variation. b. Calculate the z-score for a house that sells for

$310,000. c. Using the empirical rule, determine the range of prices

that includes 95% of the homes around the mean. d. Using Chebyshev’s Theorem, determine the range of

prices that includes at least 94% of the homes around the mean.

3.27 The following data represent the number of touchdown passes per season thrown by the Benedict Arnold of the National Football League, Brett Favre (can you tell I’m

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112 C h a p t e r 3 | Calculating Descriptive Statistics

3.4 Working with Grouped Data Back in Chapter 2, you learned about frequency distributions that were organized as grouped data instead of individual data points. We can calculate the mean, the variance, and the stan- dard deviation for this type of data using the following example.

A survey that collected demographic data from respondents asked for their ages as grouped data. This is often done to increase the likelihood of getting people to respond to surveys. People tend to feel more comfortable selecting an age category rather than owning up to their actual ages. The number of respondents in each age group is shown in Table 3.19.

{ table 3.19 | Number of Respondents and Their Age Groups

age group number of respondents 1fi2 20 to under 30 20

30 to under 40 41

40 to under 50 19

50 to under 60 9

60 to under 70 11

Total 100

The following sections will demonstrate the procedures for computing the mean, vari- ance, and standard deviation for the age of the respondents.

the Mean of Grouped Data To calculate the average age in our frequency distribution, we first need to find the midpoint for each range of ages. The midpoint is the halfway point in each age group. You can find it by taking the average of the endpoints for each class, or group. For example, the midpoint for the first group is found as follows:

20 + 30 2

= 25

Table 3.20 shows the midpoints for each class in our example.

The midpoint is the halfway point in each group and can be found by taking the average of the endpoints for each class.

a diehard Green Bay Packer fan?), during his first 18 seasons.

0 18 19 33 38 39 35 31 22

20 32 27 32 30 20 18 28 22

Verify that Chebyshev’s Theorem holds true by determin- ing the percent of observations that fall within { two and three standard deviations from the mean.

3.28 Assume the average age of an MBA student is 31.6 years old with a standard deviation of 2.8 years.

a. Determine the coefficient of variation. b. Calculate the z-score for an MBA student who is

28 years old.

c. Using the empirical rule, determine the range of ages that will include 99.7% of the students around the mean.

d. Using Chebyshev’s Theorem, determine the range of ages that will include at least 94% of the students around the mean.

e. Using Chebyshev’s Theorem, determine the range of ages that will include at least 80% of the students around the mean.

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C h a p t e r 3 | 3.4 Working with Grouped Data 113

Equation 3.21 can be used to calculate the average age of this sample of grouped data.

{ table 3.20 | The Midpoints for the Respondents’ Age Groups

age group midpoint 1mi2 20 to under 30 25

30 to under 40 35

40 to under 50 45

50 to under 60 55

60 to under 70 65

Formula 3.21 for the Sample Mean: Grouped Data

x ≈ a

i = 1 1fimi2 n

where

fi = The frequency for class i mi = The midpoint for class i

n = a k

i = 1 fi (the total number of observations)

k = number of classes Did you notice the “squiggly” equal sign in Equation 3.21? This symbol indicates that this equation is only an approximation of the average age of the respondents. By grouping the ages, we are losing information about individual respondents. Because the midpoint approximates only the ages within each class, our final result is merely an approximation of the average age of all the respondents.

The approximate sample mean for our respondents’ ages is calculated as follows:

x ≈ a

i = 1 1fimi2 n

= 12021252 + 14121352 + 11921452 + 1921552 + 11121652

20 + 41 + 19 + 9 + 11

x ≈ 500 + 1,435 + 855 + 495 + 715

100 =

4,000

100 = 40 years

The average age of the respondents is about 40 years. If our grouped data are a population, the population mean is found using Equation 3.22.

Because the midpoint estimates only the ages in each class, the weighted average is only an approximate value.

Formula 3.22 for the Population Mean: Grouped Data

m ≈ a

i = 1 1fimi2 N

Equation 3.22 is identical to Equation 3.21, except m replaces x and N replaces n.

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114 C h a p t e r 3 | Calculating Descriptive Statistics

the Variance and Standard Deviation of Grouped Data We can also determine the variability of grouped data. The sample variance for grouped data is computed using Equation 3.23.

Formula 3.23 for the Sample Variance: Grouped Data

s2 ≈ a

i = 1 1mi - x22fi n - 1

where

x = The approximate sample mean fi = The frequency for class i mi = The midpoint for class i

n = a n

i = 1 fi (the total number of observations)

k = number of classes

Table 3.21 summarizes the sample variance calculations for our respondents’ age example. Notice that we are using x = 40 from the previous section.

{ table 3.21 | Variance Calculations for the Respondents’ Ages

age mi fi 1x2 1mi - x2 1mi - x2 2 1mi − x 22fi 20–30 25 20 40 - 15 225 4,500

30–40 35 41 40 - 5 25 1,025

40–50 45 19 40 5 25 475

50–60 55 9 40 15 225 2,025

60–70 65 11 40 25 625 6,875

n = a fi = 100 a 1mi - x22fi = 14,900

s2 ≈ a

i = 1 1mi - x22fi n - 1

= 14,900

100 - 1 = 150.5

The sample standard deviation for the respondents’ ages is

s = 2s2 ≃ 2150.5 = 12.3 years If our data represent a population, the variance equation is calculated using Equation 3.24.

The standard deviation of grouped data is also just an estimate, for the same reasons the mean is.

Formula 3.24 for the Population Variance: Grouped Data

s2 ≈ a

i = 1 1mi - m22fi

Equation 3.24 is identical to Equation 3.23, except that m replaces x and N replaces n - 1. Now work the following Your Turn problem to see if you understand the procedure for

calculating the mean and standard deviation of grouped data.

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C h a p t e r 3 | 3.4 Working with Grouped Data 115

3.4 Section Problems Basic Skills 3.29 Consider the following grouped sample data:

Values Frequency

0 to under 2 4

2 to under 4 2

4 to under 6 9

6 to under 8 7

a. Calculate the approximate mean of these sample data. b. Calculate the approximate standard deviation of these

sample data.

3.30 Consider the following grouped sample data:

Values Frequency

5 to under 10 15

10 to under 15 9

15 to under 20 21

20 to under 25 24

25 to under 30 6

a. Calculate the approximate mean of these sample data. b. Calculate the approximate standard deviation of these

sample data.

3.31 Consider the following grouped sample data:

Values Frequency

10 to under 30 26

30 to under 50 35

50 to under 70 51

70 to under 90 33

a. Calculate the approximate mean of these sample data. b. Calculate the approximate standard deviation of these

sample data.

applications 3.32 A survey that collected demographic data from respon-

dents asked them to categorize their annual incomes (in thousands of dollars) into the groups in the following table. The number of respondents for each income group is also shown in the table.

Income Number of Respondents

$20 to under $30 67

$30 to under $40 111

$40 to under $50 125

$50 to under $60 21

$60 to under $70 38

a. Calculate the approximate average income of the respondents.

b. Calculate the approximate variance and standard devia- tion of the incomes of the respondents.

3.33 A walk-in clinic for emergency room services maintains records of the number of patients it treats per day. The following table shows the frequency of the patient arrivals over the course of a 150-day period:

Number of Patients per Day Frequency

20 to under 40 10

40 to under 60 16

60 to under 80 25

80 to under 100 65

100 to under 120 34

a. Calculate the approximate average number of patients per day.

b. Calculate the approximate variance and standard devia- tion of the number of patients per day.

3.34 The Olive Garden Italian Restaurant records the amount of time customers wait for a table on Saturday evenings.

The following table summarizes the GPAs of a sample of MBA students graduating from Goldey-Beacom College this year. Calculate the mean GPA of the class, along with the variance and standard deviation.

GPA Frequency

3.0 to under 3.2 6

3.2 to under 3.4 8

3.4 to under 3.6 6

3.6 to under 3.8 14

3.8 to 4.0 6

Y o u r t u r n # 7N

Answer can be found on u page 143

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116 C h a p t e r 3 | Calculating Descriptive Statistics

3.5 Measures of Relative Position In addition to central tendency, data can also be described in terms of relative position, which compares the position of one value in relation to other values in the data set. Percentiles and quartiles are the most common measures of relative position.

percentiles Percentiles measure the approximate percentage of values in the data set that are below the value of interest. Standardized test scores are often reported in terms of percentiles. For ex- ample, when my daughter Christin informed me that she scored a 36 on the MCAT, which is the admission test for medical schools, I wasn’t sure how to respond. However, when she explained to me that this score was in the 97th percentile, I then realized it was time to celebrate because she did better than at least 97% of the students who took the test. (It looks like there’s going to be another Dr. Donnelly in the family!)

The formal definition of the p th percentile of a data set (where p is any number between 1 and 100) is the value that at least p percent of the observations will fall below.

Some of my students confuse percentiles with percentages, so be careful not to make this mistake. In the MCAT example, the maximum score is a 45. Christin’s percentage score is therefore 36>45 = 80%, which is not the same as the 97th percentile.

There’s more than one way to determine percentiles manually, and the results from the various approaches differ slightly. I’ll present the same method used to determine the median, which uses the index point.

To demonstrate the manual technique for determining percentiles, we’ll use the following data. The numbers are the total seconds 15 customers of PNC Bank took to complete their ATM transactions. Let’s determine the value that represents the 70th percentile.

124 137 88 156 66 142 96 122

69 107 104 87 111 93 99

We first need to sort our data from lowest to highest, as shown in Table 3.22, and indicate their position.

Measures of relative position compare the position of one value in relation to other values in the data set.

Percentiles measure the approximate percentage of values in the data set that are below the value of interest.

The pth percentile of a data set (where p is any number between 1 and 100) is the value that at least p percent of the observations will fall below.

Percentiles and percentages are not the same thing. Be sure you understand the difference.

The median, discussed earlier in this chapter, is the 50th percentile.

The following table lists the frequency of wait times for the last 100 customers.

Number of Minutes per Customer Frequency

0 to under 10 6

10 to under 20 22

20 to under 30 10

30 to under 40 18

40 to under 50 28

50 to under 60 16

a. Calculate the approximate average wait time per customer.

b. Calculate the approximate variance and standard devia- tion of the wait time per customer.

3.35 The following data represent the number of home runs Hank Aaron hit each year during his 23-year baseball career.

13 27 26 44 30 39 40 34 45 44 24 32

44 39 29 44 38 47 34 40 20 12 10

a. Calculate the mean and standard deviation of these data.

b. Group these data into four classes with class boundar- ies of 10 to under 20, 20 to under 30, 30 to under 40, and 40 to under 50.

c. Calculate the approximate mean and standard deviation of these grouped data.

d. Explain the difference in the results of part c from those of part a.

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C h a p t e r 3 | 3.5 Measures of Relative Position 117

where

P = The percentile of interest n = The number of data values Because we have 15 ATM observations and we are looking for the 70th percentile,

i = P

100 1n2 = 70

100 1152 = 10.5

However, 10.5 is not a whole number. So what do we do? We follow the same rule that we used to determine the median:

• If i is not a whole number, round i to the next whole number. The ith position represents our value of interest.

• If i is a whole number, the midpoint between the ith and i + 1 position is our value of interest.

Because i = 10.5, the 70th percentile is the data value in the 11th position, which is 122 seconds. In other words, 122 seconds constitutes the 70th percentile. Thus, 70% of the data values in the set are below 122 seconds.

Another red flag here—some students will mistakenly claim that the 70th percentile is 11. Remember, the index point, i, is not the value of interest. It is merely the position of the value of interest in our sorted data.

Excel will calculate percentiles for us with the PERCENTILE.EXC function:

= PERCENTILE.EXC1array, k2 where

array = The data range of interest k = The percentile of interest between 0 and 1 inclusive

Figure 3.16 illustrates how this function is used to find the 70th percentile for our ATM example, with the result shown in Cell D1.

To demonstrate the use of our second index rule, we’ll find the 40th percentile from our ATM example using the following equation:

i = P

100 1n2 = 40

100 1152 = 6

Pay close attention to these two rules for finding the position of the percentile of interest.

Be sure not to confuse the value of i for the percentile. The value of i represents the position of the percentile, not the value of the percentile.

{ table 3.22 | Sorted ATM Times (in Seconds)

time position time position

66 1 107 9

69 2 111 10

87 3 122 11

88 4 124 12

93 5 137 13

96 6 142 14

99 7 156 15

104 8

Next, we need to identify the index point, , by using the Equation 3.25.

Formula 3.25 for the Index Point for the pth Percentile

i = P

100 1n2

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118 C h a p t e r 3 | Calculating Descriptive Statistics

{ figure 3.16 Excel’s PERCENTILE.EXC

Function

Because i = 6, the 40th percentile is the data value midway between the sixth (96) and seventh (99) positions in Table 3.22, as the following calculation shows:

40th percentile = 96 + 99

2 = 97.5 seconds

Notice that the percentile does not need to be an actual value in the data set. Now, let’s throw this procedure in reverse. In the previous example, we determined

the value in the data set that corresponds to a certain percentile. Now, I’ll show you how to determine the percentile rank, which identifies the percentile of a particular value within a set of data (sounds pretty exciting to me).

Table 3.23 shows the credit scores for 16 individuals. These data can also be found in the Excel file titled percentile rank.xlsx.

Excel’s value for the 70th per centile (122.4) is slightly different from the value for the 70th percentile we obtained by hand (122) because Excel uses a different method to calculate percentiles.

The percentile rank identifies the percentile of a particular value within a set of data.

{ table 3.23 | Credit Scores

name score name score

Lori 726 Marc 690

Chris 643 Kendra 747

John 777 Peter 618

Sam 695 Jordan 701

Katie 796 Teresa 660

Carole 743 David 716

Irina 642 Gia 742

Lisa 819 Ryan 682

Let’s determine the percentile rank for John, who has a credit score of 777. First, we need to sort the data from lowest to highest, as follows. The point total for John is highlighted in red.

618 642 643 660 682 690 695 701

716 726 742 743 747 777 796 819

Next, we count the number of people who have a lower credit score than John which is 13. To find the approximate percentile rank for John, we use the Equation 3.26.

Formula 3.26 for the Percentile Rank

Percentile Rank = 1Number of values below x2 + 0.5

Total number of values 11002

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C h a p t e r 3 | 3.5 Measures of Relative Position 119

{ figure 3.17 Excel’s PERCENTRANK.EXC

Function

We can use percentile ranks to compare two different values from two different data sets. Table 3.24 lists the exam scores from a math class and a physics class. A particular student scored an 88 on her math test and a 76 on her physics test. Which of her two exam scores is higher relative to those of her classmates? To answer this question, we need to find the percentile ranks.

Excel’s value for the 86th percentile is different from the value for the 84th percentile we obtained by hand because Excel uses a different method to calculate percentile ranks.

Percentile ranks are very useful when comparing values from two different data sets.

{ table 3.24 | Exam Scores

math exam scores

59 62 66 69 72 72 75 75 78 81 83 84 85

86 88 89 90 91 92 93 94 94 95 97 98

physics exam scores

47 48 53 56 57 58 60 61 61 62 63 64 71

72 74 75 76 82 89 95

There are a total of 25 math scores, of which 14 are below an 88. Therefore, the approximate math percentile rank is as follows:

Percentile Rank = 1142 + 0.5

25 11002 = 14.5

25 11002 = 58th

There are a total of 20 physics scores, of which 16 are below a 76. Therefore, the approximate physics percentile rank is as follows:

Percentile Rank = 1162 + 0.5

20 11002 = 16.5

20 11002 = 83th

Even though the student’s math score (88) was higher than her physics score (76), the student performed better on the physics test relative to her classmates because of the higher percentile rank.

In our example, the x in the equation stands for John’s credit score of 777. Because we know there are 13 people with credit scores less than 777, we can calculate John’s percentile rank as follows:

Percentile Rank = 1132 + 0.5

16 11002 = 13.5

16 11002 = 84th

John’s credit score of 777 puts him in the 84th percentile, which places him above at least 84% of this group of 16 people.

Excel will also calculate the percentile rank with the PERCENTRANK.EXC function:

= PERCENTRANK.EXC1array, x, significance2 where

array = The data range of interest x = The value for which you want the percentile rank significance = the number of decimal places displayed for the percentile rank

Figure 3.17 shows how this function is used to find the percentile rank of John. The result is shown in Cell D1.

The percentile rank must be between 0 and 100.

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120 C h a p t e r 3 | Calculating Descriptive Statistics

Quartiles Quartiles are simply specific percentiles of interest. The first quartile is the value that constitutes the 25th percentile, the second quartile is the value that constitutes the 50th percentile, and the third quartile is the value that constitutes the 75th percentile. The three quartiles are designated Q1, Q2, and Q3, respectively. The second quartile, Q2, is also the median, which was discussed earlier in the chapter.

Table 3.25 lists the number of mishandled baggage reports per 1,000 passengers from 1995 to 2012 during the month of July for the largest U.S. air carriers. These data can also be found in the Excel file titled mishandled bags.xlsx. Notice the upward trend from 2005 to 2007! This was a good time to drive to your vacation spot. Let’s find the three quartiles in these data.

The first, second, and third quartiles in a data set are the values for the 25th, 50th (median), and 75th percentiles, respectively.

{ table 3.25 | Mishandled Baggage Reports per 1,000 Passengers

year reports year reports

1995 5.18 2004 4.99

1996 5.30 2005 7.71

1997 4.82 2006 6.51

1998 5.09 2007 7.93

1999 5.75 2008 4.87

2000 5.64 2009 3.98

2001 4.48 2010 3.69

2002 3.99 2011 3.72

2003 4.37 2012 3.52

Based on: U.S. Department of Transportation data.

We’ll start by sorting our 18 values from lowest to highest:

3.52 3.69 3.72 3.98 3.99 4.37

4.48 4.82 4.87 4.99 5.09 5.18

5.30 5.64 5.75 6.51 7.71 7.93

We now can identify the index point for the first quartile. Using Equation 3.25 to determine the index point, we set P = 25 because the first quartile is the 25th percentile.

Index point for Q1: i = P

100 1n2 = 25

100 1182 = 4.5

Because 4.5 is a fraction, we round up like we did when we were trying to find the median in a data set. Doing so puts Q1 in the fifth position of our sorted data, which corresponds to 3.99.

Because the second quartile is the median, set P = 50 to find the index point for Q2.

Index point for Q2: i = P

100 1n2 = 50

100 1182 = 9

Q2 is the midpoint between the 9th (4.87) and 10th (4.99) position; therefore,

Q2 = 4.87 + 4.99

2 = 4.93

The index point for the third quartile, which is the 75th percentile, is shown as follows:

Index point for Q3: i = P

100 1n2 = 75

100 1182 = 13.5

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C h a p t e r 3 | 3.5 Measures of Relative Position 121

Q3 is in the 14th position of our sorted data; therefore, Q3 = 5.64. Thus, the three quartiles were found to be as follows:

Q1 = 3.99 Q2 = 4.93 Q3 = 5.64

Excel will calculate quartiles with the QUARTILE.EXC function

= QUARTILE.EXC1array, quart2 where

array = The data range of interest quart = 1, 2, or 3 for the first, second, or third quartile, respectively Figure 3.18 shows how this function is used to find the quartiles for our mishandled

baggage example. The results are shown in Cells E2 through E4.

{ figure 3.18 Excel’s QUARTILE.EXC Function

Once again, the value shown in Cell E4 is slightly different from the results that we calculated manually because Excel uses a different technique to determine quartiles.

The interquartile range, IQR, describes the middle 50% of a range. We can find it by discarding the values in the upper and lower quartile ranges of the distribution so that only the middle values remain. To find the interquartile range, subtract the first quartile from the third quartile, as Equation 3.27 shows.

The interquartile range, IQR, describes the middle 50% of a range.

Formula 3.27 for the Interquartile Range

IQR = Q3 - Q1

The interquartile range for our mishandled baggage data is calculated as follows:

IQR = 5.64 - 3.99 = 1.65 What remains—1.65—represents the span of the IQR, or the range in which 50% of the values fall.

This is a good time to see if you “get it” by testing your percentile skills on this next Your Turn section.

Y o u r t u r n # 8 The following table lists the U.S. and Canadian box-office revenues for the highest grossing films of all time (in millions of dollars). These data can also be found in the Excel file titled boxoffice revenues.xlsx.

Film Adjusted Revenue* ($000,000) Gone with the Wind (1939) $1,538 Star Wars: A New Hope (1977) $1,355 The Sound of Music (1965) $1,084 E.T. the Extra-Terrestrial (1982) $1,080 The Ten Commandments (1956) $998 Titanic (1997) $977 Jaws (1975) $975 Doctor Zhivago (1965) $945 The Exorcist (1973) $841 Snow White and the Seven Dwarfs (1937) $829 One Hundred and One Dalmatians (1961) $760

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122 C h a p t e r 3 | Calculating Descriptive Statistics

Film Adjusted Revenue* ($000,000) Star Wars: The Empire Strikes Back (1980) $747 Ben-Hur (1959) $746 Avatar (2009) $742 Star Wars: Return of the Jedi (1983) $716 The Sting (1973) $678 Raiders of the Lost Ark (1981) $671 Jurassic Park (1993) $656 The Graduate (1967) $651 Star Wars: The Phantom Menace (1999) $646 Fantasia (1941) $632 The Godfather (1972) $601 Based on: http://www.boxofficemojo.com.

*In U.S. dollars adjusted to the estimated 2010 average ticket price of $7.61.

Using these data, determine the following:

1. The three quartiles and the IQR 2. The 65th percentile 3. The percentile rank for the film The Sting (one of my favorites)

Answers can be found on u page 143

Box-and-Whisker plots A box-and-whisker plot is a graphical display showing the relative position of the three quartiles as a box on a number line. The display also shows the minimum and maximum values in the data set along with any extreme values, or outliers.

To illustrate the box-and-whisker plot, I’ll use the data in Table 3.26, which lists the number of visitors, in millions, for 15 U.S. National Parks during 2012. These data can also be found in the Excel file titled national parks.xlsx.

My wife, Deb, at a spot known as the Jump Off in the Great Smoky Mountain National Park.

{ table 3.26 | National Park Visitors, 2012

park number of visitors (in millions)

Great Smoky Mountains (TN, NC) 9.69

Grand Canyon (AZ) 4.42

Yosemite (CA) 3.85

Yellowstone (WY) 3.45

Rocky Mountain (CO) 3.23

Zion (UT) 2.97

Olympic (WA) 2.82

Grand Teton (WY) 2.71

Acadia (ME) 2.43

Cuyahoga Valley (OH) 2.30

Glacier (MT) 2.16

Bryce Canyon (UT) 1.39

Shenandoah (VA) 1.21

Everglades (FL) 1.14

Denali (AK) 0.39

Based on: http://www.nature.nps.gov/stats.

A box-and-whisker plot is a graphical display showing the relative position of the three quartiles as a box on a number line along with the minimum and maximum values in the data set and any outliers.

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C h a p t e r 3 | 3.5 Measures of Relative Position 123

Box-and-whisker plots are constructed by taking the following steps:

Step 1. Determine the three quartiles.

Index point for Q1: i = P

100 1n2 = 25

100 1152 = 3.75

Q1 is in the fourth position of our sorted data; therefore, Q1 = 1.39 million visitors.

Index point for Q2: i = P

100 1n2 = 50

100 1152 = 7.5

Q2 is in the eighth position of our sorted data; therefore, Q2 = 2.71 million visitors.

Index point for Q3: i = P

100 1n2 = 75

100 1152 = 11.25

Q3 is in the 12th position of our sorted data; therefore, Q3 = 3.45 million visitors. Thus, the three quartiles were found to be as follows:

Q1 = 1.39 Q2 = 2.71 Q3 = 3.45

Step 2. Draw a horizontal number line that spans the length of the data values, as shown in Figure 3.19.

Step 3. Draw a box over the number line that has vertical lines at Q1, Q2, and Q3, as shown in Figure 3.19.

0.0 4.0 6.0 8.0 10.02.0

Q1 1.39

Q2 2.71

Q3 3.45

{ figure 3.19 The Box in a Box-and-Whisker

Plot

Step 4. Determine if any outliers exist in the data. Recall that outliers are extreme values in the data set that require special consideration.

Earlier in this chapter, we identified outliers as having z-scores greater than {3 standard deviations from the mean. John Tukey (1915–2000), who was a famous American statistician, proposed defining outliers as values that are above the upper limit found in Equation 3.28 or below the lower limit determined by Equation 3.29.

Formulas 3.28 and 3.29 for the Upper and Lower Limits of Outliers

Upper Limit = Q3 + 1.51IQR2 Lower Limit = Q1 - 1.51IQR2

For a data set that is normally distributed, these equations will define approximately 1 out of 100 values as outliers.

For our National Park data, the IQR is calculated as follows:

IQR = Q3 - Q1 IQR = 3.45 - 1.39 = 2.06

This results in the following upper and lower limits:

Upper Limit = 3.45 + 1.512.062 = 6.54 Lower Limit = 1.39 - 1.512.062 = - 1.70

Paul Velleman, who was a student of John Tukey, tells a well-traveled story about the origin of these equations. When asked why the factor 1.5 is used to define outliers, Tukey responded, “Because 1 is too small and 2 is too large.” When John Tukey spoke, people listened!

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124 C h a p t e r 3 | Calculating Descriptive Statistics

Any data value below - 1.70 or above 6.54 would be considered an outlier in this example. Because the number of visitors at Great Smoky Mountain is above 6.54, 9.69 would be labeled as an outlier.

Step 5. Draw the whiskers on each end of the box as shown in Figure 3.20. The whiskers are the horizontal dotted lines that extend from each side of the box. The right whisker extends out to the highest data value in our National Park example that is not an outlier, which is 4.42 (million visitors) and which corresponds to Grand Canyon National Park. The left whisker extends to the lowest data value that is not an outlier, which is 0.39 (million visitors). It corresponds to Denali National Park.

Step 6. Identify outliers if present. If outliers exist in the data set, they are plotted with an asterisk on the number line, as shown in Figure 3.20.

{ figure 3.20 The Whiskers in a Box-and-

Whisker Plot

0.39 4.42

0.0 4.0 6.0 8.0 10.02.0

9.69

The finished box-and whisker plot for our National Park example is shown in Figure 3.21.

{ figure 3.21 Finished Box-and-Whisker Plot

0.0 4.0 6.0 8.0 10.02.0

As you can see, the box-and whisker plot provides an easy-to-understand picture of where the data are concentrated along a number line, as well as information about outliers. Now, let’s take a look at one other way to describe our data set.

the Five-number Summary We can also summarize data using a five-number summary, which consists of—you guessed it—the following five numbers for a distribution:

• The minimum value

• The first quartile

• The second quartile

• The third quartile

• The maximum value

Using the National Park data, our five-number summary would be as follows:

• 0.39

• 1.39

• 2.71

• 3.45

• 9.69

Notice that outliers are included in a five-number summary. Therefore, we include 9.69 in our summary, even though it was identified as an outlier in our box-and-whisker plot.

The following Your Turn will test your box-and-whisker skills, so I encourage you to give it a shot while all those steps are still fresh in your mind.

The five-number summary consists of a distribution’s minimum value; the values for the first, second, and third quartiles; and the maximum value.

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C h a p t e r 3 | 3.5 Measures of Relative Position 125

Y o u r t u r n # 9 Use the following credit card balances for 16 Citibank customers to construct a box-and-whisker plot and a five-number summary.

$3,414 $10,827 $8,128 $12,437

$7,960 $378 $9,970 $9,038

$19,400 $11,499 $5,030 $11,113

$8,048 $15,783 $11,349 $10,162

Answer can be found on u page 144

3.5 Section Problems Basic Skills 3.36 Consider the following data set:

38 55 76 20 30 66 17 54

a. Determine the 20th percentile. b. Determine the 40th percentile. c. Determine the 60th percentile.

3.37 Consider the following data set:

22 5 18 20 13 33 30 8 39

a. Determine the 15th percentile. b. Determine the 35th percentile. c. Determine the 85th percentile.

3.38 Consider the following data set:

71 40 45 67 5 15 43 37 13 10

Determine the percentile rank for the following values:

a. 50 b. 30 c. 15

3.39 Consider the following data set:

106 83 63 104 118 124 117 136 82 97 65 141

Determine the percentile rank for the following values:

a. 100 b. 75 c. 120

3.40 Consider the following data set:

14 24 20 18 23 20 12 16 22 18 14 26 15

a. Determine the quartiles for these data. b. Calculate the interquartile range for these data.

3.41 Identify the five-number summary for the following data set:

35 24 54 2 36 23 16 43 59 55 28 81

applications 3.42 The following table shows the estimated populations

of the 12 largest U.S. cities from the 2010 Census. These data can also be found in the Excel file titled city populations.xlsx.

City

Population (in millions)

New York, NY 8.18

Los Angeles, CA 3.79

Chicago, IL 2.70

Houston, TX 2.10

Philadelphia, PA 1.53

Phoenix, AZ 1.45

San Antonio, TX 1.33

San Diego, CA 1.31

Dallas, TX 1.20

San Jose, CA 0.95

Jacksonville, FL 0.82

Indianapolis, IN 0.82

a. Calculate the 60th percentile for the data. b. Calculate the 20th percentile for the data. c. Calculate the 85th percentile for the data.

3.43 The following table shows the total runs scored by the 16 National League teams in Major League Baseball during the 2012 season. These data can also be found in the Excel file titled MLB runs.xlsx.

Team Runs Team Runs

Brewers 776 Reds 669

Cardinals 765 Pirates 651

Rockies 758 Padres 651

Diamondbacks 734 Mets 650

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126 C h a p t e r 3 | Calculating Descriptive Statistics

3.6 Measures of Association Between Two Variables

The statistics that have been discussed in this chapter to this point have all dealt with de- scribing one variable at a time, such as Nike’s stock price or the number of televisions sold at Walmart. In this last section, you will learn about measurements that will describe the re- lationship between two variables. For example, suppose we are interested in the relationship between the number of hours that a student studies for an exam and the resulting exam score. Table 3.27 shows values for these two variables from a small sample of students.

{ table 3.27 | Exam Grades

student hours of study exam grade

1 3 86

2 7 98

3 5 92

4 4 83

5 3 78

6 2 79

{ figure 3.22 Scatter Plot of Hours and Exam

Scores

Figure 3.22 shows the scatter plot for the sample data in Table 3.27.

Team Runs Team Runs

Nationals 731 Dodgers 637

Giants 718 Cubs 613

Braves 700 Marlins 609

Phillies 684 Astros 583

a. Find the percentile rank for the runs scored by the Mets.

b. Find the percentile rank for the runs scored by the Braves.

c. Find the percentile rank for the runs scored by the Phillies.

3.44 The following data set refers to the number of customers per day at a jewelry kiosk in Christiana Mall during a 20-day period:

8 10 18 58 58 59 63 64 69 71 75 78 80 82 84 84 86 87 87 88

a. Identify the first, second, and third quartiles. b. Determine the interquartile range.

3.45 The following table shows the number of graduating students with business majors at Neumann University over the past 30 years in descending order:

269 261 242 238 225 223 217 212 208 204 199 198 197 195 195 193 193 191 189 185 181 179 177 174 174 172 171 171 170 170

a. Construct a box-and-whisker plot for these data. b. Identify the five-number summary.

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C h a p t e r 3 | 3.6 Measures of Association Between Two Variables 127

where

sxy = The sample covariance between variables x and y x = The sample mean of the x variable y = The sample mean of the y variable 1xi - x2 = The difference between each data value and the sample mean for the

x variable 1yi - y2 = The difference between each data value and the sample mean for the

y variable n = The sample size

For this example, I am making the following arbitrary assignment of variables:

x = the hours of study y = the exam score

We need to calculate the sample average for each variable. Notice that n = 6 because our sample consists of six students. Using Equation 3.1, we have:

Hours of Study: x = a

i = 1 xi

n =

3 + 7 + 5 + 4 + 3 + 2 6

= 4.0

Exam Score: y = a

i = 1 yi

n =

86 + 98 + 92 + 83 + 78 + 79 6

= 86.0

Table 3.28 shows the calculations needed for the numerator of the covariance Equation 3.30 for our exam score example.

The sample covariance, sxy, measures the direction of the linear relationship between two variables.

Notice that the data points in Figure 3.22 tend to move in a relatively straight-line pattern across the scatter plot. As a result, we can describe this relationship between hours and exam scores as linear. In the following pages, we will examine two descriptive statistics that will measure the linear relationship between variables such as those seen in Table 3.27—the sample covariance and the sample correlation coefficient.

Sample Covariance The sample covariance, sxy, measures the direction of the linear relationship between two variables by using Equation 3.30.

Formula 3.30 for the Sample Covariance

sxy = a

i = 1 1xi - x21yi - y2

n - 1

I could have also chosen to reverse these assignments of x and y—the choice is completely arbitrary.

{ table 3.28 | Covariance Calculations

xi x 1xi - x 2 yi y 1yi - y 2 1xi - x 2 1yi - y 2 3 4 - 1 86 86 0 0

7 4 3 98 86 12 36

5 4 1 92 86 6 6

4 4 0 83 86 - 3 0

3 4 - 1 78 86 - 8 8

2 4 - 2 79 86 - 7 14

a 1xi - x21yi - y2 = 64

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128 C h a p t e r 3 | Calculating Descriptive Statistics

We are now ready to plug our results into Equation 3.30:

sxy = a

i = 1 1xi - x21yi - y2

n - 1 =

6 - 1 = 12.8

Because our covariance is a positive value (12.8), we describe the relationship between hours of study and exam scores as a positive linear one. In general, a positive linear relation- ship means that as one variable increases, the second variable also tends to increase. This type of relationship fits our example well—students who study more tend to earn higher grades!

A negative covariance indicates a negative linear relationship. In this scenario, as one variable increases in value, the second variable tends to decrease. An example of this type of relationship would be price and demand for a product. As the price of the product increases, demand for that product tends to decrease.

Finally, a covariance close to zero indicates no relationship between the two variables. As the value of one variable increases, the value of second variable will have no consistent pattern.

Students may be tempted to conclude that the larger the covariance, the stronger the relationship between the two variables. Unfortunately, this statement is not true. The size of the covariance is influenced by the scale of the data. Very large data values tend to generate a large covariance. The covariance is useful in describing the direction of the relationship but not the strength of the relationship. The next descriptive statistic that we are examining, the correlation coefficient, describes both for us.

Sample Correlation Coefficient The sample correlation coefficient, rxy, measures both the strength and direction of the linear relationship between two variables and can be calculated using Equation 3.31.

The sample correlation coefficient, rxy, measures both the strength and direction of the linear relationship between two variables.

Formula 3.31 for the Sample Correlation Coefficient

rxy = sxy sxsy

where

rxy = The sample correlation coefficient sxy = The sample covariance between variables x and y sx = The sample standard deviation for the x variable sy = The sample standard deviation for the y variable

As you can see, Equation 3.31 relies on the covariance calculation that we just computed in the last section. We also need to calculate the sample standard deviations from each variable which I show here using Equation 3.6.

Hours of Study:

sx = H ani = 11xi - x22n - 1 sx = A 13 - 422 + 17 - 422 + 15 - 422 + 14 - 422 + 13 - 422 + 12 - 4226 - 1 sx = A 165 = 23.2 = 1.79

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C h a p t e r 3 | 3.6 Measures of Association Between Two Variables 129

Exam Grade:

sy = H ani = 11yi - y22n - 1 sy = A 186 - 8622 + 198 - 8622 + 192 - 8622 + 183 - 8622 + 178 - 8622 + 179 - 86226 - 1 sy = A 3025 = 260.4 = 7.77 We are now ready to calculate our sample correlation coefficient using Equation 3.31.

rxy = sxy sxsy

= 12.8

11.79217.772 = 0.920

The values of rxy range from between - 1.0 and + 1.0. When rxy is positive, the linear rela- tionship between x and y is positive, which means that as the value of x increases, the value of y tends to increase. By contrast, when rxy is negative, the linear relationship between x and y is negative. So, as the value of x increases, the value of y tends to decrease. When rxy = 0, there is no relationship between the x and y variables.

The strength of the linear relationship between x and y is measured by how close the correlation coefficient is to + 1.0 or - 1.0. Figure 3.23 illustrates the strengths of various cor- relation coefficients.

{ figure 3.23 The Strength and Direction of

Various Linear Relationships, Along with Their Correlation Coefficients

r = 1.0

r = –1.0

r = 0.6

r = 0

A B

C E x

r = –0.4

With rxy = 0.920, we can conclude that there is a strong positive relationship between the number of hours that students study and their exam score—good news for us teachers!

I do have one important word of warning here. Just because the sample correlation coefficient tells us there is a strong relationship between the variables does not prove that one variable caused the change in the other variable. To prove causation, you need to have evidence that one variable is causing the other to move, which the correlation coefficient alone does not provide. For example, if our two variables were the left and right shoe sizes of people, we would expect a very high correlation. This is not the result of one foot causing the other foot to be a certain size. These two variables just tend to move together naturally.

The following Your Turn exercise will give you the chance to calculate the covariance of correlation coefficient for a negative relationship to test your skills on this topic.

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130 C h a p t e r 3 | Calculating Descriptive Statistics

3.6 Section Problems Basic Skills 3.46 Consider the following sample data for two variables.

x 4 7 2 6 6

y 4 9 5 5 7

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

3.47 Consider the following sample data for two variables.

x 7 6 2 3

y 6 8 11 5

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

Business applications 3.48 Comcast’s customer service department asks its custom-

ers to rate their over-the-phone service on a scale of 1–20 immediately after their service has been completed. The company then matches each customer’s rating with the number of minutes the person waited on hold. The fol- lowing table shows the ratings and number of minutes on hold for 10 randomly selected customers and can also be found in the Excel file hold.xlsx.

Minutes Rating Minutes Rating

4 15 2 16

8 13 10 14

0 18 3 20

5 10 8 14

6 14 4 13

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

3.49 Suppose the EPA would like to investigate the relation- ship between the engine size of sedans and the miles per gallon (MPG) they get. The following table shows the engine size in cubic liters and rated miles per gallon for a selection of 2011 sedans and can also be found in the Excel file engine size.xlsx.

Car Engine Size MPG

1 2.0 26

2 2.2 30

3 2.4 25

4 3.3 21

5 3.0 23

6 2.0 28

7 2.4 24

8 2.0 29

9 3.2 19

Suppose Best Buy would like to examine the relationship between the price and weekly demand for a popular Epson printer. The following data represent the weekly demand for various price levels that were used for this printer.

Price and Demand Data for an Epson Printer

Price ($) Demand

70 18

75 23

80 10

85 16

90 17

95 15

100 6

a. Calculate the sample covariance for these variables. b. Calculate the sample correlation coefficient for these variables. c. Describe the relation between price and demand for this printer.

Y o u r t u r n # 1 0N

Answers can be found on u pages 144–145

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C h a p t e r 3 | 3.6 Measures of Association Between Two Variables 131

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

3.50 A regional manager at Acme Markets would like to de- velop a model to predict weekly sales of pet food based on the shelf space. The following table shows the results of data collected from nine randomly selected stores and can also be found in the Excel file Acme.xlsx.

Shelf Space (Feet)

Sales ($100)

Shelf Space (Feet)

Sales ($100)

2 3 2 4

3 3 4 5

4 6 5 7

6 5 5 12

5 18

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

3.51 The American Board of Family Medicine would like to in- vestigate the theory that a mother’s shoe size can be used to predict an infant’s birth weight. The following table records a random sample of the shoe sizes of 10 moth- ers and the birth weights of their infants and can also be found in the Excel file ABFM.xlsx.

Shoe Size Weight Shoe Size Weight

7.5 7.6 8.0 6.9

8.0 7.8 8.5 9.2

9.0 8.0 7.5 6.5

8.5 9.4 8.0 9.1

8.5 7.7 9.5 9.8

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

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132

■ Measures of central tendency describe the center point of our data set with a single value by using the mean, median, or mode.

The simple mean or average is the most common measure of central tendency and is calculated by adding all the values in our data set and then dividing the result by the number of observations.

• One drawback of the mean is that it can be heavily influenced by outliers.

The weighted mean allows you to assign more weight to certain values and less weight to others.

The median is the value in the data set for which half the observations are higher and half the obser- vations are lower.

• The median is more difficult than the mean to calculate manually, but it is not influenced by outliers.

The mode is simply the value that appears most often in a data set.

• The mode is particularly useful when it comes to describing categorical data.

• The mode might not be unique in a data set or even exist in certain data sets.

The mean, median, and mode are roughly equal to each other in symmetrical frequency distributions.

• In left-skewed distributions, the median is greater than the mean.

• In right-skewed distributions, the median is less than the mean.

■ Measures of variability determine how much spread is in your data by using the range, variance, and standard deviation.

The range is a measure of variability that is found by subtracting the lowest value from the highest value in the data set.

• The benefit of the range is that it is easy to calculate and understand but is has its drawbacks.

• It is based only on two numbers in the data set.

• It is highly affected by outliers.

The variance measures the squared difference between each data point and the mean.

The standard deviation is the square root of the variance and is in the same units as the original data.

The coefficient of variation, CV, measures the standard deviation in terms of its percentage to the mean rather than units to the mean.

• When comparing two samples, the smaller CV indicates more consistency (less variability).

The z-score identifies the number of standard deviations that a particular value is from the mean of its distribution and is positive for values above the mean and negative for values below the mean.

The empirical rule states that if a distribution follows a bell-shape, symmetrical curve centered around the mean, approximately 68%, 95%, and 99.7% of the values will fall above and below one, two, and three standard deviations around the mean, respectively.

Chebyshev’s theorem states that regardless of whether a distribution is bell-shaped or not, at least 94% of data values will fall within four standard deviations of the mean; at least 89% of data values will fall within three standard deviations of the mean; and at least 75% of data values will fall within two standard deviations of the mean.

The mean and standard deviation can be calculated for grouped data by using the midpoint of each class to approximate all the data within that class.

• Because the midpoint approximates only the data within each class, our final result is merely an approximation of the average and standard deviation.

Chapter 3 Summary

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133

Box-and-whisker plot. A graphical display showing the relative position of a distribution’s three quartiles as a box on a num- ber line, along with the minimum and maximum values.

Categorical Data. Data that are described as a category or label. Chebyshev’s Theorem. A theorem that states that regardless of

whether a distribution is bell-shaped, at least 94% of data val- ues will fall within { four standard deviations of the mean, at least 89% of data values will fall within { three standard deviations of the mean, and at least 75% of data values will fall within { two standard deviations of the mean.

Coefficient of variation. A measure of the standard deviation in terms of its percentage of the mean.

Empirical rule. A rule which states that if a distribution follows a bell-shaped, symmetrical curve centered around the mean,

approximately 68%, 95%, and 99.7% of the values fall within one, two, and three standard deviations around the mean, respectively.

Five-number summary. A list that consists of a distribution’s minimum value; first, second, and third quartiles; and maximum value.

Index point. The index point, i, marks the middle of the data values and is used to determine the position of the median in the data set.

Interquartile range, IQR. The difference between the first and third quartiles. It corresponds to the data in the middle 50% of the range.

Left-skewed distribution. The shape of the distribution when the median is higher than the mean.

■ Measures of relative position compare the position of one value in relation to other values in the data set.

■ Percentiles measure the approximate percentage of values in the data set that are below the value of interest.

■ Quartiles are simply specific percentiles of interest.

• The first quartile is the 25th percentile, the second quartile is the 50th percentile, and the third quartile is the 75th percentile.

■ The interquartile range, IQR, is the difference between the first and third quartile.

■ A box-and-whisker plot is a graphical display showing the relative position of the three quartiles as a box on a number line along with the minimum and maximum values in the data set.

■ The five-number summary consists of a distribution’s minimum value, values for the first, second, and third quartiles, and maximum value.

■ Measures of association describe the relationship between two variables.

■ The sample covariance measures the direction of the linear relationship between two variables.

• A positive covariance indicates a positive linear relationship, which means as one variable in- creases, the second variable also tends to increase.

• A negative covariance indicates a negative linear relationship, which means as one variable in- creases, the second variable tends to decrease.

■ The sample correlation coefficient measures both the strength and the direction of the linear rela- tionship between two variables.

• The correlation coefficient ranges between - 1.0 and + 1.0. • When the correlation coefficient is positive, the linear relationship between x and y is positive.

• When the correlation coefficient is negative, the linear relationship between x and y is negative.

• The strength of the linear relationship between x and y is measured by how close the correlation coefficient is to + 1.0 or - 1.0.

Chapter 3 Key Terms

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x = a

i = 1 1wixi2

a n

i = 1 wi

m = a N

i = 1 xi

x = a

i = 1 xi

Formula 3.3 for the Weighted Mean

Formula 3.2 for the Population Mean

Formula 3.1 for the Sample Mean

3.1 Central Tendency (pp 78–91)

Chapter 3 Equations

Mean. A measure of central tendency that is calculated by adding up all of the values in a data set and then dividing the result by the number of observations.

Measures of central tendency. Measures that use a single value to describe the center point of a data set.

Measures of relative position. Measures that compare the posi- tion of one value in relation to other values in a data set.

Measures of variability. Measures that determine how much of a spread there is within a data set.

Median. The value in a data set for which half the observations are higher and half the observations are lower.

Midpoint. The halfway point in a set of data. It can be found by taking the average of the endpoints for each class.

Mode. The value that appears most often in a data set. Outliers. Outliers are values that are much higher or lower than

most of the data. Percentiles. Measure the approximate percentage of values in the

data set that are below the value of interest. Percentile rank. Identifies the percentile of a particular value

within a set of data. Pth percentile. The approximate percentage of values in the data

set that are below the value of interest (where p is any number between 1 and 100).

Quartiles. The first, second, and third quartiles are the 25th, 50th (median), and 75th percentiles, respectively.

Range. A measure of variability that is found by subtracting the lowest value from the highest value in a data set.

Right-skewed distribution. The shape of a distribution when its mean is higher than its median.

Sample Correlation Coefficient, rxy. Measures both the strength and direction of the linear relationship between two variables.

Sample Covariance, sxy. Measures the direction of the linear relationship between two variables.

Standard deviation. The square root of a distribution’s variance. Variance. A measure of variation that describes the relative

distance between the data points in a set around the mean of the data set.

Weighted mean. A weighted mean allows you to assign more weight to certain values and less weight to others when calculating the mean.

z-score. A measure that identifies the number of standard deviations a particular value is from the mean of its distribution.

Formula 3.4 for the Index Point for the Median i = 0.51n2

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Formula 3.5 for the Range Range = Highest value - Lowest value

Formula 3.6 for the Sample Variance s2 = a

i = 1 1xi - x22

n - 1

Formula 3.7 for the Sample Standard Deviation s = 2s2 = H ani = 11xi - x22n - 1 Formula 3.8 for the Sample Variance (Short-Cut) s2 =

a n

i = 1 x i 2 -

a a n

i = 1 xib

n - 1

Formula 3.9 for the Sample Standard Deviation (Short-Cut) s = a ani = 1x i2 - a a n

i = 1 xib

n - 1

Formula 3.10 for the Population Variance s2 = a N

i = 1 1xi - m22

3.2 Measures of Variability (pp 91–101)

Formula 3.11 for the Population Standard Deviation s = R aNi = 11xi - m22N Formula 3.12 for the Population Variance (Short-Cut) s2 =

a N

i = 1 x i 2 -

a a N

i = 1 xib

Formula 3.13 for the Population Standard Deviation (Short-Cut)

s = a aNi = 1x i2 - a a N

i = 1 xib

N N

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3.3 Using the Mean and Standard Deviation Together (pp 101–112)

Formula 3.14 for the Sample Coefficient of Variation CV = s

x 11002

Formula 3.15 for the Population Coefficient of Variation CV = s

m 11002

Formula 3.16 for the Population z-score z = x - m s

Formula 3.17 for the Sample z-score z = x - x

Formula 3.18 for Expressing the z-Score in Terms of x x = m + zs

Formula 3.19 for Chebyshev’s Theorem

Chebyshev’s Theorem states that for any number z greater than 1, the percent of the values that fall within z standard deviations from the mean will be at least plus or minus¢1 - 1

z 2 ≤ * 100

Formula 3.20 for Expressing the z-Score (z) in Terms of the Sample (x)

x = x + zs

3.4 Working with Grouped Data (pp 112–116)

Formula 3.21 for the Sample Mean: Grouped Data x ≈ a

i = 1 1fimi2 n

Formula 3.22 for the Population Mean: Grouped Data m ≈ a

i = 1 1fimi2 N

Formula 3.23 for the Sample Variance: Grouped Data s2 ≈ a

i = 1 1mi - x22fi n - 1

Formula 3.24 for the Population Variance: Grouped Data s2 ≈ a

i = 1 1mi - m22fi

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3.5 Measures of Relative Position (pp 116–126)

Formula 3.25 for the Index Point for the pth Percentile i = P

100 1n2

Formula 3.26 for the Percentile Rank Percentile Rank = 1Number of values below x2 + 0.5

Total number of values 11002

Formula 3.27 for the Interquartile Range IQR = Q3 - Q1

Formulas 3.28, 3.29 for the Upper and Lower Limits for Outliers

Upper Limit = Q3 + 1.51IQR2 Lower Limit = Q1 - 1.51IQR2

Chapter 3 Problems 3.52 You are car shopping (lucky you!) and are considering a

Ford, a Chevy, and a Honda. You are basing your decision on the criteria in the following table. You have also ranked each of the criterions based on its importance to you, with a rank of 5 being the most important: price (5), reliability (4), looks (3), gas mileage (3), and resale value (2). You have scored each car for each criterion on a scale of 1–10 110 = highest2, as shown in the following table. Calculate the average score for each car.

Criteria Ford Chevy Honda

Price 7 8 3

Reliability 7 6 9

Looks 5 7 5

Gas mileage 4 6 8

Resale value 7 4 8

3.53 The following data represent the number of days homes were on the market before being sold in New Castle County, Delaware, in 2014:

80 74 62 79 55 56 64 44 13 74 98 87 45 142 88 47 79

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Which of these three measurements would best describe

the central tendency of the data? e. Describe the shape of this distribution.

3.54 The average number of hours that people of Brazil spend per month on social networking sites is nearly double the worldwide average, drawing the attention of companies such as Facebook. The following data represent the number of hours that a sample of people from Brazil spent on a so- cial networking site last month.

1.5 1.0 2.3 12.1 8.8 13.2 40.2 1.5 12.4

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Which of these three measurements would best describe

the central tendency of the data? e. Describe the shape of the distribution.

Formula 3.30 for the Sample Covariance sxy = a

i = 1 1xi - x21yi - y2

n - 1

Formula 3.31 for the Sample Correlation Coefficient rxy = sxy sxsy

3.6 Measures of Association Between Two Variables (pp 126–131)

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3.55 Internet service providers compete on services such as download speeds that are measured in megabits per second (Mbps). The following data represent the download speeds that 16 Comcast customers experienced recently.

6 18 7 8 8 6 10 16 11 5 14 6 12 12 16 13

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Describe the shape of the distribution.

3.56 The Fan Cost Index measures the average cost for a typical family to attend a Major League Baseball game. The index comprises the prices of two adult average-price tickets, two child average-price tickets, two small draft beers, four small soft drinks, four regular-size hot dogs, parking for one car, two game programs, and two adult-size adjustable caps (the least expensive). The Excel file Fan Cost Index lists the in- dex for each MLB team for the 2012 season. Using Excel,

a. Calculate the mean. b. Calculate the median. c. Determine the mode. d. Which of these three measurements would best describe

the central tendency of the data? e. Describe the shape of this distribution.

3.57 A country’s fertility rate can have a major long-term impact on its economic health. Low fertility rates eventually cause the average age of the population to skew higher, making it more difficult to fund programs such as social security. The following data represent a sample showing the number of children from 10 families.

1 5 3 1 2 2 1 4 0 1

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.58 The following table lists the average monthly unemploy- ment rate from July to September 2012 for various European countries.

Country Rate Country Rate

Belgium 7.4 Denmark 8.1

France 10.7 Germany 5.4

Spain 25.5 Ireland 15.0

United Kingdom 7.9 The Netherlands 5.3

Austria 4.5 Sweden 7.7

a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.59 Delta is one of the largest airlines in the world and also has one of the oldest fleets of planes. The following data represent the age, in years, of a sample of eight planes from Delta’s fleet.

17 22 8 19 20 21 7 14

a. Calculate the range for this sample. b. Calculate the variance for this sample. c. Calculate the standard deviation for this sample.

3.60 The following data represent the number of pounds lost by 10 individuals who participated in a weight loss program. A negative number represents a weight loss, whereas a positive number represents a weight gain.

- 12 -8 - 15 - 10 4 -6 -9 5 -8 - 10 a. Calculate the range. b. Calculate the variance. c. Calculate the standard deviation.

3.61 Text messages have been a major source of revenue for cell phone carriers and for the first time in history the aver- age number of text messages per month decreased from the second to the third quarter of 2012. The Excel file text messages.xlsx lists the number of text messages sent last month from a sample of 70 cell phone customers. Using Excel, determine the range, variance, and standard devia- tion for this sample.

3.62 A company operates two retail outlets at different locations. The following table lists the customer satisfaction ratings on a scale of 1–10 for each location:

Store 1 Store 2

7 8

4 7

6 10

3 8

9 8

Which location provides a more consistent level of cus- tomer satisfaction?

3.63 The following table shows the number of wins per season for the New England Patriots and Detroit Lions NFL teams from 2004 through 2012. This data can also be found in the Excel file NFL wins.xlsx.

Year New England Detroit

2004 14 6

2005 10 5

2006 12 3

2007 16 7

2008 11 0

2009 10 2

2010 14 6

2011 13 10

2012 12 4

Which team experienced more variability with games won per season?

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139

3.64 The following table lists the 20 largest companies in the world in 2013 ranked by their market capitalizations (in billions of dollars). A company’s market capitalization is defined as the number of its shares multiplied by the price per share. This data can also be found in the Excel file market cap.xlsx.

Company Market Cap ($) Company

Market Cap ($)

Apple 423 Procter & Gamble 210

Exxon 407 AT&T 203

Google 264 Pfizer 202

Berkshire Hathaway

252 Wells Fargo 189

General Electric 245 JPMorgan Chase 186

Wal-Mart 236 Coca-Cola 172

Microsoft 233 Oracle 165

IBM 227 Phillip Morris 156

Chevron 227 Merck 131

Johnson & Johnson

211 Verizon 130

a. Determine the coefficient of variation. b. Calculate the z-score for a company with a market

capitalization of $200 billion. c. Test to determine if the empirical rule holds true by

determining the percentage of observations that fall within { one standard deviation from the mean.

d. Verify that Chebyshev’s Theorem holds true by deter- mining the percentage of observations that fall within { two standard deviations from the mean.

3.65 Suppose I recorded the attendance of my students in a recent statistics class because I wanted to investigate the relationship between the number of classes they missed and their final grades. The following table shows this data for a random sample of nine students. This data can also be found in the Excel file missed classes.xlsx.

Missed Classes

Final Grade

Missed Classes

Final Grade

4 81 2 84

7 79 0 90

2 93 5 86

5 70 2 95

0 96

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

3.66 As a measure of productivity, Verizon Wireless records the number of customers each of its retail employees activates weekly. An activation is defined as either a new customer signing a cell phone contract or an existing customer

renewing a contract. The following table shows the number of weekly activations for eight randomly selected employ- ees along with their job-satisfaction levels rated on a scale of 1 - 10 110 = most satisfied2. This data can also be found in the Excel file Verizon satisfaction.xlsx.

Activations Satisfaction Activations Satisfaction

36 8.0 19 6.1

25 7.9 28 7.0

40 8.5 33 8.2

38 9.0 25 7.7

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

3.67 The Excel file labeled Customer service champs.xlsx lists the top 25 companies, with the best customer service scores ranked by Business Week in 2012.

a. Determine the coefficient of variation. b. Calculate the z-score for a company with a score of 900. c. Test if the empirical rule holds true by determining the

percentage of observations that fall within { two stan- dard deviations from the mean.

d. Verify that Chebyshev’s Theorem holds true by deter- mining the percentage of observations that fall within { three standard deviations from the mean.

3.68 The following table shows the frequency distribution for the golf scores recorded during the 2012 Masters tournament:

Score Frequency

278–282 6

283–287 12

288–292 21

293–297 16

298–302 6

303–307 1

a. Calculate the approximate average golf score for the tournament.

b. Calculate the approximate variance and standard deviation for the tournament.

3.69 The following table shows the frequency distribution for the mileage on a sample of Avis rental cars:

Mileage Frequency

5,000 to under 6,000 16

6,000 to under 7,000 11

7,000 to under 8,000 24

8,000 to under 9,000 10

9,000 to under 10,000 15

10,000 to under 11,000 9

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a. Calculate the approximate average mileage per car for the sample.

b. Calculate the approximate variance and standard devia- tion for the sample.

3.70 The following data represent the tips left by 21 customers for the waitstaff at a local restaurant:

$18.40 $10.35 $12.15 $ 7.75 $14.00 $ 9.50 $11.45 $16.00 $15.50 $12.75 $10.50 $12.40 $13.75 $10.75 $ 8.40 $17.85 $16.50 $17.60 $14.10 $10.00 $11.00

a. Calculate the 30th percentile for the data. b. Calculate the 45th percentile for the data. c. Calculate the 65th percentile for the data.

3.71 The following data represent the number of students enrolled in 20 business classes this semester at Wilmington University:

29 33 45 26 30 19 21 42 42 24 34 30 28 24 20 25 13 46 40 32

a. Calculate the 20th percentile for the data. b. Calculate the 70th percentile for the data. c. Calculate the 90th percentile for the data.

3.72 The following table lists the SAT scores for 27 college students:

845 1,040 815 1,345 1,554 668 1,165 919 1,175 629 490 1,074 387 894 1,419 941 454 673 696 974 564 1,245 1,535 665 1,471 1,188 680

a. Identify the first, second, and third quartiles. b. Determine the interquartile range.

3.73 The following data represent the number of customers per day that entered an AT&T wireless retail center in Orlando over a 30-day period.

114 106 124 84 86 104 161 183 146 123 189 138 157 166 121 153 222 103 168 147 202 115 156 183 201 202 178 122 94 227

a. Identify the first, second, and third quartiles. b. Determine the interquartile range.

3.74 The following tables record the times of swimmers in the 50-m freestyle and 100-m backstroke events at a recent meet. A particular swimmer swam the freestyle event in 33 seconds and the backstroke in 72 seconds. In which event did the swimmer perform better relative to the competition?

50-m Freestyle Times 26 29 31 32 32 33 34 35

100-m Backstroke Times 68 71 72 75 76 78 78 78 79 80

3.75 The following table shows the final exam scores for the stu- dents from my last statistics class:

78 100 88 84 73 78 70 99 92 96 87 89 81 79 80 84 86 85 79 80 72 69 89 66 76

Determine the percentile ranks for the following scores:

a. 70 b. 75 c. 82 d. 90

3.76 The following data indicate the battery life, in minutes, on a single charge, for 25 iPads:

215 219 229 230 236 239 240 244 247 255 262 264 271 279 280 282 285 285 290 296 301 310 326 327 330

a. Construct a box-and-whisker plot for these data. b. Identify the five-number summary.

3.77 The following data set represents the odometer mileage, in thousands of miles, of 30 used cars currently in inventory at Mojo Motors:

109.6 107.2 104.4 103.2 100.5 99.9 98.6 97.6 96.2 93.2 93.1 92.9 88.9 86.8 86.4 83.6 83.5 82.8 81.5 80.9 76.8 75.2 75.1 70.7 69.9 69.7 67.2 63.0 61.0 59.8

a. Construct a box-and-whisker plot for these data. b. Identify the five-number summary.

3.78 Fair Isaac, the company that developed the current credit score model used by most lenders today, would like to ex- amine the relationship between the age and credit score of an individual. The following table shows the credit scores and ages of 10 randomly selected people. This data can also be found in the Excel file Fair Isaac.xlsx.

Age Credit Score Age Credit Score

36 675 47 790

24 655 35 720

54 760 59 760

28 615 40 685

31 660 46 610

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

3.79 Suppose Consumer Reports would like to investigate the rela- tionship between the battery life, in hours, and the screen size, in diagonal inches, of a laptop computer. The following table shows a random sample of eight laptops and the battery life, in hours, and corresponding screen size, in inches, of each. This data can also be found in the Excel file laptop battery.xlsx.

Battery Life Screen Size Battery Life Screen Size

3.6 15.6 3.1 14.5

4.5 17.3 3.4 13.3

4.2 14.5 4.8 11.6

4.5 11.6 3.9 13.6

a. Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.

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Chapter 3 Solutions to Your Turn

Y o U R T U R n # 1

1. x = a x n

= 14 + 1 - 172 + 31 + 1 - 102 + 0 + 11 + 5 + 18

8 =

8 = 6.5 minutes

2. x = a wi xia wi = 1116217.22 + 180218.82 + 167216.02 + 191217.52

116 + 80 + 67 + 91 =

2,623.7

354 = 7.41

Y o U R T U R n # 2 1. The sorted ages are as follows:

26 28 31 39 43 45 45 50 57 62 i = 0.51n2 = 0.51102 = 5

The median is the midpoint between the fifth (43) and sixth (45) values.

Median = 43 + 45

2 = 44

2. The sorted SAT scores are as follows:

460 490 510 560 590 610 650 670 700 i = 0.51n2 = 0.5192 = 4.5

The median is the fifth value, which is 590.

Y o U R T U R n # 3 1. Mode = $16.74, $16.75, and $16.85. All three values appear twice. 2. Mode = 26. This value appears four times. 3. No mode exists because no value repeats in the data set.

4. Mode = Cherry. This flavor has the highest number of sales. This is not a coincidence, as it happens to be my favorite!

Y o U R T U R n # 4 1. These data are considered a sample because only a subset of adults in the

general population is included in the problem.

xi x2i 10 100

10 100

4 16

8 64

13 169

6 36

11 121

a 7

i = 1 xi = 62 a

i = 1 x i 2 = 606

a a n

i = 1 xib

= 16222 = 3,844

s2 = a

i = 1 x i 2 -

a a n

i = 1 xib

n - 1 =

606 - 3,844

7 - 1

= 606 - 549.14

6 = 9.48

s = 2s2 = 29.48 = 3.08

2. These data are considered a population because the ages of all eight employees in the bank branch are included in the problem.

xi x 2 i

32 1,024

40 1,600

27 729

29 841

34 1,156

38 1,444

46 2,116

24 576

a 8

i = 1 xi = 270 a

i = 1 x i 2 = 9,486

a a N

i = 1 xib

= 127022 = 72,900

s2 = a N

i = 1 x i 2 -

a a N

i = 1 xib

N =

9,486 - 72,900

= 9,486 - 9,112.5

8 = 46.7

s = 2s2 = 246.7 = 6.8

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Y o U R T U R n # 5

Rollins Pujols

x x2 x x2

12 144 41 1,681

25 625 49 2,401

30 900 32 1,024

11 121 37 1,369

21 441 47 2,209

8 64 42 1,764

16 256 37 1,369

23 529 30 900

Total 146 3,080 315 12,717

Rollins Pujols

sR = a ani = 1x 2i - a a n

i = 1 xib

n - 1 sP = a ani = 1x 2i - a a

i = 1 xib

n - 1

sR = H 3,080 - 11462288 - 1 sP = H 12,717 - 13152288 - 1 sR = A 3,080 - 2,664.507 sP = A 12,717 - 12,403.137 sR = 259.36 sP = 244.84 sR = 7.70 sP = 6.70

xR = a a

i = 1 xib

n =

146

8 = 18.25 xP =

a a n

i = 1 xib

n =

315

8 = 39.38

CVR = sR xR 11002 CVP =

sP xP 11002

CVR = 7.70

18.25 11002 = 42.2% CVP =

6.70

39.38 11002 = 17.0%

Albert Pujols is a more consistent home-run hitter because he has a lower CV.

Y o U R T U R n # 6

1. a) z = x - x

s =

$6,000 - $7,000 $850

= - 1.18

b) z = x - x

s =

$8,400 - $7,000 $850

= 1.65

2. a) x = x + zs x = $322 + 1121$252 = $347 x = $322 + 1 - 121$252 = $297

b) x = $322 + 1221$252 = $372 x = $322 + 1 -221$252 = $272 c) x = $322 + 1321$252 = $397 x = $322 + 1 - 321$252 = $247 3. a) x = x + zs x = $1,920 + 1221$1602 = $2,240 x = $1,920 + 1 -221$1602 = $1,600

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b) a1 - 1 z 2 b = 0.85

z 2 = 0.15

z 2 = 1

0.15 = 6.67

z = 26.67 = 2.58 x = x + zs x = $1,920 + 12.5821$1602 = $2,332.80 x = $1,920 + 1 - 2.5821$1602 = $1,507.20 c) x = $1,920 + 1421$1602 = $2,560 x = $1,920 + 1 - 421$1602 = $1,280

Y o u r T u r n # 7

GPA Midpoint

3.0 to under 3.2 3.1

3.2 to under 3.4 3.3

3.4 to under 3.6 3.5

3.6 to under 3.8 3.7

3.8 to 4.0 3.9

x ≈ a

i = 1 1fimi2 n

x ≈ 16213.12 + 18213.32 + 16213.52 + 114213.72 + 16213.92

6 + 8 + 6 + 14 + 6

= 141.2

40 = 3.53

GPA mi fi x 1mi - x2 1mi - x22 1mi - x22fi 3.0–3.2 3.1 6 3.53 - 0.43 0.1849 1.1094 3.2–3.4 3.3 8 3.53 - 0.23 0.0529 0.4232 3.4–3.6 3.5 6 3.53 - 0.03 0.0009 0.0054 3.6–3.8 3.7 14 3.53 0.17 0.0289 0.4046

3.8–4.0 3.9 6 3.53 0.37 0.1369 0.8214

n = Σfi = 40 Σ1mi - x22 fi = 2.764

s2 ≈ a k

i = 1 1mi - x22fi n - 1

= 2.764

40 - 1 = 0.0709.

s ≈ 20.0709 = 0.267 Y o u r T u r n # 8 1. n = 22

Q1: i = P

100 1n2 = 25

100 1222 = 5.5

Q1 is in the 6th position of our data sorted from lowest to highest; therefore, Q1 = $671.

Q2: i = P

100 1n2 = 50

100 1222 = 11

Q2 is the midpoint between the 11th ($747) and 12th ($760) positions; therefore,

Q2 = $747 + $760

2 = $753.5

The calculation for the third quartile index point is shown as follows:

Q3: i = P

100 1n2 = 75

100 1222 = 16.5

Q3 is in the 17th position of our sorted data; therefore, Q3 = 977. Thus, the three quartiles were found to be as follows:

Q1 = $671 Q2 = $753.5 Q3 = $977

IQR = $977 - $671 = $306

2. i = P

100 1n2 = 65

100 1222 = 14.3

The 65th percentile is the data value in the 15th position and is equal to $945.

3. There are a total of 22 films, of which 6 are below The Sting.Therefore, the approximate percentile for The Sting is as follows:

Percentile = 162 + 0.5

22 11002 = 6.5

22 11002 = 30th

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Y o U R T U R n # 9 Sort the data from lowest to highest:

Q1: i = P

100 1n2 = 25

100 1162 = 4

$378 $8,048 $10,162 $11,499

$3,414 $8,128 $10,827 $12,437

$5,030 $9,038 $11,113 $15,783

$7,960 $9,970 $11,349 $19,400

Q1 is midway between the fourth (7,960) and the fifth (8,048) positions of our sorted data.

Q1 = 7,960 + 8,048

2 = 8,004

Q2: i = P

100 1n2 = 50

100 1162 = 8

Q2 is midway between the eighth (9,970) and the ninth (10,162) positions of our sorted data.

Q2 = 9,970 + 10,162

2 = 10,066

Q3: i = P

100 1n2 = 75

100 1162 = 12

Q3 is midway between the 12th (11,349) and the 13th (11,499) positions of our sorted data.

Q3 = 11,349 + 11,499

2 = 11,424

IQR = Q3 - Q1 IQR = 11,424 - 8,004 = 3,420 Upper Limit = Q3 + 1.51IQR2 Upper Limit = 11,424 + 1.513,4202 = 16,554 Lower Limit = Q1 - 1.51IQR2 Lower Limit = 8,004 - 1.513,4202 = 2,874

There are two outliers in the data set: $378 and $19,400.

3,414 15,783

0 8 12 16 204

19,400

(000’s Dollars)

11,424

10,066

8,004

378

* *

Five-number summary: 378, 8,004, 10,066, 11,424, 19,400

Y o U R T U R n # 1 0

a) Price: x = a

i = 1 xi

n =

70 + 75 + 80 + 85 + 90 + 95 + 100 7

= 85.0

Demand: y = a

i = 1 yi

n =

18 + 23 + 10 + 16 + 17 + 15 + 6 7

= 15.0

xi x 1xi - x2 yi y 1yi - y2 1xi - x2 1yi - y2 70 85 - 15 18 15 3 - 45 75 85 - 10 23 15 8 - 80 80 85 - 5 10 15 - 5 25 85 85 0 16 15 1 0

90 85 5 17 15 2 10

95 85 10 15 15 0 0

100 85 15 6 15 - 9 - 135

a 1xi - x21yi - y2 = -225

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b) sxy = a

i = 1 1xi - x21yi - y2

n - 1 =

- 225 7 - 1

= - 37.5

sx = H ani = 11xi - x22n - 1 sx = A 170 - 8522 + 175 - 8522 + 180 - 8522 + 185 - 8522 + 190 - 8522 + 195 - 8522 + 1100 - 85227 - 1 sx = A 7006 = 2116.67 = 10.80 sy = H ani = 11yi - y22n - 1 sy = A 118 - 1522 + 123 - 1522 + 110 - 1522 + 116 - 1522 + 117 - 1522 + 115 - 1522 + 16 - 15227 - 1 sy = A 1846 = 230.67 = 5.54

rxy = sxy

sxsy =

- 37.5 110.80215.542 = - 0.627

c) We have discovered a negative relationship between price and demand for this printer. As the price increases, weekly demand tends to de- crease, which would be expected.

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C h a p t e r

I n t h I s c h a p t e r , y o u w I l l l e a r n t o :

■ Distinguish between classical, empirical, and subjective probability.

■ Identify the basic properties of probability.

■ Calculate and interpret the intersection and union of events.

■ Use the addition rule and multiplication rule for probability.

■ Interpret the meaning of conditional probability.

■ Use Bayes’ Theorem to calculate revised probabilities from prior probabilities.

■ Count the total number of possible outcomes using the fundamental counting principle.

■ Distinguish between permutations and combinations.

Introduction to Probabilities

C h a p t e r 4 M a p

4.1 An Introduction to Probabilities | 148

4.2 Probability Rules for More Than One Event | 156

4.3 Counting Principles | 176

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148 C h a p t e r 4 | Introduction to Probabilities

Organizations in today’s business world face a great deal of uncertainty when they are making decisions. Probability provides a valuable tool to quantify this uncertainty so that managers can make better decisions. Some examples of how probability is used in business follow:

• An e-commerce business would like to know the probability that a customer will make a purchase after browsing through the firm’s Web site for more than 10 minutes.

• A credit card executive would like to examine the probability that a customer with a poor credit score will be late with his or her next payment.

• A media consultant is asked to determine the probability that the viewing audience for next year’s Super Bowl will exceed 130 million people.

• A company’s sales department would like to establish the probability that a male cus- tomer will purchase an extended warranty after buying a laptop computer.

In this chapter, we will explore the topic of probability and discover its usefulness in a variety of business settings.

4.1 An Introduction to Probabilities A probability is a numerical value that indicates the chance, or likelihood, of a specific event occurring. Weather forecasters commonly use probabilities when referring to the chance of rain the following day. A probability is expressed by statisticians as a number ranging from 0 to 1. The number is translated into a percentage. For example, if an event cannot occur, then it is assigned a probability of 0, which is stated as 0%. By contrast, if the event will ab- solutely occur, the probability of it occurring is 1, or 100%.

But before we go any further, we need to tackle some new statistics jargon. The follow- ing terms are widely used when talking about probability:

• Experiment. The process of measuring or observing an activity for the purpose of collecting data. An example is rolling a single six-sided die.

• Sample space. All the possible outcomes, or results, of an experiment. The sample space for our single-die experiment is the numbers {1, 2, 3, 4, 5, 6}. Statisticians put the numbers in braces to signal that they represent the sample space.

A probability is a numerical value ranging from 0 to 1. It indicates the chance, or likelihood, of a specific event occurring. If there is no chance of the event occurring, the probability is 0. If the event is absolutely going to occur, the probability of it occurring is 1.

An outcome is just a fancy word for a result of an experiment. For example, if I roll a single die and get a six, the outcome of the experiment is a six.

Probability is probably (no pun intended) the most misunder- stood topic in the field of statistics. People often struggle to make rational decisions when dealing with probabilities be- cause it’s difficult to judge the likelihood of complex events. For example, most people (including my wife) are much more opti- mistic about winning the state lottery than the actual numbers merit. A common type of lottery is called 6/49. The goal of it is to correctly choose 6 out of 49 numbers. As you will see later in the chapter, the chance of winning this lottery is an infinitesimal 0.00000007, or nearly 1 out of 14 million. The probability is so low that it is difficult to fathom. If you purchase one lottery ticket every day of the year, you can expect to win once every 38,286 years. To give this some perspective, 38,000 years ago humans were living in caves during the Stone Age.

Let’s compare your probability of winning the 6/49 lottery to the likelihood that you will be struck by lightning. According to the

National Oceanic and Atmospheric Administration (NOAA), the federal agency that monitors atmospheric conditions, the odds of being struck by lightning in the United States on any given day are approximately 1 out of 400,000. According to these statistics, you are 35 times more likely to be struck by lightning today than win the lottery.

In spite of presenting this useful information to my wife in what I considered to be a very convincing manner, she still insisted that I go out and buy lottery tickets during a severe thunderstorm when the jackpot was unusually large. The good news is that I returned home safely without getting lit up like a Christmas tree. The bad news is that we didn’t win the lottery (big surprise), as evidenced by the fact that I haven’t quit my day job.

Stats in practice: Statistics author Sent Out by Wife to purchase Lottery tickets During thunderstorm: What’s Wrong with this picture?

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C h a p t e r 4 | 4.1 An Introduction to Probabilities 149

• Event. One or more outcomes of an experiment. The outcome, or outcomes, is a sub- set of the sample space. An example of an event is rolling a pair with two dice.

• Simple event. An event with a single outcome in its most basic form that cannot be sim- plified. An example of a simple event is rolling a five with a single die.

Table 4.1 shows some other examples of experiments and their corresponding sample spaces.

The event “rolling a pair with two dice” is not a simple event because it contains two outcomes—the numbers from each die.

{ table 4.1 | Examples of Experiments and Their Sample Spaces

experiment sample space

Flip a coin {heads, tails}

Answer a multiple-choice question {a, b, c, d, e}

Inspect a product {defective, not defective}

Draw a card from a standard deck {52 cards in the deck}

There are three methods of assigning probability that we will discuss in this chapter: classical, empirical, and subjective.

Classical probability Classical probability is used when we know the number of possible outcomes of the event of interest and can calculate the probability of that event using Equation 4.1.

Classical probability is used when we know the number of possible outcomes of the event of interest and can calculate the probability of that event using Equation 4.1.

Formula 4.1 for classical probability

P1A2 = Number of possible outcomes that constitute Event A Total number of possible outcomes in the sample space

where

P1A2 = The probability that Event A will occur For our single-die experiment, our sample space has six possible outcomes, as shown in

Figure 4.1.

{ figure 4.1 The Sample Space for a Single

Die

If the simple Event A is defined as rolling a five, which has only one possible way of oc- curring, then the probability of Event A occurring is calculated as follows:

P1A2 = 1 6

= 0.167

In other words, there is a 16.7% probability of rolling a five. This is an example of simple probability because we are considering the likelihood of a single (simple) event occurring by itself. Later in this chapter, we will expand this concept to consider two or more events occurring.

Classical probability assumes that each event in the sample space has the same likeli- hood of occurring. That is, the chance of rolling a one is the same as rolling a two, and so forth. You also need to know the total number of possible outcomes in the sample space.

Simple probability represents the likelihood of a single (simple) event occurring by itself.

Classical probability requires that you know the number of outcomes that pertain to a particular event. You also need to know the total number of possible outcomes in the sample space.

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150 C h a p t e r 4 | Introduction to Probabilities

Formula 4.2 for empirical probability

P1A2 = Frequency in which Event A occurs Total number of observations

If the sample space includes every possible simple event that can occur, then we say that set of events is collectively exhaustive (sounds like how I feel when I’m trying to make my publisher’s writing deadlines). For example, the events 1, 2, 3, 4, 5, and 6 are collectively ex- haustive when we roll a single die. When we roll the die, one of these outcomes must occur.

We can expand our classical probability example to rolling two dice together. This changes the sample space to the following one, which has 36 possible outcomes. Each pair of numbers represents the outcome of the two dice. Let’s define Event A as obtaining a sum of five. There are four ways in which a sum of five could be rolled; they are highlighted in yellow in the following sample space.

Sample Space When Rolling Two Dice to Obtain a Sum of Five

511,12 11,22 11,32 11,42 11,52 11,62 12,12 12,22 12,32 12,42 12,52 12,62 13,12 13,22 13,32 13,42 13,52 13,62 14,12 14,22 14,32 14,42 14,52 14,62 15,12 15,22 15,32 15,42 15,52 15,62 16,12 16,22 16,32 16,42 16,52 16,626

We can now calculate the probability of rolling a sum of five with our two dice. Because there are four ways to obtain a sum of five out of the 36 possible outcomes,

P1A2 = 4 36

= 0.111

In other words, there is an 11.1% chance of two dice summing to the number five.

A sample space is described as collectively exhaustive if it includes every possible simple event that can occur.

The sample space {male, female} would be an example of collectively exhaustive events because a person’s gender must be one or the other.

Try your luck on this card example to test your understanding of classical probability.

1. What is the probability of randomly selecting a spade from a standard 52-card deck? 2. What is the probability of obtaining a sum of eight when rolling two dice?

Y O u r t u r n # 1N

Answers can be found on u page 189

empirical probability With classical probability, we understand the underlying process of what we are working with, such as “There are 4 aces in a deck of 52 cards, so the probability of drawing an ace is 4/52.” When we don’t have the information needed to determine the number of outcomes associated with an event (there is no “known” sample space, in other words), we rely on empirical probability. Empirical probability involves conducting an experiment to observe the frequency with which an event occurs. We then calculate the ratio of the number of times the event occurred relative to the total number of times we observed that it both did and did not occur (Equation 4.2),

Empirical probability involves conducting an experiment to observe the frequency with which an event occurs.

Empirical probability requires that you count the frequency that an event occurs through an experiment and calculate the probability from the experiment’s relative frequency distribution. Say for example, I am a store owner and would like to know the probability that a per-

son who walks in my store will make a purchase. Classical probability is no help to me here because I don’t have any information about why someone makes a purchase. I need to use empirical probability by counting how many customers come to my store over a period of time and how many of those customers made a purchase. If 100 customers walked into my store today and 15 made a purchase, then my empirical probability for this event would be 15>100 = 0.15.

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C h a p t e r 4 | 4.1 An Introduction to Probabilities 151

On a recent trip to the Nautica Factory Outlet store in Hilton Head, South Carolina, I was presented a scratch card from the salesperson that entitled me to a discount between 5% and 50% on my purchase. My discount card turned out to be 15%. Hoping to do better, I qui- etly told my wife to pretend she didn’t know me (which she was happy to do) and to get her own card. I’m sad to report Deb came back with a whopping 5% discount. For the sake of this example, let’s suppose that when Deb approached her, the salesperson had 20 cards with the discounts shown in Table 4.2.

{ table 4.2 | Nautica Factory Outlet Discount Cards

discount (%) number of cards

5 9

10 4

15 2

20 3

30 0

40 1

50 1

Total 20

{ table 4.3 | Calculating Empirical Probabilities

discount (%) number of cards probability

5 9 9>20 = 0.45 10 4 4>20 = 0.20 15 2 2>20 = 0.10 20 3 3>20 = 0.15 30 0 0>20 = 0.00 40 1 1>20 = 0.05 50 1 1>20 = 0.05

Total 20 20,20 = 1.00

For instance, the 5% discount has a frequency of nine because there were nine of these cards in the salesperson’s hand. We can determine the probability of Deb receiving each of the discounts using the relative frequency distribution procedure discussed in Chapter 3. This gives us the empirical probabilities of Deb receiving the discounts shown in Table 4.3.

As Table 4.3 shows, the probability of Deb getting a 5% discount is 45%. What was the probability that her discount would exceed my 15%? Because there were 5 cards out of the 20 that were greater than 15%, the probability of Deb’s discount exceeding 15%, which I’ll define as Event A, was as follows:

P1A2 = 5 20

= 0.25

Oh well, it was worth a try! There is an important connection between classical and empirical probabilities that

can be explained with the following example. Suppose I flip a fair coin (one with an equal chance of observing a heads or tails) three times and observe three tails. The empirical probability for this experiment is shown in Table 4.4.

There are three cards with 20%, one card with 40%, and one card with 50%.

Empirical probability is also known as the relative frequency approach because it relies on relative frequencies that were discussed in Chapter 2.

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152 C h a p t e r 4 | Introduction to Probabilities

Even though the empirical probability of a tails in this example is 100%, we all know the classical probability is 50%. If we flipped the coin, say, 500 times, the empirical prob- ability would be very close to 50%. What we have discovered here is the law of large numbers, which states that when an experiment is conducted a large number of times, the empirical probabilities of the process will converge to the classical probabilities.

The following Your Turn problem will give you an opportunity to test your newly learned empirical probability skills.

The law of large numbers states that when an experiment is conducted a large number of times, the empirical probabilities of the process will converge to the classical probabilities.

{ table 4.4 | The Empirical Probabilities of a Coin Flip

outcome frequency probability

Heads 0 0>3 = 0.00 Tails 3 3>3 = 1.00 Total 3 3,3 = 1.00

In a recent survey of 1,025 adults conducted by the poll-taking company Gallup asked how much money they planned to spend on Christmas gifts. The results are shown in the following table:

Amount Frequency

$1,000 or more 297

$500–$999 246

$250–$499 144

$100–$249 164

Less than $100 41

No opinion 133

Total 1,025

a) What is the probability that a randomly selected person is planning to spend between $250 and $499 for Christmas?

b) What is the probability that a randomly selected person is planning to spend $250 or more for Christmas?

Y O u r t u r n # 2N

Answers can be found on u page 189

Subjective probability We use subjective probability when classical and empirical probabilities are not available. These are instances in which data or experiments are not available to calculate probabilities. Instead, we rely on our experience and judgment to estimate the probabilities.

The following examples demonstrate how subjective probability is used in the business world:

• A director of marketing estimates that there is a 20% probability that her company’s biggest competitor will reduce its prices in the next month.

• A human resource manager reports that there is a 75% chance that his company will institute a hiring freeze for the upcoming year.

• A market analyst believes that there is a 50% probability that Apple will announce a new version of the iPhone next month.

In each case, it would be very difficult to determine these probabilities using classical or empirical methods. Rather, these subjective probabilities would most likely be based on the

We use subjective probability when classical and empirical probabilities are not available and instead we rely on experience and intuition to estimate the probabilities.

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C h a p t e r 4 | 4.1 An Introduction to Probabilities 153

expertise and judgment of the individuals. We often engage in this type of assessment in our day-to-day lives, albeit more informally. Have you, perhaps, at some point in your college career mulled over the possibility of passing tomorrow’s exam after staying out late with your friends the night before? If you have, you have put subjective probability to work!

Let’s see if you can distinguish between the three different types of probabilities in the next Your Turn problem.

{ table 4.5 | Emergency Room Arrival Methods

method of arrival frequency probability

Walk-in 132 132>197 = 0.67 Ambulance 61 61>197 = 0.31 Helicopter 2 2>197 = 0.01 Prison guard 2 2>197 = 0.01 Total 197 197,197 = 1.00

Indicate whether classical, empirical, or subjective probability should be used to determine each of the following:

a. The probability that the baseball player Josh Hamilton will hit a home run during his next at bat b. The probability of drawing a jack from a deck of cards c. The probability that I will shoot lower than a 90 during my next round of golf d. The probability of winning your state’s lottery e. The probability that the price of gasoline will exceed $4.00 per gallon next summer f. The probability that I will finish writing my next textbook before my deadline

Y O u r t u r n # 3N

Answers can be found on u page 190

Basic properties of a probability Our next step is to review the “rules and regulations” that govern probability theory. The ba- sic ones are as follows:

Probability Rule 1 If P1A2 = 1, then with certainty, Event A must occur. An example is Event A = Rolling a single six-sided die and observing a 1, 2, 3, 4, 5, or 6.

Probability Rule 2 If P1A2 = 0, then with certainty, Event A will not occur. An example is Event A = Draw- ing five cards from a 52-card deck without replacing any of them and observing five aces in your hand. (A 52-card deck only has four aces. The only way this event can occur is if some- body is cheating and has another ace up his or her sleeve!)

Probability Rule 3 The probability of any event must range from 0 to 1. Probabilities can never be negative or greater than 1. For example, the probability of my wife buying a pair of shoes next month could be 0 (0%), 1 (100%), or some number in between. However, the probability could never be - 1 1 - 100%2 or 2 (200%). (My experience tells me my subjective probability of this event occurring will be closer to 100% than 0%, but I should keep that assessment to myself.)

Probability Rule 4 The sum of all the probabilities for the simple events in the sample space must be equal to 1. For example, Table 4.5 shows the way in which patients arrived at the University of Virginia Medical Center’s emergency room during a recent 24-hour period.

Note that the sum of the probabilities in the third column adds to 1.00 because every simple event in the sample space is included in the table. In other words, the only recorded methods of arrival are those listed in Table 4.5.

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154 C h a p t e r 4 | Introduction to Probabilities

Probability Rule 5 The complement to Event A is defined as all of the outcomes in the sample space that are not part of Event A. The complement is denoted as A′. Using this definition, we can state the following:

The complement to Event A is defined as all the outcomes in the sample space that are not part of Event A and is denoted as A′.

Formula 4.3 for the complement rule

P1A2 + P1A′2 = 1 or P1A2 = 1 - P1A′2

In other words, the probability of the complement of an event occurring is 100% minus the probability of the event itself occurring. We can see this graphically in Figure 4.2, which shows a Venn diagram. The box represents the entire sample space, and the shaded circle represents Event A. The area outside the circle within the box represents the complement of Event A.

{ figure 4.2 Venn Diagram Illustrating the

Complement Rule A

Table 4.6 demonstrates the complement rule by showing the results of a survey con- ducted by Impulse Research for the Downtown Cookie Company. In the survey, 1,033 adults were asked to identify their favorite cookie.

{ table 4.6 | Favorite Cookie Survey

favorite cookie frequency probability

Chocolate chip 548 0.53

Peanut butter 165 0.16

Oatmeal 155 0.15

Sugar/shortbread 114 0.11

Other 51 0.05

Total 1,033 1.00

We’ll define Event A as selecting a person whose favorite cookie was chocolate chip. This means that Event A′ represents the probability of selecting a person whose favorite cookie was not chocolate chip. From Table 4.6, we can determine P1A′2 as follows:

P1A2 = 0.53 P1A′2 = 1 - P1A2 = 1 - 0.53 = 0.47

The probability of selecting someone whose favorite cookie is not chocolate chip is 47%. Notice that the complement rule saved us some work because it is easier to calculate than adding up all the following probabilities:

P1A′2 = 0.16 + 0.15 + 0.11 + 0.05 = 0.47 Test your newfound skills in the following Your Turn box before moving on to the next

section.

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C h a p t e r 4 | 4.1 An Introduction to Probabilities 155

4.1 Section Problems Basic Skills 4.1 Identify which of the following values are valid and not

valid numbers for a probability:

a. 5

4 f. 14%

b. 0.04 g. - 3

3 c. 200% h. 0% d. 0.5 i. 1.2

e. 1

8 4.2 What is the probability of randomly selecting a diamond

from a standard 52-card deck?

4.3 Consider the experiment of rolling two dice. a. What is the probability of obtaining a sum of seven? b. What is the probability of obtaining a sum of three or

less? c. What is the probability of obtaining a sum of ten or

more?

4.4 Indicate whether classical, empirical, or subjective prob- ability should be used to determine each of the following: a. The probability that the Green Bay Packers will make

the playoffs this year. b. The probability that no students will be absent from

my next statistics class.

c. The probability that I will draw three of a kind in a five-card hand.

d. The probability that the local bakery will sell more than five loaves of rye bread tomorrow.

e. The probability that I will randomly select a red M&M from a bowl that contains five red, eight yellow, and seven green M&Ms.

applications 4.5 A customer survey asked respondents to indicate their high-

est levels of education. The only three choices in the survey were high school, college, and other. If 29% indicated high school and 53% indicated college, determine the percentage of respondents who chose the “other” category.

4.6 A survey was conducted in which 125 families were asked how many cats lived in their households. The results are as follows:

Number of Cats Number of Households

0 76

1 32

2 12

3 3

4 2

Total 125

1. Identify which of the following values are valid and not valid numbers for a probability:

a. 0.19 b. - 0.6 c. 0 d. 51% e. 1.08

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Answers can be found on u page 190

f. 124%

g. 2 3

h. 1

i. - 1 6

2. The following table shows the average change in salaries for full-time permanent employees over a recent one-year period as determined by a survey of 2,795 human resource professionals by Harris Interactive:

Change in Salary Frequency Probability

11% or more 58 0.02

4%–10% 351 0.13

1%–3% 720 0.26

No change 1,491 0.53

Decrease 175 0.06

Total 2,795 1.00

Use the complement rule to determine the probability that an employee did not take a pay cut.

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156 C h a p t e r 4 | Introduction to Probabilities

a. What is the probability that a randomly selected family has one cat?

b. What is the probability that a randomly selected family has more than one cat?

c. What is the probability that a randomly selected family has cats?

d. Is this an example of classical, empirical, or subjective probability?

4.7 A total of 200 adults responded to a survey asking them to name their favorite national park. The relative frequencies are as follows:

Park Relative Frequency (%)

Yellowstone 33

Great Smoky Mountains 19

Yosemite 19

Acadia 7

Other ?

Total 100

a. What percentage of respondents chose the “Other” category?

b. How many respondents selected Yellowstone? c. How many respondents selected either Yosemite or

Acadia? d. Is this an example of classical, empirical, or subjective

probability?

4.8 The following table shows the frequencies of executives’ salary ranges at a particular organization:

Salary Range Frequency

Under $60,000 2

$60,000 to under $70,000 5

$70,000 to under $80,000 7

$80,000 to under $90,000 4

$90,000 to under $100,000 6

$100,000 or more ?

Total 26

a. How many executives at this organization earned $100,000 or more per year?

b. What is the probability that a randomly selected executive earned $100,000 or more per year?

c. What is the probability that a randomly selected executive earned $60,000 or more per year?

d. What is the probability that a randomly selected executive earned between $70,000 and $90,000 per year?

4.9 Braun Research recently surveyed U.S. adults about the number of cups of coffee they drink per day and per week. The percentages are as follows:

Number of Cups Percentage

Two or more cups a day 33

One cup a day 19

Two cups a week 11

One cup a week 9

Don’t drink coffee 28

a. What percentage of adults drink coffee? b. What percentage of adults drink two cups or more of

coffee a week? c. What percentage of adults drink one or two cups per

week?

4.2 Probability Rules for More Than One Event

Up until now, we have focused on basic probability concepts that primarily deal with a sin- gle event. However, in the business world, situations are rarely this simple. Oftentimes, they involve two or more events that intersect with one another. For example, a banking manager might want to know the probability that a customer who has a poor credit score will default on her mortgage. In this case there are two events to consider:

Event A = Customer defaults on mortgage Event B = Customer has a low credit score

To address this type of probability, we need to explore the concepts presented in this section.

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 157

the Intersection of events To demonstrate working with probabilities of multiple events, I will use the following ex- ample. Now that my children are older and living away from home, I cherish those moments when the phone rings and I see one of their names appear on my caller ID. Experience has taught me that I can categorize these calls as either “crisis”—involving such things as a com- puter, a car, an ATM card, or a cell phone—or “noncrisis”—when the kids call just to see if I’m alive and well so that I’ll be available for their next crisis.

Table 4.7, which is called a contingency table, categorizes the last 50 phone calls by child and type of call.

A contingency table indicates the number of occurrences of events that are classified according to two categorical variables.

{ table 4.7 | A Contingency Table for Phone Calls

child crisis noncrisis total

Christin 12 7 19

Brian 10 4 14

John 6 11 17

Total 28 22 50

I received a total of 14 phone calls from Brian, of which 10 were crisis and 4 were noncrisis.

Contingency tables show the actual or relative frequency of two categorical variables at the same time. In this case, the categorical variables are the child and type of call. I’ll as- sume that this past pattern of calls will hold true in the near future. We’ll define Events A and B as follows:

Event A = The next phone call will come from Christin Event B = The next phone call will involve a crisis

We can use the contingency table to calculate the simple probability that the next phone call will come from Christin:

P1A2 = 19 50

= 0.38

The probability that the next phone call will involve a crisis would be as follows:

P1B2 = 28 50

= 0.56

These simple probabilities are also known as marginal probabilities in some settings. What about the probability that the next phone call will come from Christin and will in-

volve a crisis? This event is known as the intersection of Events A and B and is described as A ¨ B. Figure 4.3 shows a Venn diagram describing the intersection of Events A and B. It is the shaded region shared by both events.

Marginal probability is another term used for simple probability.

There were a total of 19 calls from Christin (12 crisis and 7 noncrisis) in Table 4.7.

There were a total of 28 crisis calls between all three kids in Table 4.7.

The intersection of Events A and B represents the number of instances in which Events A and B occur at the same time (that is, the same phone call is both from Christin and a crisis).

The number of phone calls from our contingency table that meet both criteria is found at the intersection of the Crisis column and the Christin row in Table 4.7, which is 12; therefore,

P1A and B2 = P1A ¨ B2 = 12 50

= 0.24

This explains why I hold my breath as I pick up the phone! The probability of the intersec- tion of two events is known as a joint probability.

The probability of the intersection of two events is known as a joint probability.

A B

A and B

{ figure 4.3 Venn Diagram Illustrating the

Intersection of A and B

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158 C H A P T E R 4 | Introduction to Probabilities

Therefore, the probability that the next phone call is either from Christin (Event A) or a crisis (Event B) is as follows:

P1A or B2 = P1A h B2 = 35 50

= 0.70

Pay close attention to the wording of a probability question. Does it include the word and or the word or? One way to distinguish between the use of and and or is to remember the following relationships that always hold true:

And S Intersection Or S Union

Our previous example shows that there is a big difference between the probability of the intersection of two events occurring and the probability of the union of two events occurring:

Intersection: P1A and B2 = P1A ¨ B2 = 12 50

= 0.24

Union: P1A or B2 = P1A h B2 = 35 50

= 0.70

{ figure 4.4 Venn Diagram Illustrating the

Union of Events A and B A B

A or B

{ table 4.8 | A Contingency Table Showing the Union of Two Events: Phone Calls

child crisis noncrisis total

Christin 12 7 19

Brian 10 4 14

John 6 11 17

Total 28 22 50

child type of call number of calls

Christin Crisis 12

Christin Noncrisis 7

Brian Crisis 10

John Crisis 6

Total = 35

Many of us college professors have been accused of tricking our poor defenseless students with the use of and and or in exams. Convince yourself that you understand the difference.

The probability of the intersection of two events occurring can never be more than the probability of the union of two events occurring.

The Union of Events The union of Events A and B represents the number of instances in which either Event A or Event B or both occur together. The union is denoted as A h B. Figure 4.4 shows a Venn diagram describing the union of Events A and B. It consists of all three shaded regions.

The union of Events A and B represents the number of instances where either Event A or B occur or both events occur together.

Using our previous example, the highlighted numbers in Table 4.8 show the instances that include either

• A call—crisis or noncrisis—from Christin (Event A)

• A crisis phone call from any of the three kids (Event B)

Note that there were a total of 35 such calls 112 + 7 + 10 + 6 = 352.

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 159

Notice that the probability of the union occurring is 70%, whereas the probability of the intersection occurring is just 24%. Why is this so? Because the union of Events A and B in- clude the intersection of Events A and B (the orange region in Figure 4.4) as well as the red and yellow regions in Figure 4.4. You will never encounter a situation in which the probabil- ity of the intersection of two events is greater than the union.

Understanding the difference between the intersection and the union of events and how to calculate them is very important. For example, suppose Comcast Cable is tracking customer complaints and would like to know the probability that an unsatisfied customer is male and has a premium channel package. This question requires the use of the intersection of events (AND). If the union of events (OR) is erroneously used, the resulting probability will be higher than the true value. Take a few minutes to test your understanding in the fol- lowing Your Turn problem before moving on.

the addition rule In the previous sections, we defined the intersection and union of events in general terms. Now it is time to use the addition rule for probabilities to calculate the probability of the union of events, that is, the probability that Event A, or Event B, or both events will occur. However, the addition rule requires us to first learn on a new concept: mutually exclusive events.

Two events are considered to be mutually exclusive if they cannot occur at the same time during the experiment. To demonstrate this, refer to Table 4.9, which shows the grade distri- bution of one of my statistics classes along with the genders of the students.

The addition rule for probabilities is used to calculate the probability of the union of events, that is, the probability that Event A, or Event B, or both events will occur.

Two events are considered to be mutually exclusive if they cannot occur at the same time during the experiment.

The following table indicates the number of customers at Walmart who purchased a digital camera with and without extended warranties. The contingency table also identifies the age group of the customers.

Age Group Warranty No Warranty Total

Less than 40 5 34 39

40 or Older 6 15 21

Total 11 49 60

What is the probability that a randomly selected customer

a. chose to purchase the warranty? b. was less than 40 years old? c. was 40 years or older and chose not to purchase a warranty? d. was less than 40 or chose not to purchase a warranty?

Y o U r t U r n # 5N

Answers can be found on u page 190

{ table 4.9 | Grade Distribution of My Statistics Class (by Gender)

grade female male total

A 6 3 9

B 4 8 12

C 0 4 4

Total 10 15 25

Consider the following two events:

Event A = Student earned an A grade Event B = Student earned a B grade

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160 C h a p t e r 4 | Introduction to Probabilities

Formula 4.5 for The Addition Rule for Events That Are Not Mutually Exclusive

P1A or B2 = P1A2 + P1B2 - P1A and B2

The addition rule depends on whether or not the two events are mutually exclusive. Events A and B would be considered mutually exclusive because a student cannot receive both an A and a B grade at the same time. This concept is shown graphically in the Venn diagram in Figure 4.5. As you can see, mutually exclusive events have no overlap in the sample space.

{ figure 4.5 Venn Diagram Illustrating

Mutually Exclusive Events A B

For mutually exclusive events, the addition rule states that the probability of two events occurring is simply the sum of their individual probabilities. Equation 4.4 shows how the ad- dition rule is written.

An example of two events that are not mutually exclusive: Event A = a randomly selected student gets an A grade on the exam and Event B = a randomly selected student is female. Both of these events can occur together.

Formula 4.4 for The Addition Rule for Mutually Exclusive Events

P1A or B2 = P1A2 + P1B2

For our grade-distribution example, the probability that a randomly selected student earned either an A or a B grade is calculated as follows:

P1A2 = 9 25

= 0.36 P1B2 = 12 25

= 0.48

P1A or B2 = P1A2 + P1B2 P1A or B2 = 0.36 + 0.48 = 0.84

Next, let’s consider the following two events from the grade example:

Event A = Student earned an A grade Event B = Student is female

Now, Events A and B are not mutually exclusive. This is because a student can receive an A grade and be female at the same time. This concept is shown graphically in the Venn dia- gram in Figure 4.6.

{ figure 4.6 Venn Diagram Illustrating

Events That Are Not Mutually Exclusive A B

You can see in Figure 4.6 that there is overlap between Events A and B (orange-shaded region). For events that are not mutually exclusive, the addition rule states that the probabil- ity of the two events occurring is the sum of their individual probabilities occurring minus the probability of the intersection of two events, as shown in Equation 4.5.

P(A and B) is shown in the orange-shaded region.

This term represents the orange-shaded region in Figure 4.6. We subtract it out of Equation 4.5 to avoid double counting this orange region. For our grade-distribution example, the probability that a randomly selected student

either earned an A (Event A) or is female (Event B) is determined as follows.

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 161

Nine of the 25 students in the class received A’s (Event A). Ten of the 25 students in the class are female (Event B). Therefore,

P1A2 = 9 25

= 0.36 P1B2 = 10 25

= 0.40

P(A and B) is the intersection of the two events: both receiving an A and being female. Our data from Table 4.8 show that six women received A’s. Thus,

P1A and B2 = 6 25

= 0.24

Now we are ready to calculate the probability that a randomly selected student either earned an A (Event A) or is female (Event B) using Equation 4.5.

P1A or B2 = P1A2 + P1B2 - P1A and B2 P1A or B2 = 0.36 + 0.40 - 0.24 = 0.52

Why do we need to subtract the intersection of Events A and B in the addition rule in Equation 4.5 when they are not mutually exclusive? Figure 4.7 shows what happens when the probabilities of the two events are added together. A total of nine students earned an A grade, of which six were female. There were a total of 10 female students, of which 6 earned an A. Adding the probabilities of Events A and B results in counting the six female students who earned an A grade twice. The overlap for these six students is evident in Figure 4.7. To avoid counting these students twice, we subtract this intersection of Events A and B in Equation 4.5.

{ figure 4.7 The Addition Rule for Events

That Are Not Mutually Exclu- sive: The Grade Distribution

Grade

A B C Total

Female

6 4 0

Male

3 8 4

Total

9 12

4 25

Before moving on, this would be a good time to convince yourself that you understand the addition rule with the following Your Turn problem.

The following table shows the number of Bank of America customers who have overdue accounts. The data are broken down by the credit card type and number of days overdue.

Card Type

Days Overdue Standard Gold Total

30 days or less 165 90 255

31–60 days 96 121 217

61–90 days 42 49 91

More than 90 days 22 15 37

Total 325 275 600

Determine the probability that a randomly selected customer

a. has a Standard card and is 31–60 days overdue b. has a Standard card or is 30 days or less overdue c. has a Gold card and is 61–90 days overdue d. is 31–60 days or is 30 days or less overdue e. is 31–60 days and is 61–90 days overdue

Y o U r t U r n # 6 N

Answers can be found on u page 190

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162 C h a p t e r 4 | Introduction to Probabilities

Conditional probability Conditional probability is the probability of Event A occurring, knowing, or given the condi- tion, that Event B has already occurred. In other words, the likelihood of Event A occurring can be affected by the fact that Event B has already occurred.

Consider the following example of a conditional probability: The SAT is a nationally administered test that many colleges use to evaluate student admissions. The math portion of the exam is scored on a scale of 0–800. Some students will enroll in a preparation course in the hopes of increasing their chances of a higher score. Table 4.10 shows the number of students in different categories of math scores and identifies the number of students who participated in a preparation class.

Conditional probability is the probability of Event A occurring, given the condition that Event B has occurred.

40 students who took the prep class scored in the 401–600 range

{ table 4.10 | Math SAT Scores with and Without a Prep Class

score range prep class no prep class total

201–400 8 53 61

401–600 40 107 147

601–800 22 20 42

Total 70 180 250

Consider the following two events:

Event A = Student scored in the 601–800 range in math Event B = Student participated in a prep class

We would like to investigate whether the prep class increased the chance of scoring in the 601–800 range. Conditional probability can help us answer this question.

Because the sample space for Event A is 250 students and the number of students who scored in the 601–800 range is 42, the simple probability, P1A2, is calculated as follows:

P1A2 = 42 250

= 0.168

There is a 16.8% chance that a randomly selected student scored in the 601–800 range. This simple probability is also known as a prior probability because it is determined without any additional information.

However, what if we had some additional information regarding the randomly selected student? Suppose we know the student participated in a prep class. We are now seeking the probability that the student scored in the 601–800 range, given he or she participated in a prep class. The notation for the probability of Event A, given the condition that Event B occurred, is shown as follows:

P1A � B2 For this conditional probability, the sample space of interest is reduced to those students who took the prep class, which is 70. Out of those 70 students, 22 scored in the 601–800 range. Therefore, the probability of a student scoring in the 601–800 range, given he or she partici- pated in a prep class, is as follows:

P1A � B2 = 22 70

= 0.314

Based on this evidence, it appears that the prep class has influenced the probability of scor- ing in the 601–800 range because

P1A2 = 0.168 P1A � B2 = 0.314

What we have learned is that students who take the prep class have a higher probability of scoring in the 601–800 range.

Prior probability is the probability of Event A occurring as determined without any additional information that could influence the event.

Students who take the prep class have nearly twice the probability of scoring in the 601–800 range than a randomly selected student from the sample space.

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 163

Equations 4.6 and 4.7 show how a conditional probability is calculated.

Formulas 4.6 and 4.7 for Calculating a Conditional Probability

P1A � B2 = P1A and B2 P1B2

P1B � A2 = P1A and B2 P1A2

Equation 4.6 requires that P1B2 7 0, whereas Equation 4.7 requires that P1A2 7 0. We can apply this equation to our example, as follows, to verify our SAT result:

P1A and B2 = 22 250

= 0.088

P1B2 = 70 250

= 0.280

P1A � B2 = P1A and B2 P1B2 =

0.088

0.280 = 0.314

Whichever way you calculate it, a conditional probability is also known as a posterior probability (I’ll resist using a “behind” joke here), which is a revision of the prior probability using additional information. In this case, the additional information is the fact that the stu- dent participated in the prep class.

Conditional probabilities have many applications in the business environment, such as the following:

• A bank would be very interested to know how probable it is that a customer with a credit- card balance of less than $3,000 will pay his entire balance in any given month.

• A car manufacturer would like to know the probability that a 40-year-old female customer will choose to purchase an extended warranty from the factory.

The following Your Turn problem will use the 2008 Summer Olympics to test your understanding of conditional probability.

This result matches the probability that we found earlier using Table 4.10.

A conditional probability is also known as a posterior probability, which is a revision of the prior probability using additional information.

The following table lists the number of medals won by particular countries during the 2012 Summer Olympics in London.

Gold Silver Bronze Total

United States 46 29 29 104

China 38 27 23 88

Russia 24 26 32 82

Other 194 222 272 688

Total 302 304 356 962

a. What is the probability that a randomly selected medal is gold? b. What is the probability that a randomly selected medal was won by Russia? c. What is the probability that a randomly selected medal was silver and won by the United States? d. What is the probability that a randomly selected medal was bronze or won by China? e. Given that a medal is gold, what is the probability that it was won by the United States? f. Given that a medal was won by Russia, what is the probability that it was Bronze?

Y o U r t U r n # 7N

Answers can be found on u page 190

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164 C h a p t e r 4 | Introduction to Probabilities

Independent and Dependent events Two events are considered independent of each other if the occurrence of one event has no impact on the occurrence of the other event. Consider the following two independent events:

Event A: Today, I join the multitudes and decide to purchase the current version of Apple’s iPhone.

Event B: Later that same day, Apple announces the development of the newest version of the iPhone with several cool features that the current version does not have.

Even though the timing of these two events may feel like the cruel work of fate, obvi- ously they are independent of one another. I would have a difficult time convincing any rational person that Apple’s management was waiting for my purchase before making this announcement.

If the occurrence of one event affected the probability of another event, we would call these events dependent. Consider the following two dependent events:

Event A: You earn an A grade on your statistics exam.

Event B: You study many hours preparing for your statistics exam.

As a teacher, I am confident that the occurrence of Event B will have an impact on the prob- ability of Event A, making these two events dependent.

To apply this concept, I’ll use the following example. My wife, Deb, is an avid tennis player, and we enjoy playing matches against each other. We do, however, have one dif- ference of opinion on the court. Deb likes to have a nice long warm-up session at the start, where we hit the ball back and forth and back and forth for what seems like an eternity to me. All during this time, a little voice in my head is saying, “Who’s winning?” and “What’s the score?” My ideal warm-up is to bend at the waist to tie my sneakers. Deb tells me that when we rush through the warm-up, she doesn’t play as well. “Hogwash!” I say, and I’ll use conditional probability to show her who’s right. Table 4.11 shows the outcomes of our last 25 matches, along with the type of warm-up before we started keeping score.

Two events are considered independent of one another if the occurrence of one event has no impact on the occurrence of the other event.

If the occurrence of one event affects the occurrence of another event, the events are considered dependent.

{ table 4.11 | A Contingency Table for Tennis Matches

warm-up time deb wins bob wins total

Short 4 11 15

Long 7 3 10

Total 11 14 25

We’ll define the following events of interest:

Event A = Deb wins the tennis match Event B = The warm-up time is long

First, using the totals columns in Table 4.11, we’ll calculate the marginal probability of Deb winning the match, P1A2. The calculation is as follows:

P1A2 = 11 25

= 0.44

Next, we’ll calculate the conditional probability of Deb winning the match, given that the warm-up period was long, P1A � B2, using Equation 4.6:

P1A and B2 = 7 25

= 0.28

P1B2 = 10 25

= 0.40

P1A � B2 = P1A and B2 P1B2 =

0.28

0.40 = 0.70

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 165

If the warm-up period has no impact on Deb’s chances of winning (the events are inde- pendent events), then I would expect the conditional probability that she will win because of a long warm-up to be equal to the marginal probability that she will win regardless of the length of the warm-up. In other words, P1A � B2 would equal P1A2.

Unfortunately for me, once again Deb seems to be right. The probability of her winning regardless of the length of the warm-up is just 44%. The probability of her winning given the condition of a long warm-up is 70%. These two events are dependent on one another. Excuse me while I go apologize to my wife.

In general, the definition of two events that are independent is shown in Equation 4.8.

Otherwise, Events A and B are dependent. Spend a few minutes on the next Your Turn problem to be sure you understand the con-

cept of independent and dependent events.

Formula 4.8 for Determining if Events A and B Are Independent

P1A � B2 = P1A2

Because this probability is not equal to 70%, Events A and B are not independent. It seems that Deb’s chance of winning is affected by the length of the warm-up.

The following table shows the number of adults (in thousands) in the United States who were employed and unemployed in 2012 along with their marital statuses.

Status Employed Unemployed Total

Married 95,886 6,480 102,366

Never Married 39,348 6,025 45,373

Total 135,234 12,505 147,739

Event A = Person is unemployed Event B = Person is married

Determine whether Events A and B are independent or dependent.

Y o U r t U r n # 8N

Answer can be found on u page 191

the Multiplication rule Earlier, I introduced the addition rule for probability to calculate the union of Events A and B. In this section, I will use the multiplication rule to determine the probability of the intersec- tion of two events occurring, or P(A and B). Recall that this situation is known as the joint probability. We can derive the multiplication rule, shown in Equation 4.9, by rearranging the conditional probability equation, Equation 4.6, and using algebra to solve for P(A and B). Likewise, the multiplication rule shown in Equation 4.10 can be derived from Equation 4.7. These equations assume that Events A and B are dependent.

The multiplication rule is used to determine the probability of the intersection (joint probability) of two events occurring, or P (A and B).

Formulas 4.9 and 4.10 for The Multiplication Rule for Dependent Events

P1A � B2 = P1A and B2 P1B2

P1A and B2 = P1B2P1A� B2 or

P1B � A2 = P1A and B2 P1A2

P1A and B2 = P1A2P1B � A2

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166 C h a p t e r 4 | Introduction to Probabilities

To demonstrate the multiplication rule for dependent events, consider a grocery store that unknowingly received a potato chip order containing an unusually high percentage of bags with salt content below quality control standards. The day you walk in the store to pur- chase two bags, there are 32 bags on the shelf, 9 of which have low salt content. What is the probability that both bags you select will have low salt content?

To answer this question, we first define the following events:

Event A = The first bag has low salt content Event B = The second bag has low salt content

The two events are dependent because if the first bag selected is removed from the shelf, the sample space for selecting the second is reduced from 32 to 31 bags, which affects the prob- abilities. We can use the multiplication rule shown in Equation 4.10 to determine the prob- ability that both bags will be low on salt, or P(A and B). First, recognize that the probability of Event A occurring is determined as follows:

P1A2 = 9 32

= 0.281

The conditional probability P1B � A2 refers to the probability that the second bag is low on salt given that the first bag selected was also low on salt. Because the first bag selected was not replaced on the shelf, there is one less low-salt bag among the remaining 31 bags. This means that

P1B � A2 = 8 31

= 0.258

Therefore, using the multiplication rule, we can tell that the joint probability that both bags selected will be low salt is as follows:

P1A and B2 = P1A2P1B � A2 P1A and B2 = 10.281210.2582 = 0.072

There is slightly more than a 7% chance that you will leave the grocery store with two bags low on salt content.

As we have explained, Events A and B are independent if P1B � A2 = P1B2. When two events are independent, the probability of them both occurring is simply the product of their individual probabilities of occurring. In this case, the multiplication rule can be simplified as shown in Equation 4.11.

Only 8 of the remaining 31 bags have low salt because the first bag selected was also low salt.

For example, the Club Bistro restaurant in Wilmington, Delaware, has observed that a customer will order the Chef’s Special for the evening 18% of the time. Assuming that cus- tomer orders are independent of one another, we can use Equation 4.11 to calculate the joint probability of the following two events occurring:

Event A = The first customer orders the Chef’s Special Event B = The second customer orders the Chef’s Special

P1A and B2 = P1A2P1B2 P1A and B2 = 10.18210.182 = 0.032

In case you were wondering, the multiplication rule for independent events is not limited to only two events. It can be used to calculate any number of independent events, as Equation 4.12 shows:

Independence means that the probability of the second person ordering the Chef’s Special is the same regardless of whether the first person ordered the Chef’s Special.

Formula 4.11 for The Multiplication Rule for Two Independent Events

P1A and B2 = P1A2P1B2

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 167

To demonstrate this rule, suppose you walk into class and your teacher surprises you with a five-question true/false pop quiz. If you randomly guess at all five answers, what is the probability you will answer all of them correctly? I’ll define the following events:

A1 = Answering question 1 correctly A2, A3, A4, and A5 = Answering questions 2, 3, 4, and 5 correctly

Because you are randomly guessing (shame on you!), you have a 50% chance of Event A1 occurring. You have the same chance (50%) of answering each of the other four questions right. Therefore,

P1of 5 correct answers2 = P1A12P1A22P1A32P1A42P1A52 P1of 5 correct answers2 = 10.5210.5210.5210.5210.52 = 0.031

You have a 3% chance of correctly guessing at all five true/false answers. As you can see, this is not a great test-taking strategy!

The multiplication rule is very helpful in a business setting because people’s intuition often leads them astray when they are assessing probabilities. To illustrate, consider the fol- lowing two statements:

Statement 1: You will complete all of the goals that were assigned to you by your boss this year.

Statement 2: You will complete all of the goals that were assigned to you by your boss this past year, and you will receive an above-average salary increase based on your performance.

Which statement has the higher probability of occurring? One research study found that most people selected choices similar to Statement 2 as being more likely. Including the above-average salary increase apparently gives Statement 2 more credibility. But now that you are well versed in the multiplication rule, I hope you see that this is not possible. To prove my point, let’s define the following events:

Event A: You will complete all of the goals that were assigned to you by your boss this year.

Event B: You will receive an above-average salary increase based on your performance.

Statement 2 corresponds to P(A and B), which, according to the multiplication rule, equals P1A2P1B � A2. Mathematically, it is impossible for P(A and B) 7 P(A), which means Statement 2 could never be more likely than Statement 1, no matter how convincing your intuition happens to be. Be sure to give the next Your Turn problem a try to make sure the multiplication rule makes sense to you.

Statement 1 = Event A occurring, while Statement 2 = Event (A and B) occurring.

1. Draw two cards from a standard 52-card deck without replacing the first card. What is the prob- ability that the first card will be a red queen and the second card will be a red card?

2. Draw two cards from a standard 52-card deck, replacing the first card after you draw it. What is the probability that the first card will be a red queen and the second card will be a red card?

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Contingency tables with probabilities So far in the chapter, we have used contingency tables to show the frequency of events. We will now convert those frequencies into probabilities. We do so by dividing each number in the table by the total number of observations. Let’s say Club Bistro has two types of entrée,

Formula 4.12 for The Multiplication Rule for n Independent Events

P1A1 and A2 and cand An2 = P1A12P1A22cP1An2

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168 C H A P T E R 4 | Introduction to Probabilities

{ table 4.12 | Frequency of Customer Satisfaction by Entrée Type

response meat fish total

Satisfied 70 100 170

Not Satisfied 10 20 30

Total 80 120 200

{ table 4.13 | Probability of Customer Satisfaction by Entrée Type

response meat fish total

Satisfied 70>200 = 0.35 100>200 = 0.50 170>200 = 0.85 Not Satisfied 10>200 = 0.05 20>200 = 0.10 30>200 = 0.15 Total 80>200 = 0.40 120>200 = 0.60 200>200 = 1.00

{ figure 4.8 A Summary of Customer

Satisfaction Probabilities by Entrée Type

Response Meat Fish Total

0.35 0.50 0.85

0.05 0.10 0.15

Total 0.40 0.60 1.00

Joint Probabilities Marginal Probabilities

Conditional Probabilities

0.35 0.40

0.50 0.60

0.05 0.40

0.10

0.875

0.833

0.125

0.167 0.60

= =

Satis�ed (A) Not Satis�ed (B)

(C ) (D)

P(A and C ) = 0.35 P(A and D) = 0.50

P(B and C ) = 0.05 P(B and D) = 0.10

P(A) = 0.85 P(B) = 0.15

P(C) = 0.40 P(D) = 0.60

P(A|C)

P(A|D)

P(B|C )

P(B|D)

P(A and C ) P(C )

P(A and D) P(D)

P(B and C ) P(C )

P(B and D) P(D)

meat and fish. Customers are individually asked after their meals if they were satisfied with them. Table 4.12 shows the contingency table for the past 200 customers.

What can we conclude about Table 4.13? Figure 4.8 summarizes the various types of probabilities discussed in this chapter using the following events:

Event A = Customer is satisfied with the meal Event B = Customer is not satisfied with the meal Event C = Customer orders a meat entrée Event D = Customer orders a fish entrée

Table 4.13 converts each frequency into a probability by dividing each number in Table 4.12 by 200, the total number of customers.

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 169

From Figure 4.8, we can conclude that the probability of a customer being satisfied with his or her meal is P1A2 = 0.85. The probability that a customer is satisfied with his or her meal and had a meat entrée is P1A and C2 = 0.35. (This is one of the joint probabilities shown in Figure 4.8.) Finally, the probability that a customer is satisfied with his or her meal, given he or she had a meat entrée, is P1A� C2 = 0.875. (This is one of the conditional probabilities shown in Figure 4.8.) We can also display the joint probabilities from Figure 4.8 using the decision tree shown in Figure 4.9.

The first two branches on the left side of the tree represent the events that a customer is satisfied with his or her meal (A) or is not satisfied with his or her meal (B). The four branches on the right side of the tree represent the joint probabilities for all the combina- tions of satisfaction and lack of satisfaction by entrée. Summing these four probabilities 10.35 + 0.50 + 0.05 + 0.102 results in 1.0, which represents the entire sample space. Note that decision trees can also be used to display marginal probabilities.

In Figure 4.9, I chose to start the tree branches on the left side with the Satisfied/Not Satisfied events. I could have just as easily chosen the Meat/Fish events on the left side of the decision tree. Either tree structure would be acceptable.

Mutually exclusive and Independent events Earlier in the chapter we discussed mutually exclusive events being events that could not occur at the same time. Using our Club Bistro example, Events A and B would be mutually exclusive because a customer could not be satisfied and not satisfied at once. Also, Events C and D are mutually exclusive because we are assuming that an entrée must be either a meat or a fish dish. Therefore, we can state the following:

P1A and B2 = 0 P1C and D2 = 0

I have one final word on mutually exclusive and independent events. Two events cannot be both independent and mutually exclusive. Think about it: Independent events can occur at the same time. One the other hand, by definition, mutually exclusive events can’t occur at the same time. Let’s look at an example:

Consider Event A (the customer is satisfied) and Event B (the customer is not satisfied.) These two events are mutually exclusive because the customer cannot be satisfied and not satisfied at the same time. As a result, these events cannot be independent and are therefore dependent on one another based on the following logic.

We know from Figure 4.3 that P1A2 = 0.85. If Events A and B are independent, then P1A� B2 should also equal 0.85. But if we are told that Event B has occurred (the customer is not satisfied), then the probability of Event A occurring (the customer is satisfied) cannot be 0.85. Instead, it must equal 0 because there is no chance the customer can also be satis- fied. Therefore, these mutually exclusive events must be dependent because

P1A� B2 ≠ P1A2 Remember, Events A and B are independent if P1A� B2 = P1A2.

{ figure 4.9 Decision Tree for Club Bistro

Satis�ed (A) P(A) = 0.85

Not Satis�ed (B) P(B) = 0.15

P(A and C ) = 0.35

P(A and D) = 0.50

P(B and C ) = 0.05

P(B and D) = 0.10

Meat (C )

Fish (D)

Meat (C )

Fish (D)

These four probabilities add up to 1.0 because they represent the entire sample space 10.35 + 0.50 + 0.05 + 0.10 = 1.02.

Decision trees are used to display marginal and joint probabilities from a contingency table.

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170 C h a p t e r 4 | Introduction to Probabilities

This argument holds true for any two mutually exclusive events. I do have one word of caution here. It is not accurate to describe the terms mutually exclusive and independent as opposites of one another. In other words, you cannot conclude that because events are not independent they will be mutually exclusive.

In this next Your Turn section, you’ll get the chance to use a contingency table and deci- sion tree to help one of Deb’s favorite stores, Costco, evaluate the spending patterns of its members.

Costco is a warehouse store that requires its customers to have a membership to purchase items. Costco offers nonbusiness customers two types of membership. Gold Star members pay an annual fee of $50, whereas Executive members pay a $100 annual fee but also earn a 2% reward on most Costco purchases. The following contingency table shows the number of customers who had check-out totals above and below $200 along with their corresponding membership type.

Membership

Check-Out Total Gold Executive

$200 or less 66 21

More than $200 15 48

Set up a contingency table similar to that in Figure 4.8 showing the following:

a. Marginal probabilities b. Joint probabilities c. Conditional probabilities d. Display these probabilities in a decision tree using check-out total as the starting branches. e. What conclusions can be made concerning the spending patterns of each membership type?

Y o U r t U r n # 1 0N

Answers can be found on u page 191

Bayes’ theorem Thomas Bayes (1701–1761) developed a mathematical rule that calculates P1A� B2 from information about P1B� A2. In other words, if we know something about the probability of Event B occurring, given that Event A occurred, we can calculate the reverse—that is, we can calculate the probability of Event A occurring, given that Event B occurs. Bayes’ Theorem has many practical applications, such as determining the probability of discovering a terrorist with airport security measures in place and evaluating the effectiveness of drug testing athletes.

Before we go any further, it’s very important to convince yourself that the two condi- tional probabilities we just discussed are not identical. I’ll use the following example to ex- plain this difference and to demonstrate Bayes’ Theorem.

I want to play golf tomorrow, but the local weather forecast calls for rain. Let’s define the following events:

Event A1 = It rains tomorrow Event A2 = It does not rain tomorrow Event B = Rain is predicted in tomorrow’s weather forecast

Events A1 and A2 are mutually exclusive because if one event occurs, the other cannot. They are also collectively exhaustive because one of them must occur tomorrow. I am interested in discovering the probability that it will rain tomorrow, given the forecast calls for rain. This is represented as the conditional probability P1A1 � B2.

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 171

To answer this question, I gathered the following information. The average number of days per year that rain is recorded in my hometown is 91. Therefore, the simple probability that rain will occur on any given day is as follows:

P1A12 = 91

365 = 0.25

Because Events A1 and A2 are mutually exclusive and collectively exhaustive, the prob- ability that it will not rain on any given day is calculated as follows:

P1A22 = 1 - 0.25 = 0.75 So far, though, we only know the probability of Events A1 and A2 occurring. What we

need to know is the probability of rain being predicted for tomorrow (Event B). Otherwise, we won’t be able to determine P1A1 � B2, which is the probability of it raining, given that rain is in the forecast.

To determine the probability of Event B—the probability of rain being predicted— I counted the number of days that it actually rained in the past year and checked the historical forecast for those days. I discovered that on 85% of the days that it actually rained, rain had been predicted. Based on this fact, I set the following conditional probability:

P1B� A12 = 0.85 P1B� A12 is the probability that rain will be in tomorrow’s forecast, given that it actually does rain tomorrow.

Notice the important distinction between the following two conditional probabilities:

P1A1 � B2 = The probability that it will rain tomorrow, given rain is in the forecast. P1B� A12 = The probability that rain will be in tomorrow’s forecast, given that it

actually does rain tomorrow.

Even though these two statements sound similar, their meanings are very different. Finally, I identified the number of days that it did not rain in the past year and checked

the historical forecast for those days. I discovered that, on 20% of the days that it did not rain, rain had been in the forecast. Based on this fact, I set the following conditional probability:

P1B� A22 = 0.20 P1B� A22 is the probability that rain will be in tomorrow’s forecast, given that it doesn’t rain tomorrow. Now I’m ready to call on Bayes’ Theorem, shown in Equation 4.13, to answer my question: What is the probability that it will rain tomorrow, given the forecast calls for rain?

The probabilities P1A12 and P1A22 are known as prior probabilities because they are determined without any other information, such as a weather forecast.

We just determined that this probability is 0.85.

This probability is what we ultimately want to determine.

Pay careful attention to the distinction between P1A1 � B2 and P1B� A12. Some of my students really struggle with this distinction.

Formula 4.13 for the Bayes’ Theorem

P1Ai� B2 = P1Ai2P1B� Ai2

P1A12P1B� A12 + P1A22P1B� A22 + g+ P1An2P1B� An2

where

Ai = The ith event of interest from a choice of n events B = An event that has already occurred In our weather example, i = 1 because we are interested in determining P1A1 � B2, and

n = 2 because the events A1 and A2 are collectively exhaustive. Carefully plug in the prob- abilities that we have acquired into Equation 4.13:

P1A1 � B2 = P1A12P1B� A12

P1A12P1B� A12 + P1A22P1B� A22 P1A1 � B2 =

10.25210.852 10.25210.852 + 10.75210.202

P1A1 � B2 = 0.2125

0.2125 + 0.15 =

0.2125

0.3625 = 0.586

It must either rain or not rain tomorrow.

Bayes’ Theorem is used to update prior probabilities with new information.

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172 C h a p t e r 4 | Introduction to Probabilities

Given that tomorrow’s forecast calls for rain, there is a 58.6% chance that it will actu- ally rain tomorrow. This probability is known as a revised, or posterior, probability. Without any weather forecast, the prior probability of rain was 25%. Now that we have a weather forecast calling for rain, we have revised this probability to 58.6%. Oh well, I better prepare myself for a soggy day on the course.

If using Equation 4.13 makes you squeamish, the revised probabilities that Bayes’ Theorem provides can also be derived in a table, as Table 4.14 shows. Columns 2 and 3 rep- resent the prior and conditional probabilities with which we started. These two columns are multiplied together to calculate the joint probabilities shown in Column 4.

When Column 4 is summed, we have the probability that Event B has occurred, which in this example is the probability that rain is in tomorrow’s forecast (36.25%).

Column 5 shows our final result, the revised probabilities. Note that this calculation is the same as Equation 4.6, which determined conditional probability earlier in the chapter.

P1A� B2 = P1A and B2 P1B2

This table summarizes the calculations for Equation 4.13.

{ table 4.15 | Credit Scores of PNC Bank Customers

event credit score percentage p 1 ai2 A1 Less than 600 25 0.25

A2 600 to 700 45 0.45

A3 More than 700 30 0.30

{ table 4.14 | An Example of Bayes’ Theorem in Table Form

(1) (2) (3) (4) (5)

event

prior prob

conditional prob

joint prob

revised prob

Ai P1Ai2 P1B� Ai2 P1Ai and B2 P1Ai� B2 A1 0.25 0.85 0.2125 0.2125>0.3625 = 0.586 A2 0.75 0.20 0.1500 0.1500>0.3625 = 0.414

P1B2 = 0.3625 Total = 1.000

Let’s look at the following example to see what happens to Bayes’ Theorem when we have more than two collectively exhaustive events. PNC Bank classifies its credit card customers into three categories by their credit scores. Table 4.15 shows the percentage of customers who fall into each group.

We’ll define Event B as follows:

Event B = Customer is late on his or her current credit card payment

We also know the following historical information:

• Customers with credit scores of less than 600 are late with their payments 40% of the time, or P1B� A12 = 0.40.

• Customers with credit scores between 600 and 700 are late with their payments 10% of the time, or P1B� A22 = 0.10. Customers with credit scores of more than 700 are late with their payments 5% of the time, or P1B� A32 = 0.05.

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 173

Suppose PNC Bank wants to use Bayes’ Theorem to determine the probability that a customer has a credit score of less than 600 given that he or she is late on his or her current credit card payment, or P1A1 � B2.

P1A1 � B2 = P1A12P1B� A12

P1A12P1B� A12 + P1A22P1B� A22 + P1A32P1B� A32 P1A1 � B2 =

10.25210.402 10.25210.402 + 10.45210.102 + 10.30210.052

P1A1 � B2 = 0.10

0.10 + 0.045 + 0.015 =

0.10

0.16 = 0.625

There is a 62.5% chance that a late payment is from a customer with a credit score of less than 600. All the revised probabilities for this example are shown in Table 4.16.

{ table 4.16 | Bayes’ Theorem in Table Form

(1) (2) (3) (4) (5)

event

prior prob

conditional prob

joint prob

revised prob

Ai P1Ai2 P1B� Ai2 P1Ai and B2 P1Ai� B2 A1 0.25 0.40 0.100 0.100>0.16 = 0.625 A2 0.45 0.10 0.045 0.045>0.16 = 0.281 A3 0.30 0.05 0.015 0.015>0.16 = 0.094

P1B2 = 0.160 Total = 1.000 The answer to PNC Bank’s question can also be found here in the table.

Now that you’re a Bayes’ Theorem expert, show off your newfound skills by complet- ing the following Your Turn problem.

Suppose that there is a 90% chance that a test for performance-enhancing drugs will provide a positive test for an athlete who actually took them. In addition, there is a 15% chance that the same test will provide a positive test for an athlete who did not take them. (This is known as a false positive.) Assume that 8% of the athlete population is currently taking performance-enhancing drugs. Using Bayes’ Theorem, determine the probability that a randomly selected athlete who tests positive is actually a user of performance-enhancing drugs.

Y o U r t U r n # 1 1N

Answer can be found on u page 192

4.2 Section Problems Basic Skills Use the following contingency table to answer questions from Problems 4.10 to 4.14.

Event A Event B

Event C 9 6

Event D 4 21

Event E 7 3

4.10 Determine the following probabilities:

a. P1A2 b. P1B2 c. P1C2 d. P1D2 e. P1E2

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174 C h a p t e r 4 | Introduction to Probabilities

4.11 Determine the following probabilities:

a. P1A and C2 b. P1A and D2 c. P1B and E2 d. P1A and B2

4.12 Determine the following probabilities:

a. P1B or C2 b. P1B or D2 c. P1A or E2 d. P1A or B2

4.13 Determine the following probabilities:

a. P1A� C2 b. P1C� A2 c. P1B� E2 d. P1E� B2

4.14 Use Bayes’ Theorem to determine P1B� D2.

applications 4.15 Consider the following experiment that involves rolling a

pair of dice:

Event A = Rolling a nine

Event B = Rolling a pair (both dice the same)

a. Are these two events mutually exclusive? Why or why not?

b. Are these two events independent? Why or why not?

4.16 Consider the following experiment that involves flipping a single coin and rolling one six-sided die:

Event A = Coin comes up tails

Event B = Rolling a four with the die

a. Are these two events mutually exclusive? Why or why not?

b. Are these two events independent? Why or why not?

4.17 Consider an experiment with a standard 52-card deck from which one card is randomly selected and not re- placed. Then a second card is randomly selected. Define the following events:

Event A = The first card is a heart

Event B = The second card is a heart

a. Are these two events mutually exclusive? Why or why not?

b. Are these two events independent? Why or why not?

4.18 A recent survey found that 75% of households had Internet access and 80% of households had cable television. Also, it was reported that 70% of the households in the survey had both Internet and cable television. Determine the probability that a randomly selected household in the survey had either Internet access or cable.

4.19 A local car dealership currently has 36 used GM, Ford, and Toyota vehicles on the lot that can be classified as either cars or trucks. The following data are available:

• Twenty-six vehicles are cars. • Eleven vehicles are GMs. • Fifteen vehicles are Fords. • Three vehicles are both Toyotas and trucks. • Fourteen vehicles are both Fords and cars.

a. What is the probability that a randomly selected vehicle is a Toyota?

b. What is the probability that a randomly selected vehicle is a truck?

c. What is the probability that a randomly selected vehicle is a either a Ford or a car?

d. What is the probability that a randomly selected vehicle is a GM truck?

e. What is the probability that a randomly selected vehicle is a Toyota, given it is a car?

f. What is the probability that a randomly selected vehicle is a truck, given it is a Ford?

g. Construct a decision tree for these events.

4.20 The following table shows the frequencies that the Green Bay Packers of the NFL called a running play or a passing play for various down and distance situations for the first five games of a recent season. (Yes, I actually counted these frequencies myself.)

Down and Distance Run Pass

First down and 10 or more yards to go 55 81

Second down and 6 or more yards to go 27 52

Second down and 5 or less yards to go 18 10

Third down and 2 or more yards to go 3 60

Third down and 1 yard to go 1 2

What is the probability that the Packers will do the following:

a. Pass the ball on any given down b. Run the ball on second down c. Pass the ball, given that it is third down and they have

1 yard to go d. Run the ball on first down e. Run the ball, given that it is second down and they

have 6 or more yards to go f. Construct a decision tree for these events.

4.21 At a local restaurant, 22% of the customers order take- out. If 8% of the customers order takeout and choose a hamburger, determine the probability that a customer who orders takeout will order a hamburger.

4.22 A recent Wall Street Journal poll found that 70% of Americans felt the U.S. government was not working well. If seven people were randomly selected, what is the probability that all seven felt the government is not work- ing well?

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C h a p t e r 4 | 4.2 Probability Rules for More Than One Event 175

4.23 The following table shows the number of people, in thou- sands, in the United States with and without health insur- ance in 2011, according to the U.S. Census Bureau:

Age Group Insured Uninsured

Under 18 years 66,835 6,902

18 to 24 years 22,491 7,649

25 to 34 years 29,690 11,529

35 to 44 years 31,588 8,399

45 to 64 years 68,544 13,382

65 years and older 40,817 690

a. What percentage of the U.S. population in 2011 did not have health insurance and was between the ages of 18 to 24 years?

b. What percentage of the U.S. population in 2011 did not have health insurance or was between the ages of 25 to 34 years?

c. What percentage of the U.S. population in 2011 did have health insurance, given they were 65 years or older?

d. Do age group and health insurance appear to be indepen- dent or dependent events? Define Event A as a person 45 to 64 years old and Event B as the person insured. What conclusions can be drawn with this information?

Based on: U.S. Census Bureau data.

4.24 Thirty-five percent of customers who purchased products from an e-commerce site had orders exceeding $100. If 20% of the customers have orders exceeding $100 and also pay with the e-commerce site’s sponsored credit card, determine the probability that a customer whose order exceeds $100 will pay with the sponsored credit card.

4.25 A single card is drawn from a standard 52-card deck. What is the probability that the card is either

a. A king, or a four, or a five b. A diamond, or a heart, or club c. A two, or a three, or a club d. A six, or a seven, or a spade, or a diamond

4.26 In 2012, 37% of taxpayers who filed their tax return elec- tronically self-prepared their taxes. If the IRS randomly selected three tax returns submitted electronically, what is the probability that all three were self-prepared?

4.27 The following table shows the number of men and women, in thousands, enrolled in degree-granting institu- tions in the United States in 2010. The men and women are categorized by their age groups.

Age Group Men Women

14 to 17 years old 92 19

18 and 19 years old 1,842 2,276

20 and 21 years old 1,931 2,120

22 to 24 years old 1,666 2,008

25 to 29 years old 1,414 1,783

30 to 34 years old 745 1,078

35 years old and over 1,354 2,587

Based on: National Center for Education Statistics

Determine the probability that a randomly selected student

a. Was a female b. Was 22 to 24 years old c. Was a woman who was 35 years old or older d. Was either a women or 25 to 29 years old e. Was a man, given that the student was 22 to 24 years old f. Was 30 to 34 years old, given that the student was a man

4.28 A hotel asked its customers to rate their stays as Excellent, Good, Fair, or Poor. The following table shows the fre- quency of each rating as well as the gender of the customer.

Rating Men Women

Excellent 3 9

Good 20 12

Fair 16 7

Poor 6 2

Determine the probability that a randomly selected customer

a. Rated his or her stay as Excellent b. Rated his or her stay as either Good or Fair c. Rated his or her stay as other than Poor d. Was a man who rated his stay as Good e. Either was a woman or rated his or her stay as Fair f. Rated her stay as Good, given the customer was a woman g. Was a woman, given the person rated the stay as Poor h. Construct a decision tree for these events.

4.29 Super D is a major video distributor that purchases blank DVDs from two sources: Disk Makers, which provides 62% of the DVDs, and Media Supply, which provides the remaining 38%. Historically, 3% of the DVDs Super D has purchased from Disk Makers have been defective, whereas 2% of the DVDs purchased from Media Supply have been defective. A randomly selected DVD is found to be defec- tive. Use Bayes’ Theorem to determine the probability that the DVD came from Disk Makers.

4.30 A sporting goods company operates retail stores in the New York and Atlanta areas. Customers were asked to rate their shopping experiences, the results of which are sum- marized in the following table:

Shopping Experience Percentage

Poor 15

Average 45

Good 40

Consider the following information:

• 67% of the customers who rated their store Poor came from the Philadelphia area.

• 36% of the customers who rated their store Average came from the Philadelphia area.

• 30% of the customers who rated their store Good came from the Philadelphia area.

Use Bayes’ Theorem to determine the probability that a customer who shopped in the Philadelphia area rated his or her experience as Good.

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176 C h a p t e r 4 | Introduction to Probabilities

4.3 Counting Principles To use classical probability, which we introduced at the beginning of this chapter, we need to be able to count the number of events of interest along with the total number of events that are possible in the sample space. For simple events, like rolling a single die, the number of possible outcomes (six) is obvious. But for more complex events, like a state lottery drawing, we need to rely on techniques known as counting principles to arrive at the correct answer. Let’s look at these techniques.

the Fundamental Counting principle After a tough round of golf on a hot afternoon, Brian, John, and I decide to revive our spirits at the ice cream store on the way home. I’m desperately trying to decide from among four flavors and three toppings in which to indulge. Let’s have

k1 = The number of ice cream flavors I’m trying to decide from among

and

k2 = The number of topping flavors I’m trying to decide from among

How many different combinations of ice cream and toppings am I faced with, assuming I can choose only one flavor and one topping? (After all, I am watching my weight.) Fortu- nately, I can use the fundamental counting principle, which tells me that if one event (my ice cream choice) can occur in k1 ways and a second event (my topping choice) can occur in k2 ways, the total number of ways both events can occur together is 1k121k22 = 142132 = 12 ways.

Now I can extend this principle to more than two events. In addition to flavors and toppings, I have another tempting choice: A choice between a small and large serv- ing. That leaves me with a mind-boggling number of combinations to choose from: 1k121k221k32 = 142132122 = 24. The combinations are summarized in Table 4.17.

The fundamental counting principle states that if there are k1 choices for the first event, k2 choices for the second event … and kn choices for the nth event, then the total number of possible outcomes is (k1)(k2)(k3)

. . . (kn).

{ table 4.17 | Ice Cream Combinations

ice cream flavors toppings size

CH = Chocolate HF = Hot Fudge LG = Large

VA = Vanilla BS = Butterscotch SM = Small

ST = Strawberry SP = Sprinkles

CF = Coffee

list of combinations (by flavor, topping, and size)

CH-HF-LG VA-HF-LG ST-HF-LG CF-HF-LG

CH-HF-SM VA-HF-SM ST-HF-SM CF-HF-SM

CH-BS-LG VA-BS-LG ST-BS-LG CF-BS-LG

CH-BS-SM VA-BS-SM ST-BS-SM CF-BS-SM

CH-SP-LG VA-SP-LG ST-SP-LG CF-SP-LG

CH-SP-SM VA-SP-SM ST-SP-SM CF-SP-SM

In general, the fundamental counting principle tells me that if there are k1 choices for the first event, k2 choices for the second event, and kn choices for the nth event, then the total number of possible outcomes is as shown in Equation 4.14.

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C h a p t e r 4 | 4.3 Counting Principles 177

where

ki = The number of choices for the ith event n = The number of events Another demonstration of the fundamental counting principle is to calculate the number

of unique combinations for a state’s automobile license plates. Suppose the state plates have three letters followed by four numbers. The state doesn’t allow the number zero and the let- ter O to be used because their resemblance to one another can cause confusion. Because we have 25 possible letters (all of the numbers of the alphabet except O) and 9 possible numbers (1–9), the total number of unique combinations is as follows:

FIRST LETTER

SECOND LETTER

THIRD LETTER

FIRST NUMBER

SECOND NUMBER

THIRD NUMBER

FOURTH NUMBER

25 25 25 9 9 9 9

25 * 25 * 25 * 9 * 9 * 9 * 9 = 102,515,625

That’s 102,515,625 possible license plates.

This number represents k1 = 25 because there are 25 possible choices for the first letter in the license plate.

Formula 4.14 for the Fundamental Counting Rule

1k121k221k32 g 1kn2

Stats in practice: running Out of Internet addresses

Every device that communicates on the Internet, whether it be a desktop computer or a smartphone, needs a unique IP address so that it can be identified. Think of the IP address as basically a postal address for your Internet device.

Those who designed the Internet back in the 1970s used the Internet Protocol Addressing Scheme version 4 (IPv4). Because computers speak a binary language, they only recognize zeros and ones. The IPv4 address format that the Internet was estab- lished on consists of a string of 32 zero/one digits. An example of an IPv4 address is as follows:

00001101.10110000.11110011.00110001

Because each digit must be a zero or one, we can use the Fundamental Counting Principle to find that there are 232 combinations or a total of

232 = 4,294,967,296

unique IP addresses available. Years ago, the founders of the Internet figured this would be an ample number of addresses. Today we know differently.

With the huge growth in devices that depend on an Internet connection over the past few years such as Apple’s iPhone and iPad, it was projected that we would run out of IP addresses with the IPv4 format in 2012. Another contributing factor to the shortage of IP addresses is the rapid increase in the demand for technology in developing countries like China and India. Lack of

IP address have many consequences; Internet connections will slow down and computers will have difficulty communicating with each other.

To address this shortage on IP addresses (pun intended), a new protocol system, known as IPv6, has been developed. This new format allows for a combination of 128 zero/one digits, as seen in the following example:

00001101.10110000.11110011.00110001. 00110000.10000001.00011000.11100011.

10100101.01101010.11001010.00001001. 10111001.11101101.01111001.01010111

Using the IPv6 format, there are a total of 2128 combinations, which provides

2128 = 340,282,366,920,938,000,000,000,000,000,000,000,000

unique IP addresses. To facilitate the transition from IPv4 to IPv6, the Internet community organized World IPv6 Launch Day on June 6, 2012. The goal of this event was to obtain a commit- ment from the Internet industry, such as equipment manufactur- ers and ISPs, to hasten the conversion to IPv6.

When I told my daughter about IPv6, her first reaction was, “They’ll never run out of all of those addresses.”

“Yeah, that’s what they said in the 1970s,” I replied.

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178 C h a p t e r 4 | Introduction to Probabilities

permutations Permutations are the number of different ways in which different objects can be arranged in or- der. In a permutation, there can be only one of each arrangement of the objects. As an example, there are six permutations for the numbers 1, 2, and 3. The permutations are as follows:

123 132 213 231 312 321

But what if we had many numbers instead of just three? How many permutations would there be? Finding all of the permutations by manually writing them down as we just did would be time consuming. Instead, we can use Equation 4.15. By definition, the number of permutations of n distinct objects is n!

Permutations are the number of different ways in which objects can be arranged in order. The number of permutations of n objects selected x at a time can be found by

nPx = n!

1n - x2!.

Let’s use our previous example to work out Equation 4.15. The number of permutations of the three distinct objects in our example is 3! Now you can plug 3! into Equation 4.15 to find the number of permutations more easily.

Using the n! permutation rule, we find the following:

3! = 132122112 = 6 There are six permutations for the numbers 1, 2, and 3. This is the same result we got by arranging the numbers manually.

Now suppose we have six numbers instead of three. Using Equation 4.15, we find that the number of permutations associated with six numbers is as follows:

6! = 162152142132122112 = 720 Let’s work another problem. Before the beginning of a professional basketball game, the

starting five players are announced one at a time. How many different ways can we arrange the order in which the players are announced? The number of permutations is calculated as follows:

5! = 152142132122112 = 120 Suppose we want to select only some of the objects in the group. If there are 12 players

on the team, how many different ways can any five players on the team be announced at the start of the game?

We can find the number of permutations for this problem using Equation 4.16.

By definition, 0! = 1.

Formula 4.15 for the Permutations of n Distinct Objects

n! = n1n - 121n - 221n - 32 g122112

Formula 4.16 for the Permutations of n Objects Selected x at a Time

nPx = n!

1n - x2!

where

n = The total number of objects x = The number of objects to be selected Using our basketball example again, if there are 12 players on the team, how many dif-

ferent ways can any 5 players on the team be announced to start the game? In this case, be- cause n = 12 and x = 5, the number of permutations is as follows:

12P5 = 12!

112 - 52! = 112211121102192182172162152142132122112

172162152142132122112 12P5 = 112211121102192182 = 95,040

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C h a p t e r 4 | 4.3 Counting Principles 179

Formula 4.17 for the Permutations of n Objects Selected x at a Time

nPx = n!

1n - x2! = n1n - 121n - 22 g1n - x + 12

I’m sure glad it’s not my job to decide the order of the announcements. It’s easier to calculate the number of permutations using Equation 4.17, which is alge-

braically the same as Equation 4.16.

I recommend using this equation to calculate permutations. It requires less effort than Equation 4.16.

This works because every value in the denominator (the bottom of the fraction) will cancel out with many values in the numerator (the top of the fraction). I will show you why with the following demonstration. Suppose we have 10 objects 1n = 102, and we are removing 2 at a time 1x = 22. How many permutations will we have?

Recall that Equation 4.16 is as follows:

nPx = n!

1n - x2! Plugging our numbers into 4.16 we get the following:

10P2 = 10!

110 - 22! = 1102192182172162152142132122112

182172162152142132122112 = 1102192 = 90

Equation 4.17 is as follows:

nPx = n!

1n - x2! = n1n - 121n - 22 g 1n - x + 12

Plugging our numbers into Equation 4.17, we get the following:

1n - x + 12 = 10 - 2 + 1 = 9 10P2 =

1n - x2! = 1102192 = 90

As you can see, Equation 4.17 has fewer calculations. Sometimes the order of events is not of consequence. We’ll discuss those cases in the

next section.

Combinations Combinations are similar to permutations, except that the order of the objects is not important. The number of combinations of n objects selected x at a time can be found using Equation 4.18.

Equation 4.17 is easier to use than Equation 4.16 because it cancels out the numbers 1 through 8 in the numerator and denominator that are present when using Equation 4.16. We are left with x number of factors, which in this case is two (10 and 9).

Combinations are the number of different ways in which objects can be arranged without regard to order. The number of combinations of n objects selected x at a

time can be found by nCx = n!

1n - x2!x!. Formula 4.18 for the Combinations of n Objects Selected x at a Time

nCx = n!

1n - x2!x!

where

n = The total number of objects x = The number of objects to be selected As with permutations, it’s easier to calculate the number of combinations using

Equation 4.19, which is algebraically the same as Equation 4.18.

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180 C h a p t e r 4 | Introduction to Probabilities

For example, in poker, 5 cards are selected randomly from a deck of 52 cards. How many five-card combinations exist?

52C5 = 52!

152 - 52!5! = 15221512150214921482

152142132122112 = 2,598,960

How many five-card permutations exist?

52P5 = 52!

152 - 52! = 15221512150214921482 = 311,875,200

There are more five-card permutations than combinations. For example, the 2 five-card groups (or hands) hands shown below have the same combination of cards but are consid- ered two different permutations. Recall that the order matters with permutations. By con- trast, with combinations the order doesn’t matter. Consequently, the two hands would be counted as only one hand, or combination, because they consist of the same cards, only in a different order:

When the order of objects is important, use permutations. When the order of objects does not matter, like a poker hand, use combinations.

Formula 4.19 for the Combinations of n Objects Selected x at a Time

nCx = n!

1n - x2!x! = n1n - 121n - 22 g 1n - x + 12

HAND 1 HAND 2

Ace of spades Ace of spades

Queen of hearts Ten of spades

Ten of spades Queen of hearts

Ten of diamonds Ten of diamonds

Three of clubs Three of clubs

Now that we know the total number of five-card combinations from a 52-card deck, we can calculate the probability of a flush, which is any five cards that are all the same suit (spades, clubs, hearts, or diamonds). For you poker veterans, I am including a royal flush and a straight flush in this calculation.

First, we need to count the number of five-card flushes of one suit, let’s say diamonds. Because there are 13 diamonds in the deck, the number of combinations of these 13 dia- monds, selected 5 at a time, is as follows:

13C5 = 13!

113 - 52!5! = 1132112211121102192 152142132122112 = 1,287

Because there are four suits in the deck, the total number of five-card flushes from any suit is 11,2872142 = 5,148. Therefore, the probability of being dealt a flush, including royal and straight, in a five-card hand is calculated as follows:

P1Flush2 = 5,148 2,598,960

= 0.00198

roughly twice in 1,000 hands of poker. Ready to deal? Combinations are also useful for calculating the probability of winning a state lottery

drawing. At the start of this chapter, I referred to a typical lottery game that requires you to pick 6 numbers out of a possible 49. The number of 6-number combinations from a pool of 49 numbers is as follows:

49C6 = 49!

149 - 62!6! = 149214821472146214521442

162152142132122112 = 13,983,816

Because the order of the numbers in a lottery does not matter, we use the combination rather than the permutation formula.

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C h a p t e r 4 | 4.3 Counting Principles 181

Because there are nearly 14 million different six-number combinations, the probability that your combination is the winner is as follows:

P1Winning a 6>49 Lottery2 = 1 13,983,816

= 0.00000007

With those chances of winning the lottery, you better not quit your day job just yet. Probability does not have a memory. The same six numbers selected in last week’s lot-

tery drawing have the exact same probability of being chosen again in this week’s lottery. That’s because the two drawings are independent events and have absolutely no influence on each other. Therefore, choosing a lottery number because it has not been selected recently does not increase your odds of winning. Sorry if I ruined your favorite strategy!

Here’s something you will like: Rather than deal with all those nasty factorial calcula- tions, we can let Excel figure out the number of permutations or combinations for us. The functions are as follows:

= PERMUT1n,x2 = COMBIN1n,x2

For example, if we type = PERMUT152,52 from our five-card poker example into Excel, the result will be 311,875,200. If we type = COMBIN152,52, the result will be 2,598,960.

Combinations are useful in business when a product or service is available with differ- ent options. For example, suppose a car model has five different option packages available. The manufacturer will probably be interested in knowing, for production reasons, how many variations of the model exist when customers choose two options.

To be sure you understand the counting rules we have discussed, work out the problems in the following Your Turn section.

This is one of the biggest misconceptions in probability, especially in gambling. No matter how many consecutive heads are observed when tossing a coin, the probability of the next toss being a heads is always 50%. The coin has no memory of previous tosses.

1. A restaurant has a menu with four appetizers, seven entrées, four desserts, and three drinks. If a meal consists of an appetizer, an entrée, a desert, and a drink, how many different meals can you order?

2. The NBA teams with the 13 worst records at the end of the season participate in a lottery to determine the order in which they will draft new players for the next season. How many different arrangements exist for the drafting order for these 13 teams?

3. A combination lock has a total of 40 numbers and will unlock with the proper three-number sequence. How many possible combinations exist?

4. I would like to select 3 paperback books from a list of 11 books to take on vacation. How many different sets of three books can I choose?

Y O u r t u r n # 1 2N

Answers can be found on u page 192

4.3 Section Problems Basic Skills 4.31 If the order of objects is of importance, how many ways

can I select eight objects three at a time?

4.32 If the order of objects is of importance, how many ways can I select nine objects two at a time?

4.33 If the order of objects is not of importance, how many ways can I select eight objects three at a time? Why is this result different from Problem 4.31?

4.34 If the order of objects is not of importance, how many ways can I select nine objects two at a time? Why is this result different from Problem 4.32?

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182 C h a p t e r 4 | Introduction to Probabilities

applications 4.35 I would like to customize a desktop computer online.

I have to choose among the following categories:

• processor (five choices) • hard drive (four choices) • memory (three choices) • monitor (four choices) • keyboard (two choices)

How many unique computers can I configure with these choices?

4.36 In a race with eight swimmers, how many ways can the swimmers finish first, second, and third?

4.37 A panel of 12 jurors needs to be selected from a group of 50 people. How many different juries can be selected?

4.38 What is the probability of being dealt a full house (three-of-a-kind and a pair) in five-card poker?

4.39 What is the probability of winning a 4>20 lottery game where 4 numbers are picked from a range of 20?

4.40 There are four teams in the NFL’s NFC North Division. How many different ways can the four teams finish at the end of the season in the standings?

4.41 An event planner needs to decide a seating arrangement for eight people at a table. Calculate the total number of possible arrangements available from which to choose.

4.42 A research study needs to select three people to partici- pate from a group of nine. Calculate the total number of unique groups that can be formed.

4.43 An executive needs to select 4 stores from a total of 10 to participate in a customer service program. Determine the number of possible groups of four stores.

4.44 Determine the number of ways in which a jury of 6 men and 6 women can be selected from a group of 13 men and 15 women.

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183

■ A probability is a numerical value ranging from zero to one that represents the likelihood of an event occurring.

• Classical probability refers to a situation in which we know the number of possible outcomes of the event of interest.

• Empirical probability observes the number of occurrences of an event through an experiment and calculates their probability from a relative frequency distribution.

• Subjective probability relies on people’s experience and intuition to estimate the probabilities.

■ If the sample space includes every possible event that can occur, then we say that the events are collectively exhaustive.

■ Simple probability represents the likelihood of a single event occurring.

■ The law of large numbers states that when an experiment is conducted a large number of times, the empirical probabilities of the process will converge to the classical probabilities.

■ A contingency table indicates the number of occurrences of events classified according to two variables.

• Marginal probabilities are simple probabilities found by using the total occurrences in the margins of a contingency table.

■ The intersection of Events A and B represents the number of instances in which Events A and B occur at the same time.

• The probability of the intersection of two events is known as a joint probability.

■ The union of Events A and B represents the number of instances in which either Event A or B occurs or both events occur together.

■ Two events are considered to be mutually exclusive if they cannot occur at the same time during an experiment.

■ Prior probability is the probability of Event A occurring without your knowing any additional information that could affect the event.

■ Conditional probability is the probability of Event A occurring when you know Event B has already occurred.

• Conditional probability is also known as a posterior probability.

• Two events are considered independent of one another if the occurrence of one event has no impact on the probability of the other event.

• If the occurrence of one event affected the probability of another event, the events are dependent events.

■ The multiplication rule is used to determine the probability of the intersection of two events, or P(A and B), otherwise known as the joint probability.

■ Thomas Bayes (1701–1761) developed a mathematical rule, known as Bayes’ Theorem, for calculating P1A � B2 from information about P1B � A2.

■ The fundamental counting principle states that if there are k1 choices for the first event, k2 choices for the second event, and kn choices for the nth event, then the total number of possible outcomes is 1k121k221k32 g1kn2.

■ Permutations are the number of different ways in which objects can be arranged in order.

■ Combinations are similar to permutations, except that the order of the objects is not important.

chapter 4 Summary

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chapter 4 Key Terms

Addition rule. A mathematical rule used to calculate the prob- ability of the union of events.

Bayes’ Theorem. A theorem used to calculate P1A� B2 from information about P1B� A2.

Classical probability. A probability that is determined by dividing the number of possible outcomes of an event by the total number of possible outcomes in the sample space.

Collectively exhaustive. Describes a set of events where at least one of the events must occur during an experiment.

Combinations. The number of different ways in which objects can be arranged without regard to order.

Complement. All of the outcomes in the sample space that are not part of the event of interest.

Conditional probability (of A given B). The probability of Event A occurring given the condition that Event B has already occurred.

Contingency table. A table that shows the number of occur- rences of events of an experiment classified by two variables.

Decision trees. Display marginal and joint probabilities from a contingency table.

Dependent events. Events wherein the occurrence of one event affects the occurrence of another event.

Empirical probability. A probability determined by counting the frequency of an event during an experiment.

Event. One or more outcomes of an experiment that are a subset of the sample space.

Experiment. The process of measuring or observing an activity for the purpose of collecting data.

Fundamental counting principle. A principle that states that if there are k1 choices for the first event, k2 choices for the sec- ond event, … and kn choices for the nth event, then the total number of possible outcomes is 1k121k221k32 g1kn2.

Independent events. Events that have no impact on the probabil- ity of each other occurring.

Intersection (of Events A and B). The number of instances in which Events A and B occur at the same time.

Joint probability. The probability of the intersection of two events.

Law of large numbers. A law that states that when an experi- ment is conducted a large number of times, the empirical probabilities of the process will converge to the classical probabilities.

Marginal probabilities. Simple probabilities found in the margins, or row and column totals, of a contingency table.

Multiplication rule. A rule used to determine the probability of the intersection of two events.

Mutually exclusive events. Events that cannot occur at the same time.

Permutations. The number of different ways in which objects can be arranged in order.

Posterior probability. A probability that is a revision of a prior probability using additional information.

Prior probability. The probability of an event occurring as determined without any additional information that could affect the event.

Probability. A numerical value ranging from zero to one that represents the likelihood of a specific event occurring.

Sample space. A set of all possible outcomes of an experiment. Simple event. An event with a single outcome in its most basic

form that cannot be simplified. Simple probability. The likelihood of a single event occurring. Subjective probability. A probability that is calculated based on

the experience and intuition of a person (or people). A subjec- tive probability is made when classical and empirical prob- abilities cannot be calculated.

Union (of Events A and B). All of the instances in which Event A or Event B or both occur.

4.1 An Introduction to Probabilities (pp 148–156)

chapter 4 Equations

Formula 4.1 for the Classical Probability P1A2 = Number of possible outcomes that constitute Event A Total number of possible outcomes in the sample space

Formula 4.2 for the Empirical Probability P1A2 = Frequency in which Event A occurs

Total number of observations

Formula 4.3 for The Complement Rule P1A2 + P1A′2 = 1 or P1A2 = 1 - P1A′2

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4.2 Probability Rules for More Than One Event (pp 156–175)

Formula 4.4 for The Addition Rule for the Mutually Exclusive Events P1A or B2 = P1A2 + P1B2

Formula 4.5 for The Addition Rule for the Events That Are Not Mutually Exclusive

P1A or B2 = P1A2 + P1B2 - P1A and B2

Formulas 4.6 and 4.7 for Calculating a Conditional Probability

P1A� B2 = P1A and B2 P1B2

P1B� A2 = P1A and B21A2

Formula 4.8 for Determining If Events A and B Are Independent P1A� B2 = P1A2

Formulas 4.9 and 4.10 for The Multiplication Rule for Dependent Events

P1A� B2 = P1A and B2 P1B2

P1A and B2 = P1B2P1A� B2 or

P1B� A2 = P1A and B2 P1A2

P1A and B2 = P1A2P1B� A2

Formula 4.11 for The Multiplication Rule for Two Independent Events

P1A and B2 = P1A2P1B2

Formula 4.12 for The Multiplication Rule for n Independent Events

P1A1 and A2 and g and An2 = P1A12P1A22 g P1An2

Formula 4.13 for the Bayes’ Theorem P1Ai� B2 = P1Ai2P1B� Ai2

P1A12P1B� A12 + P1A22P1B� A22 + g + P1An2P1B� An2

Formula 4.14 for the Fundamental Counting Rule 1k121k221k32 g1kn2

4.3 Counting Principles (pp 176–182)

Formula 4.15 for the Permutations of n Distinct Objects n! = n1n - 121n - 221n - 32 g 122112

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Formulas 4.16 and 4.17 for the Permutations of n Objects Selected x at a Time

nPx = n!

1n - x2! or

nPx = n!

1n - x!2 = n1n - 121n - 22 g1n - x + 12

Formulas 4.18 and 4.19 for the Combinations of n Objects Selected x at a Time

nCx = n!

1n - x2!x! or

nCx = n!

1n - x2!x! = n1n - 121n - 22 g1n - x + 12

4.45 Identify which of the following values are valid and not valid numbers for a probability:

a. 2.1 f. 110%

b. 4

4 g. -

3 c. 76% h. 1 d. 0.8 i. 0.007

e. 9

4.46 What is the probability of randomly selecting a heart from a standard 52-card deck?

4.47 Consider the experiment of rolling two dice.

a. What is the probability of obtaining a sum of eight? b. What is the probability of obtaining a sum of eight or

less? c. What is the probability of obtaining a sum of four or

more?

4.48 Indicate whether classical, empirical, or subjective prob- ability should be used to determine each of the following:

a. The probability that it will rain tomorrow b. The probability that you will roll doubles when throwing

two dice c. The probability that I will pass tomorrow’s statistics

exam d. The probability that I will select a Milky Way candy bar

out of a bowl of Halloween candy

4.49 A survey was conducted in which 150 households were asked how many working televisions they owned. The results are as follows:

Number of Televisions Number of Households

0 5

1 34

2 50

3 29

4 12

5 12

6 8

Total 150

a. What is the probability that a randomly selected house- hold has one television?

b. What is the probability that a randomly selected house- hold has more than two televisions?

c. What is the probability that a randomly selected house- hold has fewer than four televisions?

d. Is this an example of classical, empirical, or subjective probability?

4.50 A recent survey administered by ZoneDiet.com asked 1,000 adults, “How far are you from your ideal weight?” The results are shown in the following table:

Response Frequency

1–5 lbs 150

6–10 lbs 140

11–15 lbs 90

16 or more 380

I’m content ?

Total 1,000

chapter 4 Problems

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a. How many adults responded that they are content with their current weight?

b. What is the probability that a randomly selected adult is content with his or her weight?

c. What is the probability that a randomly selected adult is 1–5 lbs from his or her ideal weight?

d. What is the probability that a randomly selected adult is 6 lbs or more from his or her ideal weight?

4.51 The following table indicates the frequency of laptops sold per day at a local Best Buy store during a particular time period. During the period, there were never any days in which more than seven laptops were sold.

Daily Demand Frequency

0 7

1 11

2 26

3 13

4 15

5 5

6 8

7 5

a. What is the probability that three laptops will be sold tomorrow?

b. What is the probability that four or more laptops will be sold tomorrow?

c. What is the probability that either one or two laptops will be sold tomorrow?

d. What is the probability that less than two laptops will be sold tomorrow?

e. Is this an example of classical, empirical, or subjective probability?

4.52 Consider an experiment with the following two events:

A = A statistics test is scheduled for 8:00 a.m. B = Every student arrives to class on time for the test

a. Are these two events mutually exclusive? b. Are these two events independent?

4.53 Consider the following experiment—a card is chosen ran- domly from a 52-card deck, observed, and then replaced. After shuffling the deck, a second card is drawn and ob- served. Consider the following events:

A = The first card is the jack of diamonds B = The second card is the jack of diamonds a. Are these two events mutually exclusive? b. Are these two events independent?

4.54 Consider the following experiment—a card is chosen randomly from a 52-card deck, observed, and not replaced.

A second card is drawn and observed. Consider the follow- ing events:

A = The first card is the jack of diamonds B = The second card is the jack of diamonds a. Are these two events mutually exclusive? b. Are these two events independent?

4.55 A local university has a student population that is 57% male. Sixty-four percent of the students are undergraduates; 40% are both male and undergraduates.

a. What is the probability that a randomly selected student is both female and an undergraduate?

b. What is the probability that a randomly selected student is either male or an undergraduate?

4.56 The airline industry defines an on-time flight as one that arrives within 15 minutes of its scheduled time. The following table shows the number of on-time and late flights leaving Philadelphia and arriving in Orlando during a recent time period by airline:

Airline On-Time Late

Southwest 239 68

US Airways 288 130

Air Tran 180 63

a. What is the probability that a randomly selected flight was from Southwest and was on time?

b. What is the probability that a randomly selected flight was from Air Tran or was on time?

c. Given that the flight was late, what is the probability that it was from US Airways?

d. Given that the flight was from Southwest, what is the probability that it was late?

e. Construct a decision tree for these probabilities.

4.57 A test has multiple-choice questions, with four choices of answers from which to choose. After reading the first four questions, you realize you have no idea what any of the correct answers are. Assume you randomly guess at the an- swers for all four questions.

a. What is the probability that you answered the first ques- tion correctly?

b. What is the probability that you answered the first two questions correctly?

c. What is the probability that you answered the first three questions correctly?

d. What is the probability that you answered all four questions correctly?

e. What assumption are you making with these calculations?

4.58 The following table shows the number of students who earned an A, B, C, and D grade in a business statistics class along with their year in their program.

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Grade Sophomore Junior

A 14 30

B 31 38

C 20 16

D 11 5

a. What percentage of students in this class are juniors? b. What percentage of students in this class earned a

B grade? c. What percentage of students in this class were

sophomores and earned a B grade? d. What percentage of students in this class were juniors or

earned an A grade? e. What percentage of students in this class earned a C

grade, given they were sophomores? f. What percentage of students in this class were

sophomores, given they earned a C grade? g. Do course grade and year in the program appear to be

independent or dependent? Define Event A as earn- ing an A grade and Event B as a student in his or her sophomore year.

4.59 In college basketball, a turnover is defined as losing pos- session of the basketball to the opposing team without scor- ing a basket. In theory, reducing the number of turnovers increases the likelihood of winning a game. The following table categorizes the number of wins and losses by the number of turnovers made by the Villanova Wildcats during two recent seasons.

Turnovers Per Game Win Lose

5–8 6 0

9–12 17 5

13–16 22 10

17–20 7 4

21–24 0 2

Assume that a randomly selected game was chosen during one of the two seasons. What is the probability that the game

a. was won by Villanova? b. had 9–12 turnovers made by Villanova? c. was lost by Villanova, and Villanova made 13–16

turnovers? d. was won by Villanova or Villanova made 5–8 turnovers? e. was won by Villanova, given that Villanova made 9–12

turnovers? f. had 21–24 turnovers made by Villanova, given that

Villanova lost the game? g. Construct a decision tree for these probabilities. h. Do winning and turnovers appear to be independent or

dependent events for Villanova? Define Event A as a game with 13–16 turnovers and Event B as a win. What conclusions can be drawn about these events?

4.60 A snack-food company produces potato chips on three dif- ferent production lines. Line 1 produces 20% of the total output, Line 2 produces 30%, and Line 3 produces the balance. Each line is sampled to ensure the salt content is within the proper range. Historically, 4% of Line 1’s output does not meet the salt standards, 3% of Line 2’s fails to meet the standards, and 2% of Line 3’s does not meet the salt standards. A random bag of potato chips was selected and did not meet the salt standards. What is the probability that the chips were produced in Line 3?

4.61 A computer program consists of 1,000 lines of code. For the program to work properly, all 1,000 lines must be correct. Assume that the probability of any line of code being correct is 0.9999.

a. What is the probability that the program will work properly?

b. What assumptions do you need to make to determine this probability?

4.62 According to Internet World Stats, 79% of North American adults used the Internet in 2012. If five people are randomly chosen, what is the probability that all five use the Internet?

4.63 A call center for customer support records the time each customer requires for service and whether the call occurred during the week (Monday through Friday) or during the weekend (Saturday and Sunday). The results are shown in the following table:

Time (Minutes) Week Weekend

Less than 5 15 3

5 to less than 10 26 4

10 to less than 15 11 9

15 to less than 20 22 6

More than 20 10 14

Determine the probability that a randomly selected call

a. was during the week b. was less than 20 minutes c. was 10 to less than 20 minutes d. was more than 20 minutes during the weekend e. was 15 to less than 20 minutes or occurred during the

weekend f. was 5 to less than 10 minutes, given it occurred during

the week g. occurred during the week, given it was 5 to less than

10 minutes h. Construct a decision tree for these events.

4.64 Ann is the owner of Pizzas R Us, a small company that delivers pizza from two different locations. The Mayfair site handles 40% of the total orders while the Claymont location delivers the remaining orders. In its promotional materials, Pizzas R Us strives to deliver all orders within 45 minutes. Historically, the Claymont location has

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experienced 5% of its orders taking longer than 45 minutes to deliver. The Mayfair store has been late on 15% of its orders. Ann has just spoken to a customer whose order was delivered 1 hour after it was placed but did not remember which location the order originated. What is the probability that the late order came from the Mayfair store?

4.65 I would like to order a car from the factory and I have to choose among the following categories:

• exterior color (four choices) • interior color (four choices) • engine size (three choices) • transmission (two choices) • option packages (four choices)

How many unique cars can I choose from with these choices?

4.66 What is the probability of winning a 5/30 lottery game where 5 numbers are picked from a range of 30?

4.67 A multiple-choice test has 10 questions, with each question having four choices.

a. How many different ways can a student answer the 10 questions?

b. What is the probability that a student who randomly answers each question will answer all of them correctly?

4.68 How many different ways can 10 new movies be ranked first, second, and third by a movie critic?

4.69 My bank requires me to establish a four-digit pin number for my debit card.

a. How many possible pin numbers can I choose from? b. What is the probability that someone can correctly guess

my pin number?

4.70 A director for a play has to choose an actor to play a lead male role and an actress to play a lead female role. The director has to choose from among five males and seven females. Determine how many possible male–female combinations the director has to choose from for the two lead roles.

4.71 If a salesperson is responsible for eight stores in her terri- tory, determine how many different ways she can schedule a visit with five of them today.

4.72 A graduate class has 12 students with A grades and 16 students with B grades. Calculate the number of ways in which 4 A students and 5 B students can be uniquely selected.

4.73 The NFL has 32 teams that are evenly divided into two con- ferences, the NFC and the AFC. One team from each con- ference will play in the Super Bowl to determine the NFL champion.

a. How many unique pairs of teams exist that can play in the Super Bowl?

b. Do you have enough information from this problem to calculate the probability that the two Super Bowl teams will be the Green Bay Packers and the Pittsburgh Steelers? Why or why not?

chapter 4 Solutions to Your Turn

Y o u r T u r n # 1

1. P1Spade2 = 13 52

= 0.25 2. P182 = 5 36

= 0.139

Y o u r T u r n # 2

1. P1250 - 4992 = 144 1025

= 0.140 2. P1 7 2502 = 144 + 246 + 297 1025

= 687

1025 = 0.670

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a. empirical, because we have historical data for the number of home runs Josh Hamilton normally hits

b. classical, because we know the number of cards and the number of jacks in the deck

c. If I have data from my last several rounds of golf, this would be empirical; otherwise, subjective.

d. classical, because we can calculate the probability based on the lottery rules e. subjective, because I would not be collecting data for this experiment f. subjective, because I would not be collecting data for this experiment

Y O u R T u R n # 3

Y O u R T u R n # 4 1. a. valid probability b. not a valid probability because probability cannot be less than zero c. valid probability d. valid probability e. not a valid probability because probability cannot be more than one f. not a valid probability because probability cannot be more than one

g. valid probability h. valid probability i. not a valid probability because probability cannot be less than zero 2. P1No pay cut2 = 1 - 10.02 + 0.13 + 0.26 + 0.532 = 1 - 0.94 = 0.06

Y O u R T u R n # 5

a. P1warranty2 = 11 60

= 0.183

b. P1 6 402 = 39 60

= 0.65

c. P1 Ú 40 and no waranty2 = 15 60

= 0.25

d. P1 6 40 or no waranty2 = 5 60

+ 34

60 +

60 =

60 = 0.90

Y O u R T u R n # 6

a. P1Standard and 319602 = 96 600

= 0.16

b. P1Standard or 6 302 = 325 600

+ 255

600 -

165

600 =

415

600 = 0.692

c. P1Gold and 619902 = 49 600

= 0.082

d. P131960 or 6 302 = 217 600

+ 255

600 =

472

600 = 0.787

e. P131960 and 619902 = 0 600

= 0.0

Y O u R T u R n # 7

a. P1Gold2 = 302 962

= 0.314

b. P1Russia2 = 82 962

= 0.085

c. P1Silver and United States2 = 29 962

= 0.030

d. P1Bronze or China2 = 356 962

+ 88

962 -

962 =

421

962 = 0.438

e. P1United States� Gold2 = P1United States and Gold2

P1Gold2 P1United States and Gold2 = 46

962 = 0.048

P1Gold2 = 302 962

= 0.314

P1United States� Gold2 = 0.048 0.314

= 0.153

f. P1Bronze� Russia2 = P1Bronze and Russia2

P1Russia2 P1Bronze and Russia2 = 32

962 = 0.033

P1Russia2 = 82 962

= 0.085

P1Bronze� Russia2 = 0.033 0.085

= 0.388

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191

Y o u r T u r n # 1 0 a., b., c. d.

Less than $200 (A) P(A) = 0.58

More than $200 (B) P(B) = 0.42

Gold (C )

Exec (D)

Gold (C )

Exec (D)

P(A and C) = 0.44

P(A and D) = 0.14

P(B and C ) = 0.10

P(B and D) = 0.32

e. Because the conditional probabilities are P1A� C2 = 0.815, we can con- clude that Gold Star members are more likely to spend less than $200. And because P1B� D2 = 0.696, we can conclude that executive members are more likely to spend more than $200.

Y o u r T u r n # 9 1. Define A = First card is a red queen Define B = Second card is red

P1A2 = 2 52

= 0.038

P1B� A2 = 25 51

= 0.490

P1A and B2 = P1A2P1B� A2 = 10.038210.4902 = 0.0186

2. Define A = First card is a red queen Define B = Second card is red

P1A2 = 2 52

= 0.038

P1B2 = 26 52

= 0.50

P1A and B2 = P1A2P1B2 = 10.038210.502 = 0.019

P1A2 = 12,505 147,739

= 0.085

P1A � B2 = P1A and B2

P1B2 P1A and B2 = 6,480

147,739 = 0.044 P1B2 = 102,366

147,739 = 0.693

P1A � B2 = 0.044 0.693

= 0.063

P1A � B2 ≠ P1A2 Events A and B are dependent.

Y o u r T u r n # 8

Check-Out Total Gold Executive Total

0.44 0.14 0.58 0.10 0.32 0.42

Total 0.54 0.46 1.00

Joint Probabilities Marginal Probabilities

Conditional Probabilities

0.44 0.54

0.14 0.46

0.10 0.54

0.32

0.815

0.304

0.185

0.696 0.46

= =

$200 or Less (A) More than $200 (B)

(C ) (D)

P(A and C ) = 0.44 P(A and D) = 0.14

P(B and C ) = 0.10 P(B and D) = 0.32

P(A) = 0.58 P(B) = 0.42

P(C) = 0.54 P(D) = 0.46

P(A|C)

P(A|D)

P(B|C )

P(B|D)

P(A and C ) P(C )

P(A and D) P(C )

P(B and C ) P(C )

P(B and D) P(D)

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Y O u R T u R n # 1 1 Define A1 = Player used drug, A2 = Player did not use drug, and B = Player tested positive for the drug. P1A12 = 0.08 P1A22 = 1 - 0.08 = 0.92 P1B� A12 = 0.90 P1B� A22 = 0.15

P1A1� B2 = P1A12P1B� A12

P1A12P1B� A12 + P1A22P1B� A22 P1A1� B2 =

10.08210.902 10.08210.902 + 10.92210.152

P1A1� B2 = 0.072

0.072 + 0.138 =

0.072

0.21 = 0.343

Y O u R T u R n # 1 2 1. 4 * 7 * 4 * 3 = 336 different meals 2. 13! = 6,227,020,800 different ordered arrangements

3. 40P3 = 40!

140 - 32! = 140213921382 = 59,280

4. 11C3 = 11!

111 - 32!3! = 11121102192 132122112 = 165

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I n t h I s c h a p t e r , y o u w I l l l e a r n t o :

■ Identify the rules of a discrete probability distribution.

■ Calculate and interpret the mean and standard deviation of a discrete distribution.

■ Understand the characteristics of a binomial experiment.

■ Calculate binomial probabilities.

■ Calculate and interpret the mean and standard deviation of a binomial distribution.

■ Understand the characteristics of a Poisson process.

■ Calculate Poisson probabilities.

■ Approximate binomial probabilities using the Poisson distribution.

■ Calculate hypergeometric probabilities.

Discrete Probability Distributions

C h a p t e r 5 M a p

5.1 Introduction to Discrete Probability Distributions | 195

5.2 Binomial Distributions | 207

5.3 Poisson Distributions | 218

5.4 The Hypergeometric Distribution | 229

5 C h a p t e r

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194 C h a p t e r 5 | Discrete Probability Distributions

We discussed the difference between discrete and continuous data back in Chapter 2. If you are counting the number of customers visiting your store on a given day, you are working with discrete data. In this case, the values are typically whole numbers (or integers) such as 10 or 16. In most applications, discrete data cannot take on fractional values like 30.5 because having 30.5 customers visit your store in one day is not physically possible. By contrast, continuous data, which are the focus of Chapter 6, can take on any real number, including fractional values. Unlike discrete data, which are counted, continuous data are typ- ically measured. Examples are the height and weight of people, which can take on an infinite number of possible values. For example, a hypothetical statistics author’s weight could be 186.1 lbs or 186.1456 lbs. The number of possibilities is limited only by the precision of the measurement.

This chapter will show you how to determine discrete probabilities. Discrete probabili- ties are probabilities that describe the likelihood that certain discrete data will occur. For example, consider Tony’s Pizza, which offers on its menu three pizza sizes that sell for $5, $10, and $15. Table 5.1 shows the historic probabilities of customers ordering each size.

{ table 5.2 | Probabilities of Passing an Exam when Guessing at the Answers

number of test questions

probability of passing

10 0.0035

20 0.0000295

40 0.0000000028

{ table 5.1 | Discrete Probabilities for Tony’s Pizza

price ($) probability (%)

5 30

10 25

15 45

The price of the pizza would be considered discrete data because the values are limited to these three prices. Table 5.1 describes the discrete probability distribution for these sizes. We can see from these data that the $15 pizza is the most popular size, whereas the $10 is the least popular. Probabilities, in general, are often difficult to assess when we rely on our intuition or judgment. For example, you might be surprised by how low the probability is of your passing a multiple-choice test you are completely unprepared for and must guess the answers to. (Not that you would ever try such a crazy thing!) Table 5.2 summarizes the prob- abilities for such a scenario based on the number of test questions. (Assume that 70% is a passing grade.)

For example, the probability of correctly guessing at least 7 (70%) of the 10 multiple- choice questions each with four choices is only 0.0035 (less than 1%). For a 40-question test, the probability drops to an infinitesimal 0.0000000028. This translates to passing the exam once in every 357,142,857 attempts. I’m not sure your professor is going to let you try the exam that many times to pass. In this chapter, you will learn how to calculate the prob- abilities displayed in Table 5.2.

Our test-taking example brings to mind a recurring nightmare I have experienced for as long as I can remember: In the dream, I am back in college. Suddenly, I realize I’m enrolled in a math course I have never attended. I finally work up the nerve to walk into the class- room and find myself face to face with . . . an exam. At this point, I wake up in a cold sweat, thankful that I had only been dreaming. After calculating the probability of passing an exam

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C h a p t e r 5 | 5.1 Introduction to Discrete Probability Distributions 195

in this situation, you surely have come to the same conclusion that I did many years ago: It doesn’t pay to take a multiple-choice exam unprepared and by guessing at the answers. Table 5.2 provides proof.

Businesses often rely on discrete probabilities to help them make important decisions, including the following:

• Forecasting customer demand to plan for staffing and inventory needs

• Quantifying the risk of producing defective items to provide better control of a manu- facturing process

• Gauging different customer satisfaction scores to improve customer service

These and other types of discrete probabilities will be considered throughout this chapter.

5.1 Introduction to Discrete Probability Distributions

In Chapter 4, we discussed conducting experiments to acquire data. A simple experiment would be to roll a fair six-sided die (one with an equal chance of observing each number). Examples of more complex experiments businesses might conduct to calculate probabilities would be the following:

• To record the number of customers who enter a Best Buy store each hour during the business day

• To randomly select six customers who walked into an AT&T Wireless store and count the number who signed new cell phone contracts

• To request that each customer checking out of a Marriott Hotel rate his or her satisfac- tion with the stay as a 1, 2, 3, 4, or 5

Each of these experiments would generate discrete data, such as the following:

• Twenty-six customers entered the Best Buy store between 10:00 and 11:00 a.m.

• One of the six customers at the AT&T store signed a cell phone contract.

• Four is the satisfaction rating, on a scale of 1–5, provided by the last customer to check out of the Marriott Hotel.

Discrete random Variables Discrete random variables have outcomes that typically take on whole numbers as a result of conducting an experiment. The values 26, 1, and 4 are examples of values that a discrete random variable often takes on. The term random is used because the value of the variable cannot be known with certainty before the experiment. For example, it would be unreason- able to expect me (or anyone else) to accurately predict that 26 customers will arrive during the first hour of business at the Best Buy store. The value of the random variable becomes known with certainty after the experiment is conducted. By convention, the value of the ran- dom variable is often denoted by x.

Some experiments will generate continuous data and are known as continuous random variables. Continuous random variables have outcomes that take on any numerical value as a result of conducting an experiment. Examples of continuous random variables include the following:

• The length of time a customer waits in a checkout line at a Whole Foods grocery store

• The ounces of soda consumed by an adult in one month

• The weight of a tractor-trailer at a weigh station on Interstate 95

The acid test for determining whether we are dealing with discrete or continuous ran- dom variables is to ask ourselves, “How many data values can be found in a specific interval

Discrete random variables have outcomes that typically take on whole numbers as a result of conducting an experiment.

Continuous random variables have outcomes that take on any numerical value as a result of conducting an experiment.

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196 C h a p t e r 5 | Discrete Probability Distributions

of numbers?” Discrete random variables have outcomes that are limited to a finite number of values within an interval, whereas continuous random variables have an infinite number of outcomes within an interval. In the Marriot Hotel (discrete) example, there are only five pos- sible outcomes within the interval 1–5 for a customer to choose from when rating his or her satisfaction. In the Whole Foods (continuous) example, there are an infinite number of wait times within the interval 0–5 minutes. Possible wait times could be 3 minutes, 3.2 minutes, or 3.27 minutes. Because we are measuring time on a continuous scale, the only limitation in the number of values within this 0- to 5-minute interval is our measuring instrument’s level of precision.

As mentioned in Chapter 2, there are some exceptions to these guidelines, especially when the outcome of the random variable covers a wide range of values. Income values, for instance, are technically discrete data (we count our money, not measure it). But because of the very large number of possible income values, it is more practical to consider these data as continuous.

We will discuss continuous random variables in more detail in Chapter 6.

rules for Discrete probability Distributions A listing of all the possible outcomes of an experiment for a discrete random variable along with the relative frequency of each outcome is called a discrete probability distribution. To demonstrate this concept, let’s consider the Olive Garden restaurant, which records the fre- quencies of group sizes that arrive for dinner. The Olive Garden wants to know this informa- tion so that it can arrange its available seating in the most efficient manner.

Table 5.3 shows these frequencies. The “Group Size” column shows the discrete random variables in our experiment. Notice how the group sizes are all whole numbers (the Olive Garden won’t see a party of 2.6 people). The “Frequency” column displays the number of times each random variable occurred during the experiment. The “Relative Frequency” column shows the relative frequencies for each group size for the last 50 tables. Relative frequencies are a proportion of the total, so they can consist of fractions. (Recall that relative frequency distributions were discussed at length in Chapter 2.)

A listing of all the possible outcomes of an experiment for a discrete random variable along with the relative frequency of each outcome is called a discrete probability distribution.

Group size represents discrete data because there are only a finite number (five) of possible values for the outcome of this random variable within the interval 2–6.

{ table 5.3 | Olive Garden’s Probability Distribution for Group Sizes

group size frequency relative frequency

2 17 17>50 = 0.34 3 6 6>50 = 0.12 4 16 16>50 = 0.32 5 4 4>50 = 0.08 6 7 7>50 = 0.14

Total 50 1.00

The relative frequencies in Table 5.3 represent the probabilities for each outcome of the discrete random variable (group size). If the 50 tables are representative of a typical evening, we can conclude the probability is greatest (34%) that the next group arriving for dinner will consist of two people.

Figure 5.1 shows the probability distribution in graphical form. See Chapter 2 for a re- fresher on how to construct this chart in Excel.

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C h a p t e r 5 | 5.1 Introduction to Discrete Probability Distributions 197

A discrete probability distribution meets the following conditions:

• Each outcome in the distribution needs to be mutually exclusive with other outcomes in the distribution—that is, the value of the random variable cannot fall into more than one of the frequency distribution classes. For example, it is not possible for the same group to consist of both two and three people.

• The probability of each outcome, P1x2, must be between 0 and 1 (inclusive); that is, 0 … P1x2 … 1 for all values of x. In the previous example, P1x = 42 = 0.32, which falls between 0 and 1.

• The sum of the probabilities for all the outcomes in the distribution needs to add up to

1—that is, a n

i = 1 P1xi2 = 1, where n equals the total number of possible outcomes.

In Table 5.3, notice that the “Relative Frequency” (probability) column adds to 1.00. Occa- sionally, rounding errors occur when probabilities are added, causing the total to be slightly more or less than 1, such as 0.999 or 1.001. We normally assume that values such as these are close enough to 1 for our purposes.

the Mean of a Discrete probability Distribution The mean of a discrete probability distribution is simply the weighted average of the out- comes of the random variable that comprise it. We can calculate it using Equation 5.1.

{ figure 5.1 Graph Showing Olive Garden’s

Probability Distribution for Group Sizes

0% 5%

2 3 4 5 6

10% 15% 20%

30% 25%

35% 40%

Pr ob

ab ili

Olive Garden Probability Distribution

Group Size

Formula 5.1 for the Mean of a Discrete probability Distribution

m = a n

i = 1 xiP1xi2

where

m = The mean of the discrete probability distribution xi = The value of the random variable for the ith outcome

P1xi2 = The probability that the ith outcome will occur n = The number of outcomes in the distribution

Table 5.4 shows the calculations that determined the mean of our discrete probability distribution for our Olive Garden example. The third column in the table is the product of the first two columns.

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198 C h a p t e r 5 | Discrete Probability Distributions

The table shows us that the mean group size at the Olive Garden is 3.56 people. Obviously, 3.56 will never be the actual size of any one particular group arriving for dinner. Rather, it represents the average group size over the last 50 tables. The mean of a discrete probability distribution does not have to equal one of the values of the outcome for the random variables (2, 3, 4, 5, or 6 in this case).

We could have also directly calculated the mean by using Equation 5.1:

m = a n

i = 1 xiP1xi2

m = 12210.342 + 13210.122 + 14210.322 + 15210.082 + 16210.142 m = 0.68 + 0.36 + 1.28 + 0.40 + 0.84 m = 3.56 people

Another term for describing the mean of a discrete probability distribution is the expected value, E1x2. We can calculate the expected value using Equation 5.2.

The mean of a discrete probability distribution is the weighted average of the outcomes of the random variable in the distribution.

{ table 5.4 | Calculating the Mean of a Discrete Probability Distribution (for Olive Garden’s Group Sizes)

group size probability

xi P1xi2 xi P1xi2 2 0.34 0.68

3 0.12 0.36

4 0.32 1.28

5 0.08 0.40

6 0.14 0.84

Total 1.00 3.56

E1x2 = m = a n

i = 1 xiP1xi2

where

E1x2 = The expected value of the discrete probability distribution m = The mean of the discrete probability distribution xi = The value of the random variable for the ith outcome

P1xi2 = The probability that the ith outcome will occur n = The number of outcomes in the distribution

the Variance and Standard Deviation of a Discrete probability Distribution Just when you thought it was safe to get back into the water, along comes another variance and standard deviation! Fear not. If you’ve seen one variance and standard deviation calcula- tion, you’ve seen them all. Remember, we learned in Chapter 3 that the variance measures the spread of the individual values around the mean of a data set and that the standard devia- tion is the square root of the variance. We can calculate the variance for a discrete probabil- ity distribution using Equation 5.3.

Expected value is another term for the mean of a probability distribution.

Formula 5.2 for the expected Value of a Discrete probability Distribution

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C h a p t e r 5 | 5.1 Introduction to Discrete Probability Distributions 199

where

s2 = The variance of the discrete probability distribution xi = The value of the random variable for the ith outcome

P1xi2 = The probability that the ith outcome will occur n = The number of outcomes in the distribution

To demonstrate the use of these equations, we’ll use the Olive Garden distribution shown in Table 5.4. The calculations for the variance are summarized in Table 5.5. Columns 1, 2, and 3 are the results from Table 5.4. The remaining columns are found as follows:

Column 4 = Column 1 minus Column 3 S xi - m Column 5 = Column 4 squared S 1xi - m22 Column 6 = Column 5 multiplied by Column 2 S 1xi - m2 2 P1xi2

The sum of Column 6 (1.93) equals the variance of the discrete probability distribution.

Formula 5.3 for the Variance of a Discrete Probability Distribution

s2 = a n

i = 1 1xi - m22P1xi2

{ table 5.5 | Calculating the Variance of a Discrete Probability Distribution (for Olive Garden’s Group Sizes)

(1) (2) (3) (4) (5) (6)

xi P1xi2 M xi − M 1xi − M22 1xi − M22P1xi2 2 0.34 3.56 - 1.56 2.43 0.83

3 0.12 3.56 - 0.56 0.31 0.04

4 0.32 3.56 0.44 0.19 0.06

5 0.08 3.56 1.44 2.07 0.17

6 0.14 3.56 2.44 5.95 0.83

Total = 1.93

s2 = a n

i = 1 1xi - m22P1xi2 = 1.93 people squared

Recall that the standard deviation 1s2 is the square root of the variance: s = 2s2

The standard deviation of our example distribution is as follows:

s = 2s2 = 21.93 = 1.39 people A more efficient way to calculate the variance of a discrete probability distribution is to

use Equation 5.4.

The units of the variance are the squared units of the original data.

Formula 5.4 for Calculating the Variance of a Discrete Probability Distribution (Shortcut)

s2 = a a n

i = 1 x 2i P1xi2b - m2

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200 C h a p t e r 5 | Discrete Probability Distributions

where

s2 = The variance of the discrete probability distribution xi = The value of the random variable for the ith outcome

P1xi2 = The probability that the ith outcome will occur m = The mean of the discrete probability distribution n = The number of outcomes in the distribution

Table 5.6 summarizes the calculations needed to solve Equation 5.4. The second column is the square of the first column. The fourth column is the product of the second and third columns.

{ table 5.6 | Calculating the Variance of a Discrete Probability Distribution (for Olive Garden’s Group Sizes)

xi x2i P1xi2 x2i P1xi2 2 4 0.34 1.36

3 9 0.12 1.08

4 16 0.32 5.12

5 25 0.08 2.00

6 36 0.14 5.04

Total = 14.60

a n

i = 1 x 2i P1xi2 = 14.60

s2 = a a n

i = 1 x 2i P1xi2b - m2

s2 = 14.60 - 13.5622 = 14.60 - 12.67 = 1.93 Again, the variance equals 1.93, so the standard deviation for our example is again 1.39 people. However, Equation 5.4 requires a lot less effort to calculate.

As explained in Chapter 3, the standard deviation measures the dispersion of the out- come of the discrete random variables in relation to the distribution’s mean. The closer each group size is to 3.56, the lower the standard deviation is. If the individual group sizes tend to be far from 3.56, the standard deviation will be larger. If the Olive Garden discovers that on certain nights the standard deviation of the group size tends to be higher, it may want to set up tables differently on those nights than on nights when the group sizes tend to be more consistent (where the standard deviation tends to be smaller).

I can sense that look of doubt on your face, so let me convince you by changing the Olive Garden example as follows. Let’s assume all 50 groups that arrived consisted of four people. In this case, P1x = 42 = 1.0 and all of the other discrete probabilities are equal to zero, as shown in Table 5.7.

A large standard deviation indicates that the size of the party from one group to the next varies quite a bit. A small standard deviation tells us that most of the group sizes are close to the average group size (3.56).

{ table 5.7 | Calculating the Variance of a Discrete Probability Distribution (for Olive Garden’s Group Sizes)

xi x2i P1xi2 x2i P1xi2 2 4 0.0 0.0

3 9 0.0 0.0

4 16 1.0 16.0

5 25 0.0 0.0

6 36 0.0 0.0

Total = 16.0

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C h a p t e r 5 | 5.1 Introduction to Discrete Probability Distributions 201

Because all 50 tables consisted of four people, the mean of this distribution is 4.0, or m = 4.0. The variance for this new distribution is calculated as follows:

a n

i = 1 x 2i P1xi2 = 16.0

s2 = a a n

i = 1 x 2i P1xi2b - m2

s2 = 16.0 - 14.022 = 16.0 - 16.0 = 0.0 This result demonstrates that the variance of the distribution measures the degree of dispersion of the random variable from the mean of the distribution. Because all outcomes of the discrete random variable equal four and the mean is four, the variance is zero. Have I convinced you yet?

The mean and standard deviation for discrete probability distributions are useful when comparing two different distributions. For example, the two best-selling statistics reference books on the market at the time of this writing are Statistics for Dummies and The Complete Idiot’s Guide to Statistics. Both books are sold on the Amazon Web site, which provides the distribution for the number of stars reviewers gave each book. (The rating scale is from one to five stars.) Table 5.8 provides these data.

{ table 5.8 | Ratings for Two Statistics Books on Amazon

number of reviewers

number of stars dummies idiot’s guide

5 16 19

4 8 7

3 4 3

2 6 0

1 2 0

Total 36 29

First, we need to convert these frequencies to probabilities, as shown in Table 5.9.

{ table 5.9 | Probability Distribution for Two Statistics Books on Amazon

dummies idiot’s guide

stars frequency probability frequency probability

5 16 16>36 = 0.444 19 19>29 = 0.655 4 8 8>36 = 0.222 7 7>29 = 0.241 3 4 4>36 = 0.111 3 3>29 = 0.103 2 6 6>36 = 0.167 0 0>29 = 0.000 1 2 2>36 = 0.056 0 0>29 = 0.000

Total 36 1.000 29 0.999

Using Equation 5.1, we calculate the mean number of stars for each book:

Dummies:

m = a n

i = 1 xiP1xi2

m = 15210.4442 + 14210.2222 + 13210.1112 + 12210.1672 + 11210.0562 m = 2.220 + 0.888 + 0.333 + 0.334 + 0.056 m = 3.831

Occasionally, summing all of the probabilities for a discrete distribution will not add up to exactly 1.0 because of rounding.

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202 C h a p t e r 5 | Discrete Probability Distributions

Idiot’s Guide:

m = a n

i = 1 xiP1xi2

m = 15210.6552 + 14210.2412 + 13210.1032 + 12210.02 + 11210.02 m = 3.275 + 0.964 + 0.309 + 0.0 + 0.0 m = 4.548

According to these calculations, the Idiot’s Guide book has a higher average rating from re- viewers than the Dummies book.

Let’s next examine the variances and standard deviations. Table 5.10 shows the variance calculations for both books.

We calculate the standard deviations for the reviews of each book as follows:

Dummies:

a n

i = 1 x 2i P1xi2 = 16.375

s2 = a a n

i = 1 x 2i P1xi2b - m2

s2 = 16.375 - 13.83122 = 16.375 - 14.677 = 1.698 s = 2s2 = 21.698 = 1.303

Idiot’s Guide:

a n

i = 1 x 2i P1xi2 = 21.158

s2 = a a n

i = 1 x 2i P1xi2b - m2

s2 = 21.158 - 14.54822 = 21.158 - 20.684 = 0.474 s = 2s2 = 20.474 = 0.688

Because the Idiot’s Guide book has a lower standard deviation, the customer reviews for this book are more consistent than those of the Dummies book. By observing the original data in Table 5.8, you can see that the number of stars ranges from three to five for the Idiot’s Guide book, whereas the number of stars for the Dummies book varies from one to five. This is shown graphically in Figure 5.2, which compares the two distributions.

This column was calculated in Table 5.9 { table 5.10 | Variance Calculations for Two Statistics Books on Amazon

dummies idiot’s guide

xi x2i P1xi2 x2i P1xi2 P1xi2 x2i P1xi2 5 25 0.444 11.100 0.655 16.375

4 16 0.222 3.552 0.241 3.856

3 9 0.111 0.999 0.103 0.927

2 4 0.167 0.668 0.000 0.000

1 1 0.056 0.056 0.000 0.000

Total 16.375 21.158

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C h a p t e r 5 | 5.1 Introduction to Discrete Probability Distributions 203

I have one word of warning concerning this last example. Technically, the book reviewer data are ordinal data. I’m sure you remember that back in Chapter 1, we defined ordinal data as measurements that are rank-ordered with no assurance that the differences between the ranks are uniform. In our book example, we have no certainty that the difference between a two-star and three-star rating is identical to the difference between a three-star and four-star rating. By contrast, the intervals between the Olive Garden data are consistent, making the Olive Garden data interval data. Everyone agrees that the difference between a party of two and a party of three is the same as the difference between a party of four and a party of five.

Because of the ambiguity of the intervals, technically it’s not proper to calculate the mean and standard deviation of ordinal data. Nonetheless, it’s common practice when it comes to data such as book ratings. For example, Amazon graphically shows the mean num- ber of stars for books on its Web site.

Another example of this statistical “controversy” is the use of the Likert Scale, which is often used in survey questions, such as the one shown in Table 5.11.

The Dummies distribution is more spread out than the Idiot’s Guide distribution, which provides visual evidence that the Dummies standard deviation is higher.

{ figure 5.2a Ratings Distribution for the

Statistics for Dummies Book

0 0.1

1 2 3 4 5

0.2 0.3 0.4

0.6 0.5

0.7

Stars

Dummies Distribution

P ro

b ab

il it

y { figure 5.2b Ratings Distribution for The

Complete Idiot’s Guide to Statistics Book

0 0.1

1 2 3 4 5

0.2 0.3 0.4

0.6 0.5

0.7

Stars

P ro

b ab

il it

y Idiot’s Guide Distribution

{ table 5.11 | Likert Scale Example

please circle the number indicating your agreement or disagreement to the following statement:

i love statistics.

Strongly Agree 5

Agree 4

Neither Agree nor Disagree 3

Disagree 2

Strongly Disagree 1

Again, we are dealing with ordinal data here because we can’t say with certainty that the difference between “strongly disagree” and “disagree” is the same as the difference between “strongly agree” and “agree.” However, it is often assumed that the differences between the categories are close enough to consider Likert scales to be interval data.

There are plenty of other examples of this issue in the world of statistics. Anytime a researcher gathers data by asking respondents to rate something such as their satisfaction on

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204 C h a p t e r 5 | Discrete Probability Distributions

a scale of 1–5, or 1–10, or 1–100, technically the researcher is collecting ordinal data. How- ever, as we have seen already, ordinal data are often treated as interval data, which allows us to calculate means and standard deviations to our heart’s delight. Just be aware that some people in the world of statistics take offense at such liberties.

Now is your chance to make sure you have followed this discussion so far. I’m particu- larly fond of this next Your Turn question, which highlights the performance of the NFL quarterback who broke my heart, Brett Favre. After many years of playing for my beloved Green Bay Packers, Favre retired in 2008. But then in 2009, he came out of retirement (again) and even had the audacity to play for the Packer’s division rival, the Minnesota Vikings. To my chagrin, he played pretty well that year, beating Green Bay twice in the process. Revenge is sweet, however. In 2010, the Packers put Favre and the Vikings in their place, beating them twice on their way to winning the Super Bowl.

The following table shows the frequency distribution for the number of interceptions thrown per game by Brett Favre relative to the Packer’s wins and losses over the course of his career. (An interception occurs when the quarterback throws a pass that is caught by the opposing team.) Favre currently holds the all-time career interception record for the NFL.

Number of Interceptions per Game Games Won Games Lost

0 91 12

1 59 37

2 23 29

3 9 21

4 0 5

5 0 1

a. Develop a discrete probability distribution of interceptions thrown by Brett Favre for the games won. Next, develop a discrete probability distribution of interceptions thrown by Brett Favre for the games lost.

b. What was the average number of interceptions thrown by Brett Favre per game that was won? c. What was the average number of interceptions thrown by Brett Favre per game that was lost? d. Calculate the variance and standard deviation for both distributions. e. What conclusions can you draw about Brett Favre’s interceptions during his career (other than

he threw too many)?

Y o u r t u r n # 1N

Answers can be found on u page 242

the expected Monetary Value When the discrete random variable is expressed in terms of dollars, the mean of the distribu- tion is known as the expected monetary value (EMV). The EMV is commonly used to evalu- ate the profit potential of projects or the financial return of investment portfolios. We can use Equation 5.2 to calculate the EMV. The following example will demonstrate this concept.

The RAD Construction Company has been awarded a contract with the state of California to replace a bridge on Pike Creek Road. If the project is completed on time, which is defined as during the month of September, RAD expects to earn $100,000 in profit. If the project is completed before September, the state contract calls for a $20,000 bonus to be paid to RAD. If the project is completed after September, there is a $10,000 penalty. RAD estimates the probability of completing the project early, on time, and late to be 25%, 60%, and 15%, respectively. What is the EMV of the project for RAD? Table 5.12 summarizes the information that has been provided so far.

The expected monetary value (EMV) is the mean of a discrete probability distribution when the discrete random variable is expressed in terms of dollars.

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C h a p t e r 5 | 5.1 Introduction to Discrete Probability Distributions 205

We can use Equation 5.2 to calculate the EMV for this project:

EMV = m = a n

i = 1 xiP1xi2

EMV = 1$120,000210.252 + 1$100,000210.602 + 1$90,000210.152 EMV = $103,500

Once the project is complete, RAD will earn $90,000, $100,000, or $120,000, depending on the completion time. What does the $103,500 represent in the context of this example? (Don’t worry—when I ask this question of my students, I often get a blank look.) The EMV is the long-term average profit for this project. Even though the actual project will occur only once, to determine the EMV we need to consider the hypothetical case that the project occurs over and over again. There will be times the profit will be $90,000, other times it will be $100,000, and other times it will be $120,000, with the frequencies reflected in the prob- abilities shown in Table 5.12. The long-term average of all these profits, weighted with their respective probabilities, is expected to be $103,500.

See how the EMV can help you win a bet playing basketball in the following Your Turn problem.

The EMV represents the long-term average profit for this project, as if this project occurred over and over again.

{ table 5.12 | RAD Construction Example

completion net profit ($) probability

Before September 120,000 0.25

September 100,000 0.60

After September 90,000 0.15

Your best friend proposes a bet on the basketball court. You get to attempt one free throw. If you make the free throw, your friend will pay you $20. If you miss the free throw, you will pay your friend $25. Based on your past performance on the court, you estimate the probability that you will make the free throw to be 65%. Should you take the bet? Why or why not?

Y o u r t u r n # 2N

Answer can be found on u page 243

5.1 Section Problems Basic Skills 5.1 Consider the following discrete probability distribution:

Outcome Probability

0 0.06

1 0.11

2 0.24

3 0.27

4 0.20

5 ?

Determine the missing probability for the random variable that equals 5.

5.2 Consider the following discrete probability distribution:

Outcome Probability

1 0.18

2 0.25

3 0.35

4 0.22

a. Calculate the mean of this distribution. b. Calculate the standard deviation of this distribution.

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206 C h a p t e r 5 | Discrete Probability Distributions

5.3 Consider the following discrete probability distribution:

Outcome Probability

10 0.10

15 0.30

20 0.20

25 0.30

30 0.10

a. Calculate the mean of this distribution. b. Calculate the standard deviation of this distribution.

5.4 The following table provides five different distributions for the daily demand for loaves of sourdough bread at a local bakery. Identify which distributions are valid dis- crete probability distributions and which ones are not. Explain your answers.

Daily Demand

Probability Distributions

A B C D E

0 0.1 0.0 0.1 0.1 0.1

1 0.1 0.1 0.1 0.1 0.1

2 0.2 0.2 0.2 0.2 0.2

3 0.4 0.3 0.3 0.4 0.2

4 0.3 0.3 0.2 1.3 0.2

5 0.1 0.1 0.1 0.1 0.1

applications 5.5 A survey recorded the number of children 18 years old or

younger who lived in 200 households. The following table shows the results:

Number of Children Number of Families

0 88

1 40

2 42

3 20

4 7

5 3

a. Determine the mean number of children per household. b. Determine the standard deviation for the number of

children per household.

5.6 Steve Blass owns two fast-food restaurants in Delaware: one located in Stanton and another located in Newark. Steve has recorded customer satisfaction ratings for both locations on a scale of 1 to 5 15 = Most satisfied2. The following table summarizes the data:

Number of Customers

Rating Stanton Newark

1 4 12

2 12 12

3 8 18

4 36 15

5 20 18

a. Calculate the mean satisfaction rating at each location. b. Calculate the standard deviation of each distribution. c. What conclusions can be drawn from these results?

5.7 The following table shows the discrete probability distri- bution for the number of bedrooms per house in a certain community:

Number of Bedrooms Probability

3 0.23

4 0.57

5 0.14

6 0.06

a. Determine the mean number of bedrooms per house. b. Determine the standard deviation for the number of

bedrooms per house.

5.8 Kohls Candy Company developed a new consumer prod- uct that is expected to earn $4,000 in profit each year if consumer demand is low, $20,000 per year if consumer demand is moderate, and $36,000 per year if consumer demand is high. The probability of low, moderate, and high demand is 30%, 45%, and 25%, respectively. Deter- mine the EMV for the new product.

5.9 Avalon Bakery would like you to recommend how many loaves of its famous marble rye bread to bake at the beginning of the day. Each loaf costs the bakery $3.00 and can be sold for $6.00. Leftover loaves at the end of each day are donated to charity. Research has shown that the probabilities for demands of 25, 50, and 75 loaves are 35%, 25%, and 40%, respectively. Make a recom- mendation for Avalon to bake 25, 50, or 75 loaves each morning.

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C h a p t e r 5 | 5.2 Binomial Distributions 207

5.2 Binomial Distributions We are now ready to address a specific type of discrete probability distribution known as the binomial distribution. In the previous section, we defined an experiment as the process of measuring or observing an activity for the purpose of collecting data. Let’s say our experi- ment involves counting the customers who sign cell phone contracts at an AT&T Wireless store. Each customer would be considered a trial for the experiment. For this particular experiment, we have only two possible outcomes for each trial: either the customer signs a contract (a success) or the customer doesn’t sign a contract (a failure). Because we can have only two possible outcomes for each trial, this is known as a binomial experiment.

the Characteristics of a Binomial experiment Let’s say that, historically, 10% of the customers who walk into an AT&T store sign cell phone contracts. Therefore, the probability of a success, p, for any given customer is 0.10. Because there are only two outcomes possible, the probability of a failure for any given customer, q, is 0.90. For a binomial experiment, the probability of a success, p, and the prob- ability of a failure, q, must be the same for every trial in the experiment. Because only two outcomes are allowed in a binomial experiment, p = 1 - q always holds true.

The word success does not have to be assigned to the positive outcome, and the word failure need not apply to the negative one. These are merely labels that can be assigned to either outcome. Just be sure to keep the two straight:

p = The probability of a success q = The probability of a failure

Another characteristic of a binomial experiment is that each trial is independent of any other trials. In other words, the probability of the second customer signing a contract is not affected by whether the first customer signed or did not sign a contract.

Finally, a binomial experiment consists of a fixed number of trials, denoted by n. We count the number of successes in those trials, which we label as the value of x. For example, if our experiment consists of six randomly selected customers, and one of them signs a con- tract, we have the following parameters:

n = 6 x = 1

Recall that we already established that the probability of a success, p, and the probability of a failure, q, are as follows:

p = 0.10 q = 0.90

Given this information, we can calculate the probability that one customer of the six will sign a contract.

Other examples of binomial experiments include the following:

• Testing whether a part is defective or not defective after it has been manufactured

• Observing whether a customer purchases or does not purchase an extended warranty for a new laptop

• Recording whether a household watched or did not watch the series finale of ABC’s Lost

Using Binomial Distributions A binomial probability distribution allows us to calculate the probability of a specific num- ber of successes for a certain number of trials. Therefore, the random variable for this dis- tribution would be the number of successes that were observed. To demonstrate a binomial distribution, let’s use the following example.

Orange Lake Golf Resort in Orlando, Florida, provides a free 1-hour golf clinic every Monday for its guests. Historically, 20% of the guests who attend the clinic register for

A binomial experiment has the following characteristics: (1) The experiment consists of a fixed number of trials, denoted by n; (2) each trial has only two possible outcomes, a success or a failure; (3) the probability of a success and the probability of a failure are constant throughout the experiment; (4) each trial is independent of the other trials in the experiment.

Assigning the terms success and failure to outcomes is completely arbitrary. I could have easily assigned success to a customer not signing a contract and failure to a customer signing a contract.

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208 C h a p t e r 5 | Discrete Probability Distributions

where

n! = n1n - 121n - 221n - 32 c 142132122112 For our Orange Lake example, n = 3 and x = 1.

nCx = n!

1n - x2!x! = 3!

13 - 12!1! = 132122112 122112112 = 3

So far, we have calculated a binomial probability the hard way by breaking it down step by step. Let me show you a much easier way to calculate these probabilities using the

paid golf lessons during the remainder of the week. I have long suspected this is why they provide the “free” clinics. If we randomly select three attendees—call them Bob, Kate, and Roger—what is the probability that one of the attendees will pay for golf lessons this week? This is a job for the binomial probability distribution.

Let’s define Event A as follows:

Event A = Bob pays for lessons, whereas Kate and Roger do not

We can calculate the probability of Event A by identifying the following simple probabilities:

P1Bob = Yes2 = 0.20 P1Kate = No2 = 0.80 P1Roger = No2 = 0.80 If we define a success as a person paying for lessons and a failure as a person not paying

for lessons, then p = 0.20 and q = 0.80. We have three trials in this example 1n = 32 be- cause each of the three people can either pay or not pay for golf lessons. The probability that all three of these simple events will occur together (Event A) can be found using the multi- plication rule for independent events, shown as Equation 4.12 in Chapter 4.

P1A2 = pqq = pq 2 = 10.2210.822 = 0.128 However, this probability represents the likelihood that only Bob pays for lessons, not

the likelihood that any one of the three people will pay for lessons. If Kate pays for lessons, whereas Bob and Roger decline (let’s call this Event B), we have the following sequence and probability:

P1Bob = No2 = 0.80 P1Kate = Yes2 = 0.20 P1Roger = No2 = 0.80 P1B2 = qpq = pq 2 = 10.2210.822 = 0.128

Lastly, if Roger is the only person from the group paying for the lessons (Event C), we have the following sequence and probability:

P1Bob = No2 = 0.80 P1Kate = No2 = 0.80 P1Roger = Yes2 = 0.20 P1C2 = qqp = pq 2 = 10.2210.822 = 0.128

We have a total of three ways in which one person from the group will pay for lessons, with each scenario having a 12.8% chance of occurring. Therefore, the probability that exactly one person from this group of three will pay for golf lessons is the probability that Event A or Event B or Event C will occur. We can find this probability by using the addition rule for mutually exclusive events, shown as Equation 4.4 in Chapter 4.

P1A or B or C2 = P1A2 + P1B2 + P1C2 = 0.128 + 0.128 + 0.128 = 13210.1282 = 0.384

The value “3” in this analysis represents the number of ways in which three people can be selected one at a time. We explored this concept with combinations in Chapter 4. In general, the number of ways in which n objects can be selected x at a time is shown in Equation 5.5.

These three simple events are independent because we are assuming that, for example, if Bob decides to pay for lessons, the probability of Kate or Roger paying for lessons is still 20%. The decision of one person does not influence the decision of another.

Events A, B, and C are mutually exclusive because if one of these events occurs, the other two events cannot.

Formula 5.5 for Combinations of n Objects Selected x at a Time

nCx = n!

1n - x2!x!

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C h a p t e r 5 | 5.2 Binomial Distributions 209

where

P1x, n2 = The probability of observing x successes in n trials n = Number of trials x = Number of successes p = Probability of a success q = Probability of a failure

We can calculate the probability that one of our three resort guests will pay for golf les- sons this week in one step using Equation 5.6:

P1x, n2 = n!1n - x2!x! p xq n - x

P11, 32 = 3!13 - 12!1!10.202 110.8023 - 1

P11, 32 = 13212211212211211210.20210.802 2

P11, 32 = 13210.20210.642 = 0.384 What other values can the random variable x take on in this example other than 1?

Because this example includes only Bob, Kate, and Roger, the number of successes is lim- ited to the values 0, 1, 2, and 3. In general, the number of successes in a binomial experiment must range from zero to n. The number of successes can never exceed the number of trials. Let’s first look at the probability that none of the three resort guests pay for golf lessons.

P1x, n2 = n!1n - x2!x! p xq n - x

P10, 32 = 3!13 - 02!0!10.202 010.8023 - 0

P10, 32 = 13212211213212211211210.202 010.8023

P10, 32 = 11211210.5122 = 0.512 What is the probability that two of the three resort guests pay for golf lessons?

P12, 32 = 3!13 - 22!2!10.202 210.8023 - 2

P12, 32 = 13212211211212211210.202 210.8021

P12, 32 = 13210.04210.802 = 0.096 Finally, the probability that all three of the resort guests pay for golf lessons is as

follows:

P13, 32 = 3!13 - 32!3!10.202 310.8023 - 3

P13, 32 = 13212211211213212211210.202 310.8020

P13, 32 = 11210.0082112 = 0.008

The number of successes, x, can never exceed the number of trials, n, in a binomial experiment.

Remember that 0! = 1 and x0 = 1 for any value of x.

general equation for a binomial distribution, which is shown in Equation 5.6. This equation determines the probability of exactly x successes from n trials.

Formula 5.6 for the Binomial Probability Distribution

P1x, n2 = n!1n - x2!x! p xq n - x

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210 C h a p t e r 5 | Discrete Probability Distributions

Let’s revisit our AT&T Wireless store example and calculate the probability that exactly one customer of the six randomly selected will sign a cell phone contract. Remember, 10% of the customers who walk in the store sign a contract. For this example, our variables are as follows:

n = 6 x = 1 p = 0.10 q = 0.90

Using Equation 5.6, we have the following:

P1x, n2 = n!1n - x2!x! p xq n - x

P11, 62 = 6!16 - 12!1!10.102 110.9026 - 1

P11, 62 = 16215214213212211215214213212211211210.10210.902 5

P11, 62 = 16210.10210.59052 = 0.3543 There is a 35.4% chance that exactly one customer from a group of six will sign a

cell phone contract. What about the probability that none of the six customers will sign a contract?

P10, 62 = 6!16 - 02!0!10.102 010.9026 - 0

P10, 62 = 16215214213212211216215214213212211211210.102 010.9026

P10, 62 = 11211210.53142 = 0.5314 Here’s a tricky question that causes some of my students to struggle. What is the prob-

ability that more than one of the six selected customers will sign a contract? This event includes two, three, four, five, or six customers signing a contract. Using the notation that P1x2 = The probability of x successes, I’m looking for the following:

P122 + P132 + P142 + P152 + P162 In order to calculate this probability directly, I would have to employ Equation 5.6 five

separate times and add all of the results. A much smarter approach (and a lot less work) is to call on the complement rule of probability that we learned in Chapter 4. From our previous discussions about discrete probability distributions, we know the following to be true:

P102 + P112 + P122 + P132 + P142 + P152 + P162 = 1.0

Equation 4.3 stated that P 1A2 = 1 - P 1A'2. That is, the probability of Event A equals one minus the probability that Event A does not occur (its complement).

Notice that the probability of 0 or 1 or 2 or 3 successes sums to 1.0, as shown in the follow- ing equation, because we have three trials in our experiment.

P1x = 0 or 1 or 2 or 32 = 0.512 + 0.384 + 0.096 + 0.008 = 1.0 When we conduct our experiment, the value of our random variable, x, must be equal to

0, 1, 2, or 3. This is consistent with the discrete probability distribution rule that states that the sum of the probabilities for all the random variables in the distribution must sum to 1. Figure 5.3 shows this binomial distribution graphically.

{ figure 5.3 The Golf Lesson Binomial

Distribution

0 1 2 3 4

0.1 0.2 0.3 0.4

0.6 0.5

Number of Golfers

Golf Lesson Binomial Distribution

Pr ob

ab ili

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C h a p t e r 5 | 5.2 Binomial Distributions 211

where

m = The mean of the binomial distribution n = The number of trials p = The probability of a success For our AT&T example, the mean of the distribution is as follows:

m = np = 16210.12 = 0.6 customer In other words, out of six randomly selected customers, we expect an average of 0.6 (10% of them) to sign a new contract. The mean of the binomial distribution represents the long-term average number of successes to expect based on the number of trials conducted.

To demonstrate the mean of this binomial distribution, let’s consider five different groups that consist of six customers. Suppose that zero customers signed a contract from the first group of six. The second group of six had one customer sign, whereas the third and fourth groups also had zero customers sign a contract. From the last (fifth) group of six cus- tomers, two signed a contract. The mean number of customers who signed a contract from each group is calculated as follows:

0 + 1 + 0 + 0 + 2 5

= 0.6 customer

The standard deviation for a binomial probability distribution can be calculated using Equation 5.8.

By simply rearranging the above equation with a little bit of algebra, I can answer my ques- tion with fewer calculations.

P122 + P132 + P142 + P152 + P162 = 1.0 - P102 - P112 P122 + P132 + P142 + P152 + P162 = 1.0 - 0.5314 - 0.3543 = 0.1143

There is an 11.4% probability that more than one of our six customers will sign a cell phone contract.

the Mean and Standard Deviation of a Binomial Distribution Earlier in the chapter, we calculated the mean and standard deviation of a discrete prob- ability distribution. The mean for a binomial probability distribution can be calculated using Equation 5.7.

Formula 5.7 for Calculating the Mean of a Binomial Distribution

m = np

Formula 5.8 for Calculating the Standard Deviation of a Binomial Distribution

s = 2npq where

s = The standard deviation of the binomial distribution n = The number of trials p = The probability of a success q = The probability of a failure

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212 C h a p t e r 5 | Discrete Probability Distributions

For our AT&T example, the standard deviation of the distribution is calculated as follows:

s = 2npq = 216210.1210.92 s = 20.54 = 0.735 customer

This value represents the variation that we would see in the number of successes over six tri- als, assuming the chance of a success is 10%. The larger the standard deviation is, the more variation we would expect in the number of successes.

Take a few minutes to see if you can correctly determine some binomial probabilities in the next Your Turn problem.

The standard deviation of a binomial distribution represents the variation that we would see in the number of successes over n trials, assuming the chance of a success is p.

Suppose a recent survey conducted by Cafe Napoli found that 84% of their customers were satisfied with their most recent meal.

a. Calculate the probability that exactly four of the next five customers will be satisfied. b. Calculate the probability that less than four of the next five customers will be satisfied. c. Calculate the mean for this distribution. d. Calculate the standard deviation for this distribution. e. What assumptions need to be made to validate your results?

Y o U r t U r n # 3N

Answers can be found on u page 243

Binomial probability tables As the number of trials increases in a binomial experiment, calculating probabilities using Equation 5.6 will really drain the batteries in your calculator and possibly even your brain. An easier way to arrive at these probabilities is to use a binomial probability table, which is provided in Table 1 in Appendix A in the back of this textbook. Table 5.13 is an excerpt from this appendix.

The probability table is organized by values of n, the total number of trials. The number of successes, x, makes up the rows of each section, and the probability of success, p, makes up the columns. Notice that the sum of each block of probabilities for a particular value of p adds to 1.0.

Let’s go back to our Orange Lake Golf Resort example, where three resort guests at- tended the free golf clinic on Monday. Recall that 20% of the attendees pay for golf lessons later in the week. The probabilities that we calculated earlier for each value of x are high- lighted in yellow in Table 5.13.

The highlighted values represent the probabilities that 0, 1, 2, or 3 people from 3 randomly selected people will pay for golf lessons at Orange Lake.

{ table 5.13 | Binomial Probabilities from Table 1 in Appendix A p

n x 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

2 0 0.8100 0.7225 0.6400 0.5625 0.4900 0.4225 0.3600 0.3025 0.2500

1 0.1800 0.2550 0.3200 0.3750 0.4200 0.4550 0.4800 0.4950 0.5000

2 0.0100 0.0225 0.0400 0.0625 0.0900 0.1225 0.1600 0.2025 0.2500

3 0 0.7290 0.6141 0.5120 0.4219 0.3430 0.2746 0.2160 0.1664 0.1250

1 0.2430 0.3251 0.3840 0.4219 0.4410 0.4436 0.4320 0.4084 0.3750

2 0.0270 0.0574 0.0960 0.1406 0.1890 0.2389 0.2880 0.3341 0.3750

3 0.0010 0.0034 0.0080 0.0156 0.0270 0.0429 0.0640 0.0911 0.1250

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C h a p t e r 5 | 5.2 Binomial Distributions 213

It’s now time to revisit the question I posed at the start of the chapter: What is the prob- ability that a student will pass a 10-question, multiple-choice test by randomly guessing at each answer? We assumed each question had four choices, so if a student were guessing, the success of correctly answering a question would be 25%. Therefore, p = 0.25, which would mean q = 0.75. Each question represents a trial, so the number of trials, n, equals 10. We also assumed that 70% was a passing grade, which would mean the student would need to answer 7, 8, 9, or 10 questions correctly to pass.

Table 5.14 shows the section of Appendix A that we need to answer this question. The probability of earning a score of 70% or better is the sum of the four probabilities under the p = 0.25 column that are highlighted in yellow.

{ table 5.14 | Binomial Probability Tables from Appendix A

n x 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

10 0 0.3487 0.1969 0.1074 0.0563 0.0282 0.0135 0.0060 0.0025 0.0010

1 0.3874 0.3474 0.2684 0.1877 0.1211 0.0725 0.0403 0.0207 0.0098

2 0.1937 0.2759 0.3020 0.2816 0.2335 0.1757 0.1209 0.0763 0.0439

3 0.0574 0.1298 0.2013 0.2503 0.2668 0.2522 0.2150 0.1665 0.1172

4 0.0112 0.0401 0.0881 0.1460 0.2001 0.2377 0.2508 0.2384 0.2051

5 0.0015 0.0085 0.0264 0.0584 0.1029 0.1536 0.2007 0.2340 0.2461

6 0.0001 0.0012 0.0055 0.0162 0.0368 0.0689 0.1115 0.1596 0.2051

7 0.0000 0.0001 0.0008 0.0031 0.0090 0.0212 0.0425 0.0746 0.1172

8 0.0000 0.0000 0.0001 0.0004 0.0014 0.0043 0.0106 0.0229 0.0439

9 0.0000 0.0000 0.0000 0.0000 0.0001 0.0005 0.0016 0.0042 0.0098

10 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0003 0.0010

Notice that the probability of answering exactly nine correctly is 0.0000 according to Table 5.14. If we use Equation 5.6, the actual probability for this event is 0.000029. How- ever, because we have space for only four decimal places in the table, the probability was rounded to 0.0000. The same situation holds true in this table for the probability of answer- ing exactly 10 questions correctly.

Using the yellow highlighted numbers in Table 5.14, we can determine the probability of passing the multiple-choice (MC) test by guessing as follows:

P1Passing MC Test2 = 0.0031 + 0.0004 = 0.0035 Notice that this is the probability I reported in Table 5.2.

Table 5.14 also has highlighted (in green) the probabilities needed to determine the chance of passing a 10-question true/false (TF) test, under the p = 0.50 column.

P1Passing TF Test2 = 0.1172 + 0.0439 + 0.0098 + 0.0010 P1Passing TF Test2 = 0.1719

This probability is much higher than those for the multiple-choice test because the prob- ability of success is higher 1p = 0.502 when there are only two choices for each question. In other words, the student has a 50% chance of answering one of the true/false questions right but only a 25% chance of answering one of the multiple-choice questions right when guessing.

We saved ourselves a lot of work in this last example by using Table 1 in Appendix A rather Equation 5.6. However, the tables are limited to specific combinations of values for n, x, and p. As a result, some binomial probabilities cannot be determined using the tables.

The next Your Turn problem will give you the chance to test your understanding of the binomial probability tables.

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214 C h a p t e r 5 | Discrete Probability Distributions

Using excel and phStat to Calculate Binomial probabilities Another convenient way to calculate binomial probabilities is to rely on Excel’s BINOM.DIST function which has the following characteristics:

= BINOM.DIST1x, n, p, cumulative2 where

x = Number of successes n = Number of trials p = Probability of a success cumulative = FALSE, if you want to determine the probability of exactly x successes occurring cumulative = TRUE, if you want to determine the probability of x or fewer successes occurring

Consider the following example. According to First American CoreLogic, 45% of the mortgages in Florida were underwater in 2012. The term underwater refers to a situation in which borrowers owe more on their mortgages than their properties are worth. If we ran- domly select three Florida mortgages, what is the probability that exactly two of them are underwater?

We can use Table 5.13 to see that this probability equals 0.3341.We can also use Excel’s BINOM.DIST function to give us the same result.

= BINOM.DIST12, 3, 0.45, FALSE2 = 0.3341 We specify FALSE because we want to determine the probability of exactly two mortgages from our sample being underwater. Excel will also calculate the probability that two or fewer mortgages from our sample will be underwater.

= BINOM.DIST12, 3, 0.45, TRUE2 = 0.9089 We specify TRUE because we want to determine the probability of two or fewer mortgages from our sample being underwater. The probability of this event is 0.9089, which can be confirmed using Table 5.13 and the following equation:

P1x … 22 = P102 + P112 + P122 P1x … 22 = 0.1664 + 0.4084 + 0.3341 = 0.9089

In other words, there is nearly a 91% chance that zero, one, or two mortgages from our Florida sample of three will be underwater. This is also known as the cumulative probability because it sums up all the individual probabilities of two or less.

Finally, we can calculate binomial probabilities using the Excel Add-In PHStat for Windows. See the textbook’s Web site (www.pearsonhighered.com/donnelly) for Excel for Mac instructions. The following steps demonstrate this technique for our Florida mortgage example:

1. Go to Add-Ins + PHStat + Probability & Prob. Distributions + Binomial, as shown in Figure 5.4.

It is estimated that 40% of drivers passing through a particular tollbooth use E-ZPass, an electronic toll collection system. Use the binomial tables in Appendix A to determine the probability of the following:

a. Exactly three of the next seven cars will use the E-ZPass system b. At least five of the next seven cars will use the E-ZPass system

Y o U r t U r n # 4N

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C h a p t e r 5 | 5.2 Binomial Distributions 215

{ figure 5.4 PHStat’s Function for

Calculating Binomial Distributions (Step 1)

2. Fill in the Binomial Probability Distribution dialog box as shown in Figure 5.5. Click OK.

{ figure 5.5 PHStat’s Function for

Calculating Binomial Distributions (Step 2)

Figure 5.6 shows the binomial probabilities that we found in Table 5.13 as well as the mean and standard deviation for this distribution. Notice that cumulative probabilities are also displayed in the output. For example, the probability that one or fewer mortgages from our sample of three are underwater is calculated as follows:

P1x … 12 = P102 + P112 = 0.57475

{ figure 5.6 Final Binomial Probabilities

Calculated Using PHStat

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216 C h a p t e r 5 | Discrete Probability Distributions

{ figure 5.7 Final Binomial Probabilities

Calculated Using PHStat: A Histogram

From this information, we can easily determine the probability that more than one mort- gage from our sample of three are underwater by recognizing that this is the complement of the previous probability:

P1x 7 12 = 1 - P1x … 12 P1x 7 12 = 1 - 0.57475 = 0.42525

This probability is also provided by PHStat in Cell F15 in Figure 5.6. As an added bonus, PHStat will also display the histogram for a binomial distribution.

Figure 5.7 shows the histogram for our Florida mortgage example.

The x-axis shows the number of mortgages that are underwater (successes) while the y-axis denotes the probability for observing the exact number of successes in the sample.

One benefit of using Excel and PHStat to determine binomial probabilities is that you are not limited to the values of n, x, and p shown in the binomial table in Appendix A. Excel’s BINOM.DIST function allows you to use any value between zero and one for p and any value for n that you choose. Obviously, the value of x—the number of successes—will always be limited to whole numbers ranging from zero to n.

Use the next Your Turn example to determine the probability that customers will pur- chase insurance for their cell phones.

Cell phone retailers offer insurance plans for customers that provide protection if the phone is lost or accidently damaged. Sarah is a sales representative for Verizon Wireless, and historically has sold insur- ance plans to 53% of her customers who purchase new cell phones. Suppose that Sarah has sold cell phones to 16 customers so far this week. Answer the following questions using Excel and PHStat.

a. What is the probability that exactly 10 customers purchased insurance plans this week from Sarah?

b. What is the probability that fewer than 10 customers purchased insurance plans this week from Sarah?

c. What is the probability that more than five customers purchased insurance plans this week from Sarah?

d. Construct a histogram for this binomial probability distribution.

Y o U r t U r n # 5N

Answers can be found on u pages 243–244

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C h a p t e r 5 | 5.2 Binomial Distributions 217

5.2 Section Problems Basic Skills 5.10 Consider a binomial probability distribution with

p = 0.60 and n = 7. Determine the following probabilities:

a. P1x = 22 b. P1x … 12 c. P1x 7 52

5.11 Consider a binomial probability distribution with p = 0.35 and n = 8. Determine the following probabilities:

a. exactly three successes b. less than three successes c. six or more successes

5.12 Consider a binomial probability distribution with p = 0.25. Determine the probability of exactly four suc- cesses for the following conditions:

a. n = 5 b. n = 7 c. n = 10

5.13 Consider a binomial probability distribution with p = 0.65 and n = 15. Determine the mean and standard deviation of this distribution.

applications 5.14 If a student estimated that the probability of correctly

answering each question in a multiple-choice question is 80%, use the binomial tables to determine the probability of earning at least a 70% grade on a 20-question exam.

5.15 A recent survey taken by Adecco of 1,047 employees found that 28% of them would lay off their bosses if they could. Answer the following questions based on a random sample of 10 employees:

a. What is the probability that exactly three employees would lay off their bosses?

b. What is the probability that three or fewer employees would lay off their bosses?

c. What is the probability that five or more employees would lay off their bosses?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.16 A hotel claims that 90% of its customers are very satisfied with its service. Answer the following questions based on a random sample of eight customers:

a. What is the probability that exactly seven customers are very satisfied?

b. What is the probability that more than seven customers are very satisfied?

c. What is the probability that less than six customers are very satisfied?

d. Suppose that of the eight customers selected, four re- sponded that they are very satisfied. What conclusions can be drawn about the hotel’s claim?

5.17 The budgeting Web site HelloWallet reported that 25% of U.S. households have withdrawn money from a 401(k) or other retirement account for needs other than retirement in 2013. A random sample of 11 U.S. households was selected.

a. What is the probability that exactly two households withdrew funds from a retirement account for needs other than retirement?

b. What is the probability that less than three households withdrew funds from a retirement account for needs other than retirement?

c. What is the probability that more than five households withdrew funds from a retirement account for needs other than retirement?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.18 According to a survey by Apartments.com, 35% of apart- ment renters in 2013 had owned a home at one point. Consider a random sample of nine apartment renters.

a. What is the probability that exactly three renters from this sample previously owned a home?

b. What is the probability that less than four renters from this sample previously owned a home?

c. What is the probability that exactly six or seven renters from this sample previously owned a home?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.19 According to Catalyst, based on a survey of Fortune 500 companies, 14% of Executive Officer positions in 2012 were female. Answer the following questions based on a random sample of 12 corporate officers:

a. What is the probability that one corporate officer was female?

b. What is the probability that less than four corporate of- ficers were female?

c. What is the probability that more than two corporate officers were female?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.20 An e-commerce Web site claims that 6% of people who visit the site make a purchase. Answer the following questions based on a random sample of 15 people who visited the Web site.

a. What is the probability that none of the people will make a purchase?

b. What is the probability that less than three people will make a purchase?

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218 C h a p t e r 5 | Discrete Probability Distributions

5.3 Poisson Distributions Now that we have mastered the binomial probability distribution, we are ready to move on to the next discrete distribution, the Poisson.

A Poisson distribution is useful for calculating the probability that a certain number of events will occur over a specific interval of time or space. For example, we could use the distribution to determine the likelihood that 10 customers will walk into a store during the next hour or that two car accidents will occur at a busy intersection this month. So, let’s grab some crepes and croissants and learn about some math cooked up by the French.

the Characteristics of a poisson process In our last section, we defined a binomial experiment as one in which the number of suc- cesses are counted over a specific number of trials. The result of each trial is either a success or a failure. A Poisson process counts the number of occurrences of an event over a period of time, area, distance, or any other type of measurement.

In a binomial experiment, the value of the random variable (number of successes) was limited to the number of trials. By contrast, the Poisson process can have any number of outcomes. For instance, the number of customers who walk into our local convenience store during the next hour could be zero, one, two, three, or so on. The outcome for the random variable for the Poisson distribution would be the actual number of occurrences—in this case, the number of customers arriving during the next hour.

The mean for a Poisson distribution is the average number of occurrences that would be expected over the unit of measurement. For a Poisson process, the mean has to be the same for each equal interval of measurement. For instance, if the average number of customers walking into the store each hour is 11, this average needs to apply to every 1-hour increment.

Another characteristic of a Poisson process is that the number of occurrences during one interval is independent of the number of occurrences in other intervals. In other words, if six customers walk into the store during the first hour of business, this would have no effect on the number of customers arriving during the second hour.

Finally, the intervals that are defined in the Poisson process cannot overlap. Using our previous example, if we are interested in counting the number of customers arriving during 1-hour periods, the time segments cannot overlap one another. For example, using the 1-hour intervals 1:00–2:00 p.m. and 1:30–2:30 p.m would violate this requirement.

Examples of the random variables that could be used in a Poisson probability distribu- tion include the following:

• The number of cars that arrive at a tollbooth over a specific period of time

• The number of typographical errors found in a manuscript

• The number of customers who call to complain about a service problem this month

• The total number of home runs hit in Major League Baseball games today

This probability distribution is named after Simeon Poisson, a French mathematician who developed the distribution during the early 1800s.

A Poisson process has the following characteristics: (1) The experiment consists of counting the number of occurrences of an event over a period of time, area, distance, or any other type of measurement; (2) the mean of the Poisson distribution has to be the same for each equal interval of measurement; (3) the number of occurrences during one interval has to be independent of the number of occurrences in any other interval; (4) the intervals that are defined in the Poisson process cannot overlap.

A Poisson process does not count successes and failures like the binomial. Rather, it counts the number of occurrences of a particular event over a specific interval such as time or distance.

c. What is the probability that more than one person will make a purchase?

d. Construct a histogram for this distribution. e. Suppose that out of the 15 customers, 4 made a

purchase. What conclusions can be drawn about the e-commerce site’s claim?

5.21 According to IDC, 70% of smartphones in the world use Google’s Android operating system in the fourth quarter of 2012. Answer the following questions based on a ran- dom sample of 14 smartphone users.

a. What is the probability that exactly nine people from this sample have a smartphone using the Android oper- ating system?

b. What is the probability that all 14 people from this sample have a smartphone using the Android operating system?

c. What is the probability that 11 or fewer people from this sample have a smartphone using the Android oper- ating system?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

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C h a p t e r 5 | 5.3 Poisson Distributions 219

Now that you understand the basics of a Poisson process, let’s calculate some probabil- ity distributions.

Using poisson probability Distributions If a random variable follows a pattern consistent with a Poisson probability distribution, we can calculate the probability of a certain number of occurrences over a given interval. To make this calculation, we need to know the average number of occurrences for the event over the interval. To demonstrate the use of the Poisson probability distribution, I’ll use this example.

The following story is true, and the names have not been changed because nobody in this story is innocent. For many years, when my sons Brian and John were teenagers they would accompany me on a golf pilgrimage to Myrtle Beach, South Carolina. On our last night one particular year, we were browsing through Martin’s Golf Superstore. Brian somehow convinced me to purchase a used, fancy, brand-name golf club that he swore he absolutely had to have in order to reach his full potential as a golfer. Even used, the club cost more than any club I had ever purchased new.

Early the following morning, we packed our bags, checked out of the hotel, and drove to our final round of golf, which I had cleverly planned to be along our route back home. On the first tee, Brian pulled out his new, used prize possession and proceeded to hit a “duck hook,” which is a golfer’s term for a ball that goes very short and very left, often into a bunch of trees, rarely to be seen again. I smiled nervously at Brian and tried to convince my- self that he’d be fine on the next hole. After hitting duck hooks on holes two, three, and four, I found myself physically restraining Brian from throwing his new, used, prized possession into the lake.

After our round was over, I patiently drove back to Myrtle Beach to return the club, add- ing an hour to what would have been a 10-hour car ride. At Martin’s, the woman cheerfully said she would take the club back as long as I had the receipt (cue scary music). I vaguely remembered putting the receipt someplace “special” just in case I would need it, but after packing, checking out, and playing golf, I would have had a better chance of discovering a cure for cancer than remembering where I had put that piece of paper.

Not being one to give up easily, I marched back to the car and started unpacking every- thing. After a short while, during which time I had spread out our dirty underwear and socks all over the parking lot, the same woman walked out to tell me the store would gladly refund my money without the receipt if I would just pack up my things and put them back in the car. Apparently, I was scaring some customers away. (I recommend you consider this strategy when you’re trying to return something without the darned receipt.)

Anyway, let’s assume that during Brian’s brief career as a golfer, the mean number of golf balls he will lose during a round is 4.2. Because this value is an average, it does not have to be a whole number. We will also assume that the actual number of Brian’s lost golf balls during one round follows the Poisson distribution.

We can use the Poisson probability distribution to calculate the probability that Brian will lose x number of golf balls during his next round with Equation 5.9.

To test this assumption that the number of golf balls that Brian loses during one round follows the Poisson distribution, we’ll need to perform a goodness-of- fit test, a procedure that will be discussed in Chapter 12.

Formula 5.9 for the Poisson Probability Distribution

P1x2 = l xe - l

where

x = The number of occurrences of interest over the interval l = The mean number of occurrences over the interval e ≈ 2.71828 P1x2 = The probability of exactly x occurrences over the interval

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220 C h a p t e r 5 | Discrete Probability Distributions

We can now calculate the probability that Brian will lose exactly two golf balls during his next round. With l = 4.2 and x = 2, Equation 5.9 becomes as follows:

P122 = l xe - l

x! =

14.22212.71828 - 4.22 2!

P122 = 117.64210.0149962 2

= 0.1323

In other words, Brian has a 13.2% chance of losing exactly two golf balls during his next round. We can also calculate the cumulative probability that Brian will lose no more than two golf balls using the following equations:

P1x … 22 = P102 + P112 + P122

P102 = 14.2 0212.71828 - 4.22

0! =

11210.0149962 1

= 0.0150

P112 = 14.2 1212.71828 - 4.22

1! =

14.2210.0149962 1

= 0.0630

P1x … 22 = 0.0150 + 0.0630 + 0.1323 = 0.2103 There is a 21% chance that Brian will lose no more than two golf balls during his next round.

I have one final question: What is the probability that Brian will lose four or more golf balls during his next round? You can express this probability as follows:

P1x Ú 42 = P142 + P152 + g Notice from this equation that the probability of x being greater than or equal to four in-

cludes an infinite number of events. Mathematically, with a Poisson distribution, there is no upper limit to the number of occurrences during an interval. If Brian is having a particularly bad round of golf, he could lose a lot of balls with no limit other than his patience. (Unfor- tunately, I have personally observed this on occasion.) Because we cannot add all the prob- abilities for an infinite number of occurrences, we need to take another approach. From our rules of discrete probability distributions, we know the following:

P102 + P112 + P122 + P132 + P142 + P152 + g = 1.0 With a little algebra, we can rearrange the previous equation as follows:

P142 + P152 + g = 1.0 - P102 - P112 - P122 - P132 Therefore, the probability that Brian will lose four or more golf balls can be expressed in the following manner:

P1x Ú 42 = 1.0 - P102 - P112 - P122 - P132 We have previously used the complement rule of probability discussed in Chapter 4 to

solve the problem of summing an infinite number of occurrences. Note that we already cal- culated the probabilities on the right side of the equation except for P(3). Like we did before, we can use the Poisson distribution equation to find P132:

P132 = 14.2 3212.71828 - 4.22

3! =

174.088210.0149962 132122112 = 0.1852

Now we can plug in the values of all of the probabilities and calculate the answer:

P1x Ú 42 = 1.0 - P102 - P112 - P122 - P132 P1x Ú 42 = 1.0 - 0.0150 - 0.0630 - 0.1323 - 0.1852 = 0.6045

There is a 60.45% chance that Brian will lose four or more golf balls during his next round. I see we will be visiting the golf store very soon to buy more golf balls! Not Martin’s Golf Superstore, however. Apparently, the Donnelly family has been given a lifetime ban after the suitcase “incident.”

There’s one more cool feature of the Poisson distribution: As Equation 5.10 shows, the variance of the distribution is the same as the mean.

I recommend carrying out the intermediate Poisson calculations out to six decimal places to avoid rounding errors in the final results. Check with your instructor about rounding your final results.

There is no upper limit to the number of occurrences for the Poisson distribution. Mathematically, there are an infinite number of terms in this equation.

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C h a p t e r 5 | 5.3 Poisson Distributions 221

Using a poisson Distribution to Calculate the probability of arrivals One of the more common uses of the Poisson distribution is to determine the probability of customer arrivals. For example, a local PNC Bank branch has a drive-through window. On average, 12 customers per hour arrive at the window, and we’ll assume these arrivals follow the Poisson distribution. What is the probability that exactly four customers will arrive dur- ing the next 30 minutes?

To answer this question, we need to adjust the average number of arrivals per hour to a 30-minute interval. If the bank averages 12 customers arriving per hour, it would average 6 customers every 30 minutes, so l = 6.0. To find the probability that 4 customers will arrive during the next 30 minutes, we use Equation 5.9 with x = 4:

P1x2 = l xe - l

P142 = 16 4212.71828 - 62

4! =

11,296210.0024792 142132122112 = 0.1339

How about the probability that two customers will arrive during the next 15 minutes? Some students will tell me that this probability is the same as our last example, which was four customers over 30 minutes (0.1339). I’ll show you that this is not the case. First, we need to convert the arrival average to a 15-minute interval. If we average six customers ar- riving every 30 minutes, then we average three customers every 15 minutes, which makes l = 3.0. To find the probability that two (half of the four for the 30-minute interval) cus- tomers will arrive during the next 15 minutes, we have the following:

P122 = 13 2212.71828 - 32

2! =

19210.0497872 122112 = 0.2240

You can see this probability is not the same as our previous example. I’m sure you’re itching to see if you understand how to calculate a Poisson probability distribution—the next Your Turn problem will give you a chance.

Suppose the average number of complaints received by Christiana Hospital from patients is 7.4 every four weeks. Assuming the number of complaints per month follows the Poisson distribution, determine the following probabilities:

a. Exactly three complaints during the next four weeks b. Three or fewer complaints during the next four weeks c. Five or more complaints during the next four weeks d. Exactly three complaints during the next two weeks

Y o U r t U r n # 6N

Answers can be found on u page 244

Formula 5.10 for the Variance of a Poisson Distribution

s2 = l

So, for this distribution there are no nasty variance calculations such as those we dealt with in previous chapters. For our golf ball example, the variance is as follows:

s2 = l = 4.2

The standard deviation then is calculated as follows:

s = 2l = 24.2 = 2.05 The variance of the Poisson distribution is equal to the mean.

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222 C h a p t e r 5 | Discrete Probability Distributions

The probability table is organized by values of l, the average number of occurrences. The probability that Brian will lose two golf balls during his next round when he averages losing 4.2 balls is highlighted in yellow (0.1323). This value was calculated earlier using the Poisson distribution equation. If you add up the probabilities in a column for a particular value of l, they will sum to 1. As with the binomial tables, one limitation of using the Poisson tables is that you are restricted to using only the values of l that are shown in the table.

The next Your Turn problem will give you the chance to test your understanding of the Poisson probability tables.

poisson probability tables Just like binomial distributions, for certain mean values, Poisson probability distributions can be looked up in a table. You can find the Poisson distribution table in Table 2 in Appendix A of this book. Table 5.15 is an excerpt from this appendix for the probabilities from our golf ball example.

{ table 5.15 | Poisson Probability Tables from Appendix A

x 4.10 4.20 4.30 4.40 4.50 4.60 4.70 4.80 4.90 5.00

0 0.0166 0.0150 0.0136 0.0123 0.0111 0.0101 0.0091 0.0082 0.0074 0.0067

1 0.0679 0.0630 0.0583 0.0540 0.0500 0.0462 0.0427 0.0395 0.0365 0.0337

2 0.1393 0.1323 0.1254 0.1188 0.1125 0.1063 0.1005 0.0948 0.0894 0.0842

3 0.1904 0.1852 0.1798 0.1743 0.1687 0.1631 0.1574 0.1517 0.1460 0.1404

4 0.1951 0.1944 0.1933 0.1917 0.1898 0.1875 0.1849 0.1820 0.1789 0.1755

5 0.1600 0.1633 0.1662 0.1687 0.1708 0.1725 0.1738 0.1747 0.1753 0.1755

6 0.1093 0.1143 0.1191 0.1237 0.1281 0.1323 0.1362 0.1398 0.1432 0.1462

7 0.0640 0.0686 0.0732 0.0778 0.0824 0.0869 0.0914 0.0959 0.1002 0.1044

8 0.0328 0.0360 0.0393 0.0428 0.0463 0.0500 0.0537 0.0575 0.0614 0.0653

9 0.0150 0.0168 0.0188 0.0209 0.0232 0.0255 0.0281 0.0307 0.0334 0.0363

10 0.0061 0.0071 0.0081 0.0092 0.0104 0.0118 0.0132 0.0147 0.0164 0.0181

11 0.0023 0.0027 0.0032 0.0037 0.0043 0.0049 0.0056 0.0064 0.0073 0.0082

12 0.0008 0.0009 0.0011 0.0013 0.0016 0.0019 0.0022 0.0026 0.0030 0.0034

13 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0011 0.0013

14 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 0.0003 0.0003 0.0004 0.0005

15 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002

While working on research for my Ph.D. dissertation back in the Stone Age, I spent time at a DuPont facility in Buffalo, New York, that produced Tedlar. Tedlar is a thin film used to protect a variety of surfaces. It is produced in sheets that resemble plastic wrap. To inspect the Tedlar for defects, three-foot squares of it are placed on light panels and examined. Assume the number of defects in a three-foot square follows the Poisson distribution. The average number of defects per square is 0.05. Use the Poisson probabilities in Appendix A to answer the following:

a. What is the probability that there is exactly one defect in the square? b. What is the probability that there is a defect in the square?

Y o U r t U r n # 7N

Answers can be found on u page 244

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C h a p t e r 5 | 5.3 Poisson Distributions 223

Using excel and phStat to Calculate poisson probabilities You can also conveniently calculate Poisson probabilities using Excel. The built-in POISSON.DIST function has the following characteristics:

= POISSON.DIST1x, l, cumulative2 where

cumulative = FALSE, if you want to determine the probability of exactly x occurrences cumulative = TRUE, if you want to determine the probability of x or fewer occurrences

We can use Excel’s POISSON.DIST function to calculate the probability that Brian will lose exactly two golf balls during his next round of golf when he averages losing 4.2 balls.

= POISSON.DIST12, 4.2, FALSE2 = 0.1323 Excel will also calculate the cumulative probability that Brian will lose two balls or less dur- ing his next round.

= POISSON.DIST12, 4.2, TRUE2 = 0.2102 Earlier in the chapter, we calculated this probability as being 0.2103. This (very) slight

difference is due to the fact that Excel uses more significant figures in the Poisson calcula- tions than we used with our calculator.

We can also calculate Poisson probabilities using the Excel Add-In PHStat for Windows. See the textbook’s Web site for Excel for Mac instructions. The following steps demonstrate this technique for our golf ball example.

1. Go to Add-Ins + PHStat + Probability & Prob. Distributions + Poisson, as shown in Figure 5.4.

2. Fill in the Poisson Probability Distribution dialog box as shown in Figure 5.8 on the next page. Click OK.

Is this something that could happen once in a million times? Is it something that could happen once in a thousand times, or once every 5,000 times? What exactly are the risks involved?

—President Barack Obama, following the BP oil spill in the Gulf of Mexico

President Obama’s question was on the minds of many people following BP’s disastrous oil spill in the Gulf of Mexico begin- ning in the spring of 2010. Increasing oil prices in the last decade have made it economically feasible to drill in deep waters. During this time, the number of deepwater wells in the Gulf of Mexico increased from 17 to 130. But just how safe is deepwater drilling?

We can use the Poisson distribution to determine the probabil- ity of future oil spills, but to do so, we need to assign a value for l, which represents the average number of major oil spills per year. Given the fact that we have experienced one major spill during the 10-year period when deepwater drilling became

more prevalent, a rough estimate for l could be set at 0.1 spills per year. Using this assumption, what is the probability that a major oil spill will occur during the next 20 years? First, we need to convert l to a 20-year interval by multiplying 0.1 by 20. With l = 2.0 spills per 20-year period and x = 1, we have the following:

P112 = l xe-l

x ! =

12.01212.71828-2.02 1!

P112 = 12.0210.1353352

1 = 0.2707

Based on our assumptions, there is a 27% chance of another major deepwater oil spill occurring during the next 20 years. I have to admit that this conclusion is based on a very small amount of data; nevertheless, I imagine this is not a statistic that President Obama would be thrilled to hear.

Stats in practice: Using the poisson Distribution to estimate the probability of an oil Spill

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224 C h a p t e r 5 | Discrete Probability Distributions

Figure 5.9 shows the Poisson probabilities that we saw in Table 5.15. Isn’t technology wonderful?

{ figure 5.8 PHStat’s Function for

Calculating a Poisson Distribution

{ figure 5.9 Final Poisson Probabilities

Calculated Using PHStat

Notice that we also have the cumulative probabilities displayed in Column D in Figure 5.9. For example, we calculated earlier that the probability of Brian losing two or fewer balls in his next round was as follows:

P1x … 22 = P102 + P112 + P122 P1x … 22 = 0.0150 + 0.0630 + 0.1323 = 0.2103

This result is shown in Figure 5.9 in Cell D10. The slight difference in probabilities is due to rounding.

We also calculated the probability that Brian will lose four or more balls on his next round as being 0.6045. You can find this value in Figure 5.9 as well, in Cell G12.

As we discovered earlier, PHStat can also display the histogram for our Poisson distri- bution; Figure 5.10 shows the histogram for our golf ball example.

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C h a p t e r 5 | 5.3 Poisson Distributions 225

One benefit of using Excel to determine Poisson probabilities is that you are not limited to the values of l shown in the Poisson table in Appendix A. Excel’s POISSON.DIST func- tion allows you to use any value for l.

Use the next Your Turn problem to see if you understand how to use Excel and PHStat to calculate Poisson probabilities.

{ figure 5.10 Final Poison Probabilities

Displayed Using PHStat: A Histogram

As a young engineer, I spent several years working for Diamond Shamrock Corporation at its chlorine plant in Delaware City, Delaware. Each month, we would track the number of recordable injuries, which are defined by the Occupational Safety and Health Administration as injuries to employees that require medical treatment beyond routine first aid. Assume the number of recordable injuries each month fol- lows a Poisson distribution with an average of 1.4 injuries per month. Use Excel to answer the following questions:

a. What is the probability that exactly one recordable injury will occur next month? b. What is the probability that less than two recordable injuries will occur next month? c. What is the probability that four or more recordable injuries will occur next month? d. Construct a histogram of the probabilities for this Poisson distribution using PHStat.

Y o U r t U r n # 8N

Answers can be found on u page 245

Using poisson Distributions to approximate Binomial Distributions We can calculate binomial probabilities using the Poisson distribution when the following conditions are present:

• When the number of trials, n, is greater than or equal to 20

and

• When the probability of a success, p, is less than or equal to 0.05

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226 C h a p t e r 5 | Discrete Probability Distributions

where

x = Number of successes n = The number of trials p = The probability of a success You might be asking yourself at this moment just why you would want to do this. The

answer is because the Poisson formula has fewer computations than the binomial formula and, under the stated conditions, the distributions are very close to one another. I’m all for taking the easy way when I have a choice.

I’ll demonstrate this point with an example. Avon Products ships boxes to its representa- tives that contain the products that the representatives ordered. Historically, 2% of the boxes do not contain the correct products. Twenty boxes are randomly selected and inspected to ensure they have the proper contents. We want to determine the probability that exactly two of these boxes do not have the correct items. This is a binomial experiment with n = 20, x = 2, and p = 0.02. Using Equation 5.6, we can calculate the binomial probability as follows:

P1x, n2 = n!1n - x2!x! p xq n - x

P12, 202 = 20!120 - 22!2! 10.022 210.98220 - 2

P12, 202 = 1202119211821172 g 132122112!11821172 g 132122112122112 10.022 210.98218

P12, 202 = 12021192122112 10.0004210.6951352 = 0.0528

Now, I’ll use Equation 5.11, the Poisson approximation to the binomial distribution, to prove my point:

np = 120210.022 = 0.4

P1x2 = 1np2 x2.71828-1np2

P122 = 10.42 22.71828-10.42

2! =

10.16210.6703202 2

= 0.0536

In my opinion, these probabilities are close enough (5.3% versus 5.4%), and fewer calcula- tions were needed using Equation 5.11. If you need further evidence, Figures 5.11 and 5.12 show the histograms for each distribution.

Formula 5.11 for Using the Poisson Equation to Calculate Binomial Probabilities

P1x2 = 1np2 x 2.71828 - 1np2

In other words, a distribution with a large number of trials and a small probability for suc- cess in those trials can be considered a Poisson distribution. Binomial probabilities can be calculated using Equation 5.11 under these circumstances.

If you need to calculate binomial probabilities with the number of trials, n, greater than or equal to 20 and the probability of a success, p, less than or equal to 0.05, you can use the equation for the Poisson distribution to approximate the binomial probabilities.

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C h a p t e r 5 | 5.3 Poisson Distributions 227

{ figure 5.11 The Binomial Probability

Distribution When n = 20 and p = 0.02

Even to a skeptic, these two distributions look very much alike. So, my advice to you is to use the Poisson equation if you’re faced with calculating binomial probabilities when n Ú 20 and p … 0.05.

The following Your Turn problem will give you the chance to use this approximation technique to predict the probability that your next fast-food order will not be correct.

{ figure 5.12 The Poisson Probability

Distribution When the Mean = 0.4

A fast-food manager claims that only 3% of the orders that are filled are inaccurate. Use the first 20 orders of the day to answer the following:

a. Use Equation 5.6 to calculate the probability that exactly three orders from the first 20 are not correct.

b. Use Equation 5.11 to calculate the probability that exactly three orders from the first 20 are not correct.

c. How do the two probabilities compare? d. Suppose that of the first 20 orders of the day, four were not correct. What conclusions can be

drawn about the manager’s claim?

Y o U r t U r n # 9N

Answers can be found on u page 245

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228 C h a p t e r 5 | Discrete Probability Distributions

5.3 Section Problems Basic Skills 5.22 Consider a Poisson probability distribution with l = 3.0.

Determine the following probabilities.

a. P1x = 32 b. P1x … 22 c. P1x 7 42

5.23 Consider a Poisson probability distribution with l = 4.7. Determine the following probabilities.

a. exactly five occurrences b. more than six occurrences c. three or fewer occurrences

5.24 Consider a Poisson probability distribution. Determine the probability of exactly four occurrences for the following conditions:

a. l = 2.0 b. l = 3.0 c. l = 4.0 d. What conclusions can be made about how these prob-

abilities change with l?

5.25 Consider a Poisson probability distribution with l = 8.5. Determine the mean and standard deviation of this distribution.

applications 5.26 Assume the number of births in a local hospital follows a

Poisson distribution and averages 2.6 per day.

a. What is the probability that no births will occur today? b. What is the probability that less than four births will

occur today? c. What is the probability that more than one birth will

occur today?

5.27 Sara is a salesperson for Camera’s Etc. which is a retailer for high-end digital cameras. Historically, Sara has averaged selling 2.1 extended warranties per day for cameras that she sells. Assume the number of camera warranties that Sara sells per day follows the Poisson distribution.

a. What is the probability that Sara will sell five extended warranties tomorrow?

b. What is the probability that Sara will not sell an extended warranty tomorrow?

c. What is the probability that Sara will sell more than two extended warranties tomorrow?

d. What is the standard deviation for this distribution?

5.28 A particular intersection in Delaware is equipped with a surveillance camera. The number of traffic tickets issued to drivers passing through the intersection follows the Poisson distribution and averages 4.5 per month.

a. What is the probability that seven traffic tickets will be issued at the intersection next month?

b. What is the probability that less than three traffic tick- ets will be issued at the intersection next month?

c. What is the probability that five or more traffic tickets will be issued at the intersection next month?

5.29 A customer support center for a computer manufacturer receives an average of 1.5 phone calls every 5 minutes. Assume the number of calls received follows the Poisson distribution.

a. What is the probability that no calls will arrive during the next 5 minutes?

b. What is the probability that three or more calls will arrive during the next 5 minutes?

c. What is the probability that three calls will arrive during the next 10 minutes?

d. What is the probability that no more than two calls will arrive during the next 10 minutes?

5.30 Production records indicate that 1.5% of the light bulbs produced in a facility are defective. A random sample of 22 light bulbs was selected.

a. Use the binomial distribution to determine the prob- ability that two defective bulbs are found.

b. Use the Poisson approximation to the binomial distri- bution to determine the probability that two defective bulbs are found.

c. How do these two probabilities compare?

5.31 Research in Motion (RIM) developed the Blackberry cell phone that was widely adopted by businesses for their employees. However, with the increase in popu- larity of Apple’s iOS and Google’s Android operating systems for mobile devices, sales of Blackberry devices have declined recently. The market share for devices using RIM’s operating system was estimated to be 1.6% in 2013. A random sample of 25 mobile devices was selected.

a. Use the binomial distribution to determine the prob- ability that less than three devices from this ample use the RIM operating system.

b. Use the Poisson distribution to determine the probabil- ity that less than three devices from this ample use the RIM operating system.

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C h a p t e r 5 | 5.4 The Hypergeometric Distribution 229

5.4 The Hypergeometric Distribution One of the key features of the binomial and Poisson distributions was that the events are in- dependent of one another. The underlying assumption with these distributions is that you are sampling with replacement or from a very large population without replacement so that the probabilities of a success or number of occurrences remains constant. However, sometimes samples are randomly selected from a small population without being replaced. Under these conditions, the probabilities of success change repeatedly because the sample space becomes smaller after each selection. Have no fear, worried student, when you are faced with these odds. This is a job for the hypergeometric distribution (cue the superhero music)!

Calculating probabilities for a hypergeometric Distribution The best way to explain the “superpowers” of a hypergeometric distribution is to start with an example: A Pennsylvania retailer currently employs 22 sales associates, of which 8 are more than 55 years in age. (We’ll refer to the people over 55 as the “older employees” in this problem to differentiate them from younger employees.) Due to a recent downturn in the economy, the retailer decided to reduce its sales force by seven employees. Of the seven peo- ple laid off, five were older employees. If the employees who were laid off were randomly selected, what is the probability that five of them would be from the older group? This infor- mation would be useful for a lawyer who is investigating a possible age discrimination case.

I’ll define a success as an older employee being selected to be laid off. This example in- volves sampling without replacement because once an employee is selected to be laid off, he or she can’t be selected again to be laid off. As a result, the probability of future employees being selected changes due to the reduction in the sample space. To demonstrate, let’s define the following event:

Event A = Randomly selected laid@off employee is older than 55

Because 8 of the 22 employees are older than 55, the probability of a success is then as follows:

P1A2 = 8 22

= 0.364

However, the probability isn’t constant throughout the sampling process. Let’s assume the first employee laid off was older than 55. What is the probability that the next laid-off employee is older than 55? Well, because we are sampling without replacement, the sample space has been reduced from 22 to 21. Because one of the older employees has been previ- ously selected, the number of remaining older employees is down to seven. Therefore, the probability that an older employee will be selected to be laid off (a success) is now given as follows:

P1A2 = 7 21

= 0.333

It is important to realize that the probability of success in our example changes during the sampling process, which violates the requirements for a binomial probability distribu- tion. Therefore, the binomial distribution is not appropriate here, and we need the hypergeo- metric distribution to step in and save the day (sort of like Underdog).

For events that have only two possible outcomes, such as success or failure, the hyper- geometric probability is calculated using Equation 5.12.

The hypergeometric distribution is used when samples are taken from a small population without being replaced.

Remember, a success does not need to be defined as a positive outcome.

Because the first employee selected was older and we are sampling without replacement, only 7 of the remaining 21 employees are older.

Formula 5.12 for the Hypergeometric Distribution

P1x2 = N - RCn - x #

RCx

NCn

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230 C h a p t e r 5 | Discrete Probability Distributions

where

N = The population size R = The number of successes in the population n = The sample size x = The number of successes in the sample We have seen the term nCx in both this chapter and in Chapter 4. It represents the num-

ber of combinations of n objects selected x at a time. In this example, the objects are em- ployees. In Chapter 4, I showed you the following short-cut method for calculating combinations (Equation 4.19):

nCx = n!

1n - x2!x! = n1n - 1211n - 222c1n - x + 12

Based on our definition of a success, we have the following variables for our example:

N = The total number of employees = 22 R = The number of older employees = 8 n = The number of employees laid off = 7 x = The number of older employees laid off = 5

If we plug these data into Equation 5.12, we can find the probability that five of the seven laid-off employees will be older than 55 if they are chosen randomly:

P1x2 = N - RCn - x #

RCx

NCn

P152 = 22 - 8C7 - 5 #

8C5

22C7 = 14

C2 # 8C5 22C7

We still need to calculate the value of each of the three combinations. Using Equation 4.19, we can start with the result of 14C2:

nCx = n!

1n - x2!x! = n1n - 121n - 22 g 1n - x + 12

14C2 = 14!

114 - 22!2! 1n - x + 12 = 114 - 2 + 12 = 13

14C2 = 11421132 122112 =

182

2 = 91

There are 91 ways in which we can choose two employees from a group of 14. In our example, there were 14 employees who were younger than 55, and two of them were laid off.

Next, we obtain the value of 8C5 as follows:

8C5 = 8!

18 - 52!5! 1n - x + 12 = 18 - 5 + 12 = 4

8C5 = 182172162152142 152142132122112 =

6,720

120 = 56

There are 56 ways in which we can choose five employees from a group of eight. In our example, there were eight employees older than 55, and five of them were laid off.

Finally, we obtain the value of 22C7 as follows:

22C7 = 22!

122 - 72!7! 1n - x + 12 = 122 - 7 + 12 = 16

22C7 = 1222121212021192118211721162

172162152142132122112 = 859,541,760

5,040 = 170,544

This value represents the number of ways that younger employees can be selected to be laid off.

This value represents the number of ways that older employees can be selected to be laid off.

This value represents the total number of ways all 22 employees can be selected 7 at a time. In other words, this is the entire sample space.

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C h a p t e r 5 | 5.4 The Hypergeometric Distribution 231

There are 170,544 ways in which we can choose seven employees randomly from a group of 22. In our example, there were 22 employees in the population, and seven of them were being laid off.

We are now ready for our final result (drum roll, please). The probability that five of the seven laid-off employees will be older than 55 if they are chosen randomly is calculated as follows:

P152 = 14C2 #

8C5

22C7 =

19121562 1170,5442 = 0.0299

What we have learned is that there is less than a 3% chance that five older employees will be in this group of seven who are laid off, if the laid-off employees were randomly selected. It sounds like the lawyers might have a field day with this retailer if they file an age-discrimination case. Results such as these are often used in courts to help substantiate the claims of plaintiffs.

Just like the distributions we discussed previously in this chapter, the hypergeometric distribution also has a mean and a standard deviation, shown with Equations 5.13 and 5.14, respectively.

Formula 5.13 for the Mean of the Hypergeometric Distribution

m = nR

Formula 5.14 for the Standard Deviation of the Hypergeometric Distribution

s = A nR1N - R2N 2 A N - nN - 1 where

N = The population size R = The number of successes in the population n = The sample size For our retailer example, the mean of the distribution represents the expected number of

older employees to be laid off such that the proportion of laid-off employees who are older matches the proportion of older employees in the sales force as a whole. Using Equation 5.13, we have the following:

m = nR

N =

172182 22

= 2.55

If seven employees are laid off randomly, we would expect 2.55 of them to be older based on the number of older employees in the sales force.

The standard deviation of this distribution is found using Equation 5.14:

s = B nR1N - R2N 2 B N - nN - 1 = B 172182122 - 8212222 B 22 - 722 - 1 s = A 784484 A 1521 = 21.619820.7143 = 11.273210.8452 = 1.076

If we randomly select seven employees to be laid off over and over again, s = 1.076 repre- sents the variation we expect to see in the number of older employees in the laid-off group.

Now that we have drained our calculator batteries with all these combination calcula- tions, let me show you a simpler approach with Excel in the following section.

The fact that five older employees were laid off, which is nearly twice as high as expected, confirms the low probability (3%) that laid- off employees were chosen randomly. I smell a lawsuit!

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232 C h a p t e r 5 | Discrete Probability Distributions

Using excel and phStat to Calculate hypergeometric probabilities A much more convenient way to calculate hypergeometric probabilities is to rely on Excel’s HYPGEOM.DIST function which has the following characteristics:

= HYPGEOM.DIST1x, n, R, N, cumulative2 where

N = The population size R = The number of successes in the population n = The sample size x = The number of successes in the sample cumulative = FALSE, if you want to determine the probability of exactly x occurrences cumulative = TRUE, if you want to determine the probability of x or fewer occurrences We can use Excel’s HYPGEOM.DIST function to calculate the probability that exactly

five of the seven laid-off employees will be older than 55 years.

= HYPGEOM.DIST15, 7, 8, 22, FALSE2 = 0.0299 Excel will also calculate the cumulative probability that five or fewer of the seven laid-

off employees will be older than 55 years.

= HYPGEOM.DIST15, 7, 8, 22, TRUE2 = 0.9977 As you can see, this function returns the result 0.0299, matching the probability from

our manual calculations with a whole lot less work! Finally, we can calculate hypergeometric probabilities using the Excel add-in PHStat for

Windows. See the textbook’s Web site for Excel for Mac instructions. The following steps demonstrate this technique for our retailer example:

1. Go to Add-Ins + PHStat + Probability & Prob. Distributions + Hypergeometric, as shown in Figure 5.4.

2. Fill in the Hypergeometric Probability Distribution dialog box as shown in Figure 5.13. Click OK.

{ figure 5.13 Using PHStat to Calculate

Hypergeometric Probabilities

Figure 5.14 shows the hypergeometric probabilities for all the possible values of x (0 to n). Remember, x represents the number of successes from the sample, and n represents the size of the sample. As a result, the probabilities in Column C sum to 1.0. Note that the probability in Cell C15 represents the probability that our sample consisted of five older employees 1x = 52 and that this value matches what we manually calculated (0.0299).

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C h a p t e r 5 | 5.4 The Hypergeometric Distribution 233

PHStat will also display the histogram for a hypergeometric probability. Figure 5.15 shows the histogram for our retailer example.

{ figure 5.14 Final Hypergeometric

Probabilities Calculated Using PHStat

{ figure 5.15 Final hypergeometric

Probabilities Calculated Using PHStat: A Histogram

An assembly process has just produced 16 laptop computers. Three of them have a defect of which the manufacturer is unaware. A school just placed an order for 5 of the 16 computers.

a. Determine the probability that two of the laptops delivered to the school will have the defect. b. Calculate the mean and standard deviation for this distribution. c. Confirm your answer in part a with Excel and PHStat.

Y o U r t U r n # 1 0N

Answers can be found on u pages 245–246

5.4 Section Problems Basic Skills 5.32 Consider a hypergeometric probability distribution with

n = 5, R = 6, and N = 12. Calculate the following probabilities:

a. P1x = 32 b. P1x = 22 c. P1x … 12 d. Calculate the mean and standard deviation of this

distribution.

5.33 Consider a hypergeometric probability distribution with n = 3, R = 5, and N = 9. Calculate the following probabilities:

a. P1x = 02 b. P1x 7 12 c. P1x 6 32 d. Calculate the mean and standard deviation of this

distribution.

Now it’s your turn to see what you have learned in this section. Give this next Your Turn problem your full attention to be sure you can handle a hypergeometric probability distribu- tion on your own.

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234 C h a p t e r 5 | Discrete Probability Distributions

5.34 An urn contains six blue balls and eight yellow balls. If four balls are selected randomly without being replaced, what is the probability that of the balls selected, three of them will be blue and one of them will be yellow?

applications 5.35 I find a bowl of Christmas M&M’s (only red and green) that

Deb has hidden from me in the house one day. I quickly grab a handful before being caught. Before shoving them into my mouth to get rid of the evidence, I notice that I have selected six red M&M’s and no green ones. Let’s say there were 8 green and 10 red M&M’s in the bowl before my discovery. What is the probability of this event occurring?

5.36 White boxers are dogs that have a genetic disposition for going deaf within the first year after they are born. Suppose a litter of nine white boxer puppies contained two dogs that would eventually experience deafness. A family randomly selected two puppies from this litter to take home as family pets. Determine the following probabilities:

a. None of the two puppies selected will experience deafness.

b. One of the two puppies selected will experience deafness.

c. Both puppies selected will experience deafness. d. Calculate the mean and standard deviation of this

distribution. e. Verify these results with Excel.

5.37 A political committee consists of seven Democrats and five Republicans. A subcommittee of six people needs to be formed from this group. Determine the probability that this subcommittee will consist of the following:

a. Two Democrats and four Republicans if they were ran- domly selected

b. Four Democrats and two Republicans if they were ran- domly selected

c. Three Democrats and three Republicans if they were randomly selected

d. Calculate the mean and standard deviation for the distribution which defines selecting a Republican as a success.

e. Verify these results with Excel.

5.38 I have a statistics class of 20 students, of whom 12 are female. I randomly select three students to make a group presentation to the class. What is the probability of the following occurring?

a. All of the students in the group are male. b. One of the students in the group is male. c. Two of the students in the group are male. d. All three students in the group are female. e. Calculate the mean and standard deviation for the

distribution which defines selecting a male as a success.

f. Verify these results with Excel.

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235

■ Discrete random variables have outcomes that take on whole numbers when an experiment is conducted.

■ A discrete probability distribution meets the following conditions:

• Each outcome in the distribution needs to be mutually exclusive.

• The probability of each outcome, P1x2, must be between 0 and 1 (inclusive). • The sum of the probabilities for all the outcomes in the distribution needs to add up to 1.

■ The mean of a discrete probability distribution is a weighted average of the outcomes of the random variables in the distribution.

■ The expected value is another term for the mean of a probability distribution.

■ The expected monetary value (EMV) is the dollar outcome that can be expected based on a probability distribution.

■ The standard deviation of a discrete probability distribution measures the dispersion of the outcomes of the discrete random variables in relation to the distribution mean.

■ A binomial experiment has the following characteristics:

• The experiment consists of a fixed number of trials denoted by n.

• Each trial has only two possible outcomes, a success or a failure.

• The probability of success and the probability of failure are constant throughout the experiment.

• Each trial is independent of any other trial in the experiment.

■ The number of successes in a binomial experiment must range from zero to n.

■ The mean of the binomial distribution represents the long-term average number of successes to expect based on the number of trials.

■ The standard deviation of the binomial distribution represents the variation we would see in the number of successes over n trials, assuming the chance of a success is p.

■ A Poisson process has the following characteristics:

• The experiment consists of counting the number of occurrences of an event over a period of time, area, distance, or any other type of measurement.

• The mean of the Poisson distribution has to be the same for each interval of measurement.

• The number of occurrences during one interval is independent of the number of occurrences in any other interval.

■ Mathematically, with a Poisson distribution, there is no upper limit to the number of occurrences during the interval.

■ The variance of the Poisson distribution is equal to the mean.

■ You can use the equation for the Poisson distribution to approximate the binomial probabilities when the number of trials is greater than or equal to 20 and the probability of a success is less than or equal to 0.05.

■ When samples are randomly selected from a finite population without being replaced, the hypergeo- metric distribution is the appropriate choice rather than calculating a binomial distribution.

• Under these conditions, the probabilities of success change repeatedly because the sample space becomes smaller after each selection.

• The mean of a hypergeometric distribution represents the expected number of successes such that the proportion of successes in the sample matches the proportion of successes in the population.

Chapter 5 Summary

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236

Binomial experiment. An experiment with the following char- acteristics: (1) The experiment consists of a fixed number of trials denoted by n; (2) each trial has only two possible outcomes, a success or a failure; (3) the probability of suc- cess and the probability of failure are constant throughout the experiment; (4) each trial is independent of any other trial in the experiment.

Continuous random variables. Have outcomes that take on any numerical value, including fractions, as a result of conducting an experiment.

Discrete probability distribution. A listing of all the possible outcomes of an experiment for a discrete random variable along with the relative frequency, or probability, of each outcome.

Discrete random variables. Have outcomes that take on whole numbers as a result of conducting an experiment.

Expected monetary value (EMV). The dollar outcome that can be expected based on a probability distribution. Specifically,

the EMV is the weighted dollar average for the distribution if the dollar values are associated with discrete probabilities.

Expected value. Another term for the mean of a discrete prob- ability distribution.

Hypergeometric distribution. A probability distribution that oc- curs when samples are randomly selected from a finite popu- lation without being replaced. Under these conditions, the probabilities of success change repeatedly because the sample space becomes smaller after each selection.

Poisson process. An experiment with the following character- istics: (1) The experiment involves counting the number of occurrences of an event over a period of time, area, distance, or any other type of measurement; (2) the mean of the Poisson distribution has to be the same for each interval of measure- ment; (3) the number of occurrences during one interval is in- dependent of the number of occurrences in any other interval.

236

Chapter 5 Equations

5.1 Introduction to Discrete Probability Distributions (pp 195–206)

Formula 5.1 for the Mean of a Discrete Probability Distribution m = a n

i = 1 xiP1xi2

Formula 5.2 for the Expected Value of a Discrete Probability Distribution E1x2 = m = a n

i = 1 xiP1xi2

Formula 5.3 for the Variance of a Discrete Probability Distribution s 2 = a

i = 1 1xi - m22P1xi2

Formula 5.4 for the Variance of a Discrete Probability Distribution (Shortcut) s2 = a a n

i = 1 x 2i P1xi2b - m2

Formula 5.5 for Combinations of n Objects Selected x at a Time nCx = n!

1n - x2!x!

5.2 Binomial Distributions (pp 207–218)

Formula 5.6 for the Binomial Probability Distribution P1x, n2 = n!

1n - x2!x!p xq n - x

Chapter 5 Key Terms

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Formula 5.7 for Calculating the Mean of a Binomial Distribution m = np

Formula 5.8 for Calculating the Standard Deviation of a Binomial Distribution s = 2npq Formula 5.9 for the Poisson Probability Distribution P1x2 = l

xe - l

Formula 5.10 for the Variance of a Poisson Distribution s2 = l

Formula 5.11 for Using the Poisson Equation to Calculate Binomial Probabilities P1x2 = 1np2 x 2.71828 - 1np2

Formula 5.12 for the Hypergeometric Distribution P1x2 = N - RCn - x #

RCx

NCn

Formula 5.13 for the Mean of the Hypergeometric Distribution m = nR

5.4 The Hypergeometric Distribution (pp 229–234)

5.3 Poisson Distributions (pp 218–228)

Formula 5.14 for the Standard Deviation of the Hypergeometric Distribution s = A nR1N - R2N2 A N - nN - 1 Chapter 5 Problems

5.39 AIG, an international insurance and financial services organization, recently announced that the performance of employees would be rated on a scale from 1 to 4. The top 10% of employees would receive a 1 rating, 20% of the employees would be rated with a 2, and 50% would be rated with a 3. The remaining employees would receive a 4 rating. The rating scheme is known as a forced-ranking system. It was begun after the government bailed out AIG and forced the company to link compensation to perfor- mance. Suppose the Wilmington, Delaware, facility of AIG has 450 employees. Determine the number of employees who can receive each rating.

5.40 The following distribution list displays the number of no-shows for a particular Delta Airlines flight and the

historical probability that each number of no-shows will occur. A no-show is a person who has a reservation for a flight but fails to arrive prior to its departure.

No-Shows Probability

0 0.23

1 0.32

2 0.22

3 0.15

4 0.08

a. Determine the mean number of no-shows for this flight. b. Determine the standard deviation for the number of no-

shows for this flight.

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5.41 A survey was conducted to determine the number of cats living in individual households. The results are summarized in the following table:

Number of Cats Probability

0 0.37

1 0.28

2 0.21

3 0.10

4 0.04

a. Determine the mean number of cats per household. b. Determine the standard deviation for the number of cats

per household.

5.42 I am currently teaching two statistics classes, one at 8:00 a.m. and the other at 10:00 a.m. The following table summarizes my attendance records by showing the probability of the number of absent students per class:

Probability

Number of Absent Students 8 a.m. Class 10 a.m. Class

0 0.06 0.16

1 0.19 0.32

2 0.25 0.25

3 0.20 0.16

4 0.21 0.06

5 0.09 0.05

a. Calculate the mean number of students absent for each class. b. Calculate the standard deviation for the number of

students absent for each class. c. Are these results to be expected?

5.43 Travel insurance reimburses travelers for the cost of their trips if the trips are canceled for a variety of reasons. If a $10,000 policy costs $400 and you estimate that there is a 3% chance your trip will be canceled, should you purchase the insurance?

5.44 An investor has two portfolios to consider. Portfolio A is more heavily invested in bonds, whereas Portfolio B con- tains more stocks. The following table provides the annual returns for both portfolios under three different market conditions as well as the probability for each market con- dition in the upcoming year. Which portfolio should the investor choose to maximize her annual return?

Market Conditions

Portfolio Weak Normal Strong

A ($) -1,500 2,500 4,000 B ($) -4,000 3,000 6,000 Probabilities (%) 25 60 15

5.45 Tees R Us, which manufactures and sells T-shirts for sport- ing events, is providing shirts for an upcoming tournament. Each shirt will cost $10 to produce and will be sold for $16. Any unsold shirts left over at the end of the tournament can be sold for $5 apiece in the near future. Tees R Us assumes the demand for the shirts will be 1,000, 2,000, 3,000, or 4,000. The company also estimates that the probabilities of each of these sales level occurring will be 15%, 25%, 30%, and 30%, respectively. Determine the expected monetary value of the project if Tees R Us chooses to print 3,000 shirts for the tournament.

5.46 The following table shows Bob’s Bookstore’s estimated demand for a new calendar. The bookstore needs to decide whether to order 100, 200, or 300 calendars for the start of the year. Each calendar costs the store $6 to purchase and can be sold for $14. The store can sell any unsold calendars back to its supplier for $2 each. Determine the number of calendars Bob’s Bookstore should order to maximize its expected monetary value.

Demand Probability

100 0.25

200 0.40

300 0.35

5.47 At the start of each Super Bowl, the NFL championship game between the winners of the NFC and AFC, a coin toss determines the team that will receive the opening kickoff. The NFC team won the coin toss 14 straight years from 1998 until 2011. Determine the probability of one team winning this coin toss 14 straight times.

5.48 A recent survey found that only 45% of Americans were sat- isfied with their current jobs. Answer the following questions based on a random sample of 15 American workers:

a. What is the probability that exactly six employees were satisfied with their current jobs?

b. What is the probability that exactly nine employees were satisfied with their current jobs?

c. What is the probability that either four or five employees were satisfied with their current jobs?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.49 A foreclosure is a legal process that a lender uses to recover the balance of a mortgage owed by a homeowner who has stopped making payments. An unusually high volume of foreclosures occurred during the most recent recession, and as a result, errors were made in the foreclosure process to a significant number of borrowers. For example, auditors found that 24% of foreclosures processed by PNC Financial Services had errors that would have required compensation for the borrowers. A random sample of five foreclosures was selected from PNC Financial Services.

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a. What is the probability that exactly three foreclosures from this sample will have an error that would have re- quired compensation?

b. What is the probability that none of the foreclosures from this sample will have an error that would have re- quired compensation?

c. What is the probability that two or fewer foreclosures from this sample will have an error that would have re- quired compensation?

d. Construct a histogram for this distribution.

5.50 The Center for Disease Control and Prevention estimated that 37% of Americans received a flu shot during the winter of 2012–2013. A random sample of seven Americans was selected.

a. What is the probability that exactly four people from this sample received a flu shot that winter?

b. What is the probability that more than two people from this sample received a flu shot that winter?

c. What are the mean and standard deviation for this distribution?

d. Construct a histogram for this distribution.

5.51 A golfer claims that 80% of the time, he will score less than 90 on the golf course he normally plays. Consider his next 10 rounds of golf.

a. What is the probability that he will score lower than 90 on all 10 rounds?

b. What is the probability that he will score lower than 90 on eight or more rounds?

c. What is the probability that he will score lower than 90 on either six or seven rounds?

d. Construct a histogram for this distribution. e. Suppose that out of the 10 rounds, he scores lower than

90 on five of them. What conclusions can be drawn about the golfer’s claim?

5.52 In 2012, 79% of Southwest Airlines flights arrived at their destinations on time. Suppose a random sample of 12 Southwest Airlines flights was selected.

a. What is the probability that all 12 flights were on time? b. What is the probability that more than nine flights were

on time? c. What is the probability that seven or fewer flights were

on time? d. What are the mean and standard deviation for this

distribution? e. Construct a histogram for this distribution.

5.53 According to the NPD Group, 70% of the basketball shoes purchased by American men are never worn on a basketball court. Answer the following questions based on a random sample of 13 men who purchased basketball shoes:

a. What is the probability that 12 men will not use their shoes on a basketball court?

b. What is the probability that more than 10 men will not use their shoes on a basketball court?

c. What is the probability that less than 12 men will not use their shoes on a basketball court?

d. Construct a histogram for this distribution. e. Suppose that out of the 13 men, 5 will not use their

shoes on a basketball court. What conclusions can be drawn about the claim by the NPD Group?

5.54 According to the University Business Officers, 86% of first-time, full-time freshman at private colleges were the recipients of some form of financial aid in 2011. A random sample of six freshmen in 2011 was selected.

a. What is the probability that all six freshmen from this sample received financial aid in 2011?

b. What is the probability that four or more freshmen from this sample received financial aid in 2011?

c. What is the probability that less than five freshmen from this sample received financial aid in 2011?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.55 Regifting is defined as the act of taking a gift that has been given to you and presenting it as a gift to someone else. Ac- cording to survey conducted by American Express regard- ing consumer spending, 79% of the respondents felt that it was socially acceptable to regift during the holiday season (big sigh of relief). Suppose a random sample of 10 holiday shoppers was selected.

a. What is the probability that nine shoppers from this sample felt that regifting was socially acceptable?

b. What is the probability that more than seven shop- pers from this sample felt that regifting was socially acceptable?

c. What is the probability that less than 10 shoppers from this sample felt that regifting was socially acceptable?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.56 When iPhone was first released in 2007, Google Maps was the default mapping software that was included in Apple’s mobile operating system. In September 2012, Apple re- leased an upgraded operating system for the iPhone which replaced Google Maps with its own newly developed Apple Maps. Unfortunately, Apple Maps was plagued with many mapping errors and did not contain all of the features found in Google Maps. In a December 2012 survey of iPhone users, 45% of the respondents said that they “hated Apple Maps.” Using the binomial tables, answer the following questions based on a random sample of 20 iPhone users:

a. What is the probability that 12 iPhone users from this sample hated Apple Maps?

b. What is the probability that more than 12 iPhone users from this sample hated Apple Maps?

c. What is the probability that less than 10 iPhone users from this sample hated Apple Maps?

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d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.57 According to the Pew Research Center, 23% of American adults read an e-book during 2012. A random sample of nine adults was selected.

a. What is the probability that exactly two adults from this sample read an e-book last year?

b. What is the probability that less than four adults from this sample read an e-book last year?

c. What is the probability that five or more adults from this sample read an e-book last year?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.58 According to the Graduate Management Admission Council, applications to two-year MBA programs have been on the decline during the past few years. A survey during 2012 reported that 62% of U.S. schools experienced fewer appli- cations when compared to 2011. A random sample of eight U.S. schools with a two-year MBA program was selected.

a. What is the probability that exactly two schools from this sample experienced a decline in applications in 2012?

b. What is the probability that more than five schools from this sample experienced a decline in applications in 2012?

c. What is the probability that three or fewer schools from this sample experienced a decline in applications in 2012?

d. What are the mean and standard deviation for this distribution?

e. Construct a histogram for this distribution.

5.59 A first draft of a 400-page manuscript has 36 typos that are spread randomly across the pages. Assume the number of typos per page follows the Poisson distribution.

a. What is the probability that there is one typo in the first 10 pages?

b. What is the probability that there is one typo in the first 20 pages?

c. What is the probability that there are four typos in the first 50 pages?

d. What is the probability that there are 12 typos in the first 110 pages?

5.60 Methodist Hospital serves as a major trauma center for Indianapolis with one of the largest emergency depart- ments in the country, averaging 11.0 ED patients per hour. Assume the number of ED patients arriving per hour follows the Poisson distribution.

a. What is the probability that six ED patients will arrive during the next hour?

b. What is the probability that three ED patients will arrive during the next 30 minutes?

c. What is the probability that seven or more ED patients will arrive during the next hour? (I suggest using Excel or the Poisson tables for this question.)

d. What is the probability that less than 10 ED patients will arrive during the next hour? (I suggest using Excel or the Poisson tables for this question.)

5.61 In college basketball, a turnover is defined as losing posses- sion of the basketball to the opposing team. During a recent basketball season, the Villanova Wildcats averaged 13 turn- overs per game. Assume the number of turnovers per game follows the Poisson distribution.

a. What is the probability that Villanova will have eight turnovers during a game?

b. What is the probability that Villanova will have less than 10 turnovers during a game? (I suggest using Excel for this problem.)

c. What is the probability that Villanova will have more than 15 turnovers during a game? (I suggest using Excel for this problem.)

d. What is the probability that Villanova will have four turnovers during the first half of a game?

5.62 Customers arrive at a local ice cream stand at an average rate of 16 per hour and follow a Poisson distribution.

a. What is the probability that two customers will arrive during the next 15 minutes?

b. What is the probability that four customers will arrive during the next 15 minutes?

c. What is the probability that six customers will arrive during the next 30 minutes?

d. What is the probability that six customers will arrive during the next 45 minutes?

5.63 According to the Department of Transportation, United Airlines lost an average of 3.87 bags per 1,000 passengers during 2012. Assume the number of lost bags follows the Poisson distribution.

a. What is the probability that, over the next 1,000 United passengers, nobody will lose their bag?

b. What is the probability that, over the next 1,000 United passengers, less than three will lose their bag?

c. What is the probability that, over the next 500 United passengers, nobody will lose their bag?

d. What is the probability that, over the next 500 United passengers, less than three will lose their bag?

5.64 According to the National Hurricane Center, the average number of hurricanes during the Atlantic season, which runs from June 1 to November 30, was 6.5 hurricanes per year from 1982 to 2012. Assume the number of hurricanes per year follows the Poisson distribution.

a. What is the probability that one hurricane will occur in the Atlantic next year?

b. What is the probability that less than three hurricanes will occur in the Atlantic next year?

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c. What is the probability that more than four hurricanes will occur in the Atlantic next year?

d. What is the standard deviation for this distribution?

5.65 Harv& Marv’s Outback Alaska provides personalized 4-hour or 8-hour whale watching tours in Juneau Alaska. (Count Deb and I as two very satisfied customers.) Suppose the number of whales spotted during a 4-hour tour follows the Poisson distribution and averages 5.3.

a. What is the probability that exactly three whales will be spotted during the next 4-hour tour?

b. What is the probability that less than four whales will be spotted during the next 4-hour tour?

c. What is the probability that more than five whales will be spotted during the next 4-hour tour?

d. What is the probability that exactly six whales will be spotted during the next 8-hour tour?

5.66 According to Experian, 2.5% of car-loan payments were 30 days past due in 2012, the lowest percentage since 2007. Lenders prefer auto loans to home mortgages because, when borrowers get into financial trouble, they tend to pay the auto loan first because cars are more easily seized by banks. A random sample of 24 car loans was randomly selected.

a. Use the binomial distribution to determine the probabil- ity that exactly one loan is 30 days past due from this sample.

b. Use the binomial distribution to determine the probabil- ity that exactly four loans are 30 days past due from this sample.

c. Use the Poisson distribution to determine the probabil- ity that exactly one loan is 30 days past due from this sample.

d. Use the Poisson distribution to determine the probability that exactly four loans are 30 days past due from this sample.

5.67 According to hybridcars.com, the market share of hybrid cars in the United States in February 2013 is 3.4%. Con- sider a random sample of 20 cars in the United States.

a. Use the binomial distribution to determine the probabil- ity that there are no hybrid cars from this sample.

b. Use the binomial distribution to determine the prob- ability that there are less than two hybrid cars from this sample.

c. Use the Poisson distribution to determine the probability that there are no hybrid cars from this sample.

d. Use the Poisson distribution to determine the probability that there are less than two hybrid cars from this sample.

5.68 The U.S. Census reports that 3.5% of households have at least three generations living under one roof. Consider a random sample of 24 households.

a. Use the binomial equation to determine the probability that three households have at least three generations liv- ing in the same house.

b. Use the Poisson equation to determine the probability that three households have at least three generations liv- ing in the same house.

c. How do these two probabilities compare?

5.69 A grocery store has 18 two-liter bottles of diet cola on the shelf, four of which have exceeded their shelf-life date. You randomly select three bottles of diet cola without check- ing the expiration date. Determine the probability of the following:

a. None of the bottles have exceeded their expiration date. b. At least two of the bottles have exceeded their expiration

date. c. Calculate the mean and standard deviation of this

distribution. d. Verify your results with Excel.

5.70 An IRS agent has 19 tax returns on her desk, of which 14 were filed using the short form (1040A). She randomly chooses eight returns to audit. Determine the probability of the following:

a. All eight were filed using the short form. b. At least six were filed using the short form. c. Explain why it is not possible for none of the tax returns

to be filed using the short form. d. Calculate the mean and standard deviation of this

distribution. e. Verify your results with Excel.

5.71 My department at school is comprised of 10 faculty and 6 administrators. I need to organize a committee of six peo- ple from the department. If I randomly select these commit- tee members, determine the probability that the committee consists of the following:

a. Two faculty and four administers b. Four faculty and two administers c. All faculty and no administrators d. Calculate the mean and standard deviation for the distri-

bution which defines selecting a faculty as a success. e. Verify your results with Excel.

5.72 Winner Automotive has a current inventory of 14 used cars and 8 used trucks. Management needs to select 10 used vehicles to participate in a special used vehicle sale at the local shopping mall. If the vehicles are selected randomly, determine the probability of the following occurring:

a. Four trucks have been selected. b. Four cars have been selected. c. Explain why these probabilities are so different. d. Calculate the mean and standard deviation for the distri-

bution which defines selecting a truck as a success. e. Verify your results with Excel.

5.73 Packer Fan Tours is the official tour company for the Green Bay Packers of the NFL. One of the events in its package is to sponsor a reception the night before a game for fans

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that is attended by three of the players from the team. There are 53 players on the Green Bay Packers’ roster, of which 22 are starters. Assume that the three players attending the reception this week are chosen randomly. Determine the probability of the following occurring:

a. None of the players at the reception are starters. b. All of the players at the reception are starters. c. Two of the players at the reception are starters. d. Calculate the mean and standard deviation of this

distribution. e. Verify your results with Excel.

5.74 In a standard 52-deck of cards, there are 12 face cards con- sisting of four jacks, queens, and kings. You are randomly dealt a five-card hand from the deck. What is the probabil- ity that your hand will contain the following?

a. No face cards b. One face card c. Two face cards d. Four face cards e. Calculate the mean and standard deviation of this

distribution. f. Verify your results with Excel.

Chapter 5 Solutions to Your Turn

Y o u r T u r n # 1 a.

Inter Wins Freq

Wins Probability

Losses Freq

Losses Probability

0 91 91>182 = 0.500 12 12>105 = 0.114 1 59 59>182 = 0.324 37 37>105 = 0.352 2 23 23>182 = 0.126 29 29>105 = 0.276 3 9 9>182 = 0.049 21 21>105 = 0.200 4 0 0>182 = 0.000 5 5>105 = 0.048 5 0 0>182 = 0.000 1 1>105 = 0.010 Total 182 0.999 105 1.000

b. Wins:

m = a n

i = 1 xiP1xi2

m = 10210.52 + 11210.3242 + 12210.1262 + 13210.0492 + 14210.02 + 15210.02 m = 0.0 + 0.324 + 0.252 + 0.147 + 0.0 + 0.0 m = 0.723

c. Losses:

m = a n

i = 1 xiP1xi2

m = 10210.1142 + 11210.3522 + 12210.2762 + 13210.22 + 14210.0482 + 15210.0102 m = 0.0 + 0.352 + 0.552 + 0.6 + 0.192 + 0.05 m = 1.746

Brett Favre averaged 0.723 interceptions per game during games that his team won and 1.746 interceptions per game during games that his team lost.

Wins Losses

xi x2i P1xi2 x2i P1xi2 P1xi2 x2i P1xi2 0 0 0.500 0.000 0.114 0.000

1 1 0.324 0.324 0.352 0.352

2 4 0.126 0.504 0.276 1.104

3 9 0.049 0.441 0.200 1.800

4 16 0 0.000 0.048 0.768

5 25 0 0.000 0.010 0.250

Total 1.269 4.274

Wins:

a n

i = 1 x i 2P1xi2 = 1.269

s2 = a a n

i = 1 x i 2P1xi2b - m2

s2 = 1.269 - 10.72322 = 1.269 - 0.523 = 0.746 s = 2s2 = 20.746 = 0.864 Losses:

a n

i = 1 x i 2P1xi2 = 4.275

s2 = a a n

i = 1 x i 2P1xi2b - m2

s2 = 4.274 - 11.74622 = 4.274 - 3.049 = 1.225 s = 2s2 = 21.225 = 1.107 e. Favre averaged one more interception per game during games that his team

lost compared with those that his team won. His interceptions per game were more consistent during wins than they were during losses based on the stan- dard deviations of the two distributions.

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Y o u r T u r n # 2

EMV = m = a n

i = 1 xiP1xi2

EMV = 1$20210.652 + 1 - $25210.352 EMV = $4.25

Because the EMV is positive, you should take the bet.

Y o u r T u r n # 3

Y o u r T u r n # 4

a. P1x,n2 = n!1n - x2!x! p xq n - x

P14,52 = 5!15 - 42!4!10.842 410.1625 - 4

P14,52 = 152142132122112 11214213212211210.842

410.1621

P14,52 = 15210.4979210.162 = 0.3983 There is a 39.8% chance that exactly four out of five customers will be satisfied.

b. P1x,n2 = n!1n - x2!x! p xq n - x

P15,52 = 5!15 - 52!5!10.842 510.1625 - 5

P15,52 = 152142132122112

11215214213212211210.842 510.1620

P15,52 = 11210.41822112 = 0.4182 P102 + P112 + P122 + P132 = 1.0 - P142 - P152 P102 + P112 + P122 + P132 = 1.0 - 0.3983 - 0.4182 = 0.1835 There is an 18.35% chance that less than four out of five customers will be

satisfied. c. m = np = 15210.842 = 4.2 d. s = 2npq = 215210.84210.162 = 0.820 e. The probability of each customer being satisfied is the same and the satis-

faction of customers is independent of one another.

n x 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 7 0 0.4783 0.3206 0.2097 0.1335 0.0824 0.0490 0.0280 0.0152 0.0078

1 0.3720 0.3960 0.3670 0.3115 0.2471 0.1848 0.1306 0.0872 0.0547

2 0.1240 0.2097 0.2753 0.3115 0.3177 0.2985 0.2613 0.2140 0.1641

3 0.0230 0.0617 0.1147 0.1730 0.2269 0.2679 0.2903 0.2918 0.2734

4 0.0026 0.0109 0.0287 0.0577 0.0972 0.1442 0.1935 0.2388 0.2734

5 0.0002 0.0012 0.0043 0.0115 0.0250 0.0466 0.0774 0.1172 0.1641

6 0.0000 0.0001 0.0004 0.0013 0.0036 0.0084 0.0172 0.0320 0.0547

7 0.0000 0.0000 0.0000 0.0001 0.0002 0.0006 0.0016 0.0037 0.0078

a. There is a 29.03% chance that three of the next seven cars will use E-ZPass. b. The probability that at least five cars will use E-ZPass is

0.0774 + 0.0172 + 0.0016 = 0.0962.

Y o u r T u r n # 5 a. Use = BINOM.DIST110, 16, 0.53, FALSE2 in Excel to find the probability = 0.1510. b. Use = BINOM.DIST19, 16, 0.53, TRUE2 in Excel to find the probability = 0.6932. c. P1x 7 52 = 1 - P1x … 52 = 1 - BINOM.DIST15, 16, 0.53, TRUE2 = 1 - 0.0674 = 0.9326

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Y o u r T u r n # 6 a. l = 7.4 x = 3

P132 = l xe - l

x! =

17.43212.71828 - 7.42 3!

P132 = 1405.224210.0006112

132122112 = 0.0413

b. P1x … 32 = P102 + P112 + P122 + P132 P102 =

17.40212.71828 - 7.42 0!

= 11210.0006112

1 = 0.0006

P112 = 17.41212.71828 - 7.42

1! =

17.4210.0006112 1

= 0.0045

P122 = 17.42212.71828 - 7.42

2! =

154.76210.0006112 2

= 0.0167

P1x … 32 = 0.0006 + 0.0045 + 0.0167 + 0.0413 = 0.0631

c. P1x Ú 52 = 1.0 - P102 - P112 - P122 - P132 - P142 P142 =

17.44212.71828 - 7.42 4!

= 12,998.6576210.0006112

142132122112 = 0.0763 P1x Ú 52 = 1.0 - 0.0006 - 0.0045 - 0.0167 - 0.0413 - 0.0763 = 0.8606

d. l = 7.4

2 = 3.7 x = 3

P132 = l xe - l

x! =

13.72312.71828 - 3.72 3!

P132 = 150.653210.0247242

132122112 = 0.2087

Y o u r T u r n # 7

x 0.005 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0 0.9950 0.9900 0.9802 0.9704 0.9608 0.9512 0.9418 0.9324 0.9231 0.9139

1 0.0050 0.0099 0.0196 0.0291 0.0384 0.0476 0.0565 0.0653 0.0738 0.0823

2 0.0000 0.0000 0.0002 0.0004 0.0008 0.0012 0.0017 0.0023 0.0030 0.0037

3 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001

a. P112 = 0.0476 b. The probability that there is a defect is P1x Ú 12 = 1.0 - P102 = 1.0 - 0.9512 = 0.0488.

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Y o u r T u r n # 8 a. = POISSON.DIST11, 1.4, FALSE2 = 0.3452 b. P1x 6 22 = P1x … 12 = POISSON.DIST11,1.4,TRUE2 = 0.5918 c. P1x Ú 42 = 1 - P1x … 32 = 1 - POISSON.DIST13,1.4,TRUE2 = 0.0537 d.

Y o u r T u r n # 9

a. P1x,n2 = n!1n - x2!x! p xq n - x

P13,202 = 20!120 - 32!3!10.032 310.97220 - 3

P13,202 = 1202119211821172 g 132122112 1172 g 132122112132122112 10.032

310.97217

P13,202 = 120211921182 132122112 10.000027210.5958262 = 0.0183

b. np = 120210.032 = 0.6

P1x2 = 1np2x2.71828 - 1np2

P132 = 10.6232.71828 - 10.62

= 10.216210.5488122

132122112 = 0.0198

c. 1.83% and 1.98% are very close to one another.

d. P142 = 10.6242.71828 - 10.62

= 10.1296210.5488122

142132122112 = 0.0030

If the true probability of not correctly filling an order is 3%, there is only a 0.3% chance that 4 orders from the first 20 are incorrect. Therefore, it is doubtful that the true probability of not correctly filling an order is 3%.

Y o u r T u r n # 1 0 a. N = The total number of computers = 16 R = The number of defective computers = 3 n = The number of computers ordered by the school = 5 x = The number of defective computers delivered to the school = 2

P1x2 = N - RCn - x #

RCx

NCn

P122 = 16 - 3C5 - 2 # 3C2

16C5 =

13C3 # 3C2 16C5

nCx = n!

1n - x2 !x!

= n1n - 121n - 22 g 1n - x + 12

13C3 = 13!

113 - 32!3! 1n - x + 12 = 113 - 3 + 12 = 11

13C3 = 113211221112 132122112 =

1,716

6 = 286

3C2 = 3!

13 - 22!2! 1n - x + 12 = 13 - 2 + 12 = 2

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3C2 = 132122112 122112 =

2 = 3

16C5 = 16!

116 - 52!5! 1n - x + 12 = 116 - 5 + 12 = 12 13C3 =

11621152114211321122 152142132122112 =

524,160

120 = 4,368

P122 = 13C3 # 3C2

16C5 =

12862132 4,368

= 0.1964

b. m = nR

N =

152132 16

= 0.9375

sA nR1N - R2N 2 A N - nN - 1 = A 152132116 - 3211622 A16 - 516 - 1 sA195256 A1115 = 20.761720.7333 = 10.873210.8562 = 0.747

c. = HYPGEOM.DIST = 12, 5, 3, 162 = 0.1964

The probability that x = 4 and x = 5 equals zero because there are only three defective computers in the population.

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I n t H I s c H A P t e R , Y o U w I l l l e A R n t o :

■ Understand the properties of continuous random variables.

■ Identify the characteristics of the normal probability distribution.

■ Calculate normal probabilities.

■ Interpret z-scores for the normal probability distribution.

■ Use the normal distribution to approximate the binomial distribution.

■ Identify the characteristics of the exponential probability distribution.

■ Calculate exponential probabilities.

■ Identify the characteristics of the uniform probability distribution.

■ Calculate uniform probabilities.

Continuous Probability Distributions

C H A P t E R 6 m A P

6.1 Continuous Random Variables | 248

6.2 Normal Probability Distributions | 250

6.3 Exponential Probability Distributions | 269

6.4 Uniform Probability Distributions | 273

6 C H A P t E R

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248 C h a p t e r 6 | Continuous Probability Distributions

Our focus in Chapter 5 was on discrete data. The values for discrete data are typically whole numbers such as 0, 1, or 2. They are found as a result of counting observations while con- ducting an experiment. By contrast, a continuous data value can take on any real number, including fractional values. These data are usually measured rather than counted. Examples of continuous data include time, distance, and weight. This chapter explores the probability of continuous data values occurring. Discrete probability distributions work well when you have small counts of observations; however, businesses are often faced with large quantities that they must measure—hence the need for continuous probability distributions.

Consider the following example: During the winter of 2009–2010, the Philadelphia area—the largest metropolitan area near my home in Delaware—experienced record- breaking snowfall amounts. Because we measure snowfall levels with values that can include decimal points, the data are considered continuous. Between 1884, the first year snowfall amounts were recorded, and 2012, Philadelphia averaged 22.3 in. of snow per winter. During the 2009–2010 winter season, Philadelphia received 78.7 in. of snow. Two major storms in February alone resulted in unprecedented snowfall levels, wreaking havoc with both govern- ments and businesses in the Northeast.

The cost of removing the extraordinary amount of snow in the Philadelphia area that year led to budget shortfalls. Airlines estimated that the winter storms in the Northeast cor- ridor cost $100 million due to flight delays and cancellations. General Motors announced that February 2010 car sales in the Northeast United States were down 22% because of these weather incidents. Apparently, selling cars that are buried under three feet of snow is dif- ficult. Later in the chapter, you will learn how to calculate the probability of certain snowfall amounts based on historical data.

As I was examining the 128 years of Philadelphia snowfall data for this chapter, my at- tention was drawn to the one year in which there was no snowfall: the winter of 1972–1973. Coincidentally, in the fall of 1972, as an ambitious college student I decided to invest my en- tire savings into a used truck and snowplow. My dreams of rolling in money gradually faded away as week after snowless week went by that winter.

After Philadelphia’s record-setting winter in 2010, I suggested to Deb, my wife, that we purchase a snow blower for next season. She just laughed and reminded me of how well my last snow-removal investment had worked out. (Although I explained to her how low the probability was of no snowfall occurring and cleverly reminded her that probability has no memory—a concept discussed in Chapter 4—there doesn’t appear to be a snow blower in my near future.)

6.1 Continuous Random Variables Experiments that involve continuous data generate continuous random variables, which are outcomes that take on any numerical value in an interval, including numbers with decimal points. Tracking the number of calories consumed daily by a middle-aged statistics author who is working late at night on a continuous probability chapter is an example of such an experiment. Some examples of experiments in the business environment that generate continuous random variables are as follows:

• Measuring the weight of a randomly selected box of Cheez-It crackers at a manufacturing plant

• Recording the flight time of a American Airlines flight bound from Philadelphia to Orlando

• Recording the amount of time a customer spends on the phone with a cable company’s customer service representative

These experiments could generate continuous random variables, such as the following:

• 16.12 ounces of Cheez-It crackers in a selected box

• 2.4 hours of flight time for a American Airlines flight bound from Philadelphia and Orlando

• 6.49 minutes spent by a customer on the phone with a customer service representative

Continuous random variables are outcomes that take on any numerical value in an interval, as determined by conducting an experiment.

The road behind your author’s home after a major snowstorm.

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C h a p t e r 6 | 6.1 Continuous Random Variables 249

The number of ounces of Cheez-Its and hours of flight time are continuous “random” variables because they cannot be known with certainty before conducting the experiments. The values 16.12 and 2.4 are the values that these random variables have taken on during the experiment. The purpose of this chapter is to identify the probability that specific continuous random variables will occur using continuous probability distributions.

The value of a random variable is often denoted by x. When we discussed discrete random variables in Chapter 5, we calculated the probability that x would equal a specific value. For instance, if we flip a coin once, the probability of observing exactly one head is P1x = 12 = 0.5. However, continuous random variables can take on an infinite number of values. The flight time from Philadelphia to Orlando can be 2.4, 2.45, 2.451, or 2.4519 hours, for example.

The value of a continuous random variable depends on the level of precision in its mea- surement. Keep in mind that we aren’t counting flights that are late or that arrive on time like we did with discrete probability distributions. We are measuring an interval. Because there are an infinite number of possible times, the probability of one specific flight time occurring is theoretically equal to zero or P1x = 2.42 = 0. Some of my students struggle with this concept, so let me try to explain further. Because there are an infinite number of possible flight times, the probability of any one specific flight time occurring is so small; it is essen- tially equal to zero. We can use Equation 4.2 from Chapter 4 to calculate this probability as follows:

P1x = 2.42 = Frequency in which x = 2.4 occurs in the interval Total number of possible values of x in the interval

P1x = 2.42 = 1 ∞

≈ 0.000

This holds true for any specific flight time, not just x = 2.4 hours. However, we can calculate the probability that a continuous random variable will fall within a specific inter- val, such as P1x … 2.42. Calculating this type of probability will be the focus of much of this chapter.

Also, because P1x = 2.42 = 0, we can also claim that P1x … 2.42 = P1x = 2.42 + P1x 6 2.42 P1x … 2.42 = 0 + P1x 6 2.42 P1x … 2.42 = P1x 6 2.42

For continuous distributions, the probability that x is less than or equal to a value is the same as the probability that x is less than that value.

Continuous probability distributions can have a variety of shapes, depending on their data values. Figure 6.1 shows the shapes of three common continuous distributions that will be discussed in this chapter.

Continuous variables can take on values between whole integers that contain decimal points.

The probability of a continuous random variable being a specific value is always equal to zero.

{ figure 6.1 Common Continuous

Probability Distributions

f (x)

(a) Normal Distribution (b) Exponential Distribution (c) Uniform Distribution

The normal probability distribution seen in Figure 6.1 is useful when the data tend to fall into the center of the distribution and when very high and very low values are fairly rare. You will see this type of distribution in a quality control setting, such as filling boxes of Frosted Flakes cereal.

The exponential distribution is used to describe data where lower values tend to domi- nate and higher values don’t occur very often. The time between customer arrivals is a com- mon application for the exponential distribution.

Because it is so common, the normal probability distribution is a statistician’s workhorse. It will be used to calculate many types of inferential statistics throughout the rest of this textbook.

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250 C h a p t e r 6 | Continuous Probability Distributions

Finally, the uniform distribution describes data where all the values have the same chance of occurring. Download times on the Internet could be an application where the uniform distribution would be useful.

Keep in mind that this is not an exhaustive list of continuous distributions. There are many others that could be used in a business setting. The following section will introduce the most common type of continuous probability distribution, the normal distribution.

6.2 Normal Probability Distributions As we saw in Figure 6.1, the normal probability distribution is useful when the majority of the data tend to be in the middle of the distribution. We will use this distribution repeatedly throughout this text, so it would be a good idea for you to become very familiar with it by studying the following sections.

Characteristics of the Normal probability Distribution A continuous random variable that follows the normal probability distribution has several distinctive features. Let’s say the number of minutes a customer spends on the phone with customer service for a cable company follows the normal distribution with a mean of 12 minutes 1m2 and a standard deviation of 2 minutes 1s2. The probability distribution for such a random variable is shown in Figure 6.2.

{ figure 6.2 A Normal Distribution with

a Mean of 12 and a Standard Deviation of 2

0 3 6 9 12 15 18 21 24

Minutes

0.5 0.5

= 2

f (x)

We can make the following observations about the normal probability distribution like the one shown in Figure 6.2:

• The distribution is bell-shaped and symmetrical around the mean.

• Because the shape of the distribution is symmetrical, the mean and median are the same value, in this case 12 minutes (see Chapter 2).

• Random variables around the mean, where the curve is the tallest, have a higher like- lihood of occurring than values toward the ends of the distribution, where the curve is shorter.

• The total area under the curve is always equal to 1.0.

• Because the distribution is symmetrical around the mean, the area to the left of the mean equals 0.5, as does the area to the right of the mean.

• The left and right ends of the normal probability distribution extend indefinitely, never quite touching the horizontal axis.

The standard deviation plays an important role in the shape of the curve. Looking at Figure 6.2, we can see that nearly all the phone calls would last between 6 and 18 min- utes. Now look at Figure 6.3, which shows the effect of changing the standard deviation from 2 minutes (blue curve) to 3 minutes (red curve). The mean of both of these normal distributions is still 12 minutes.

Normal probability distribution. A con tinuous distribution that is bell shaped and symmetrical around its mean.

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C h a p t e r 6 | 6.2 Normal Probability Distributions 251

In each of the previous figures, the characteristics of the normal probability distribution hold true. In each case, the values of m, the mean, and s, the standard deviation, completely describe the shape of the distribution.

The mathematical expression that describes the shape of normal curves like those shown in the last three figures is known as the normal probability density function. It is shown in Equation 6.1.

A distribution’s mean (m) and standard deviation (s) completely describe its shape.

Normal probability density function. The mathematical expression that describes the shape of the normal probability distribution.

{ figure 6.3 The Effect of Changing a

Normal Distribution’s Standard Deviation from 2 to 3 (While Leaving the Mean at 12)

0 3 6 9 12 15 18 21 24

Minutes

27 µ

f (x)

= 3

= 2

{ figure 6.4 Normal Distributions with

Means of 12 and 21 and a Standard Deviation of 2

0 3 6 9 12 15 18 21 24

Minutes

27 µ µ

= 2

f(x)

In this figure, you see a red curve, with a standard deviation of 3 1s = 32. Notice that it is more spread out around the mean than the blue curve, which has a standard deviation of 2 1s = 22. The red curve shows that almost all of the call times are between 3 and 21 min- utes. What this shows us is that a smaller standard deviation results in a skinnier curve that is tighter and taller around the mean. By contrast, a larger s (standard deviation) makes for a fatter curve that is more spread out and not as tall.

Figure 6.4 shows the impact of changing the mean of the distribution from 12 to 21 minutes, leaving the standard deviation at 2 minutes. Notice that the curve is simply shifted to the right.

Formula 6.1 for the normal probability Density Function

f1x2 = 1 s12p e - 11>2231x - m2>s42

where

e = 2.71828 p = 3.14159 m = The mean of the distribution s = The standard deviation of the distribution x = Any continuous number of interest

Fortunately for us, we will not have to use this intimidating equation to calculate prob- abilities. Instead, we can calculate the probability of normal distributions using tables and Microsoft Excel. Let’s move on to the next section, and I will show you how this is done. Hint: It involves converting a normal distribution to the standard normal distribution, dis- cussed next.

A smaller standard deviation results in a skinnier curve that is tighter and taller around the mean. By contrast, a larger s (standard deviation) makes for a fatter curve that is more spread out and not as tall.

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252 C h a p t e r 6 | Continuous Probability Distributions

Calculating probabilities for Normal Distributions Using the Standard Normal probability tables Let’s return to our previous example where the time customers spend on the phone for service follows the normal distribution with a mean of 12 minutes and a standard deviation of 3 min- utes. What is the probability that the next customer who calls will spend 14 minutes or less on the phone? To answer this question, we need to revisit a concept introduced in Chapter 3. The z-score, shown in Equation 6.2, determines the number of standard deviations that a particu- lar value, x, is from the mean of its distribution.

The z-score determines the number of standard deviations that a particular value, x, is from the mean of its distribution.

Formula 6.2 for the z-score

z = x - m s

where

x = The continuous data value of interest m = The distribution’s mean s = The distribution’s standard deviation

Using Equation 6.2, the z-score for x = 14 is

z14 = x - m s

= 14 - 12

3 = 0.67

In other words, a customer who spends 14 minutes on the phone is 0.67 standard de- viations from the distribution mean of 12 minutes. The z-score is analogous to expressing the original data in different units, such as converting two feet to 24 in. In our example, the 14-minute time is being converted to a z-score of 0.67. However, the z-score itself has no units even though the original values for x will normally be expressed in units such as min- utes, dollars, years, and pounds. Table 6.1 shows the z-scores for this example for various call times.

Table 6.1 highlights the following features of z-scores:

• z-Scores are negative for values of x that are less than the distribution mean.

• z-Scores are positive for values of x that are more than the distribution mean.

• The z-score at the mean of the distribution equals zero.

The z-score determines the number of standard deviations that a particular value, x, is from the mean of its distribution.

{ table 6.1 | z-Scores for Times Spent on the Phone with Customer Service

minutes

x M x − M S z = 1x − M2 ,S 6 12 -6 3 - 2.00

8 12 - 4 3 - 1.33

10 12 - 2 3 - 0.67

12 12 0 3 0.00

14 12 2 3 0.67

16 12 4 3 1.33

18 12 6 3 2.00

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C h a p t e r 6 | 6.2 Normal Probability Distributions 253

When the original random variable, x, follows the normal distribution, z-scores also follow a normal distribution with m = 0 and s = 1, which is shown in Figure 6.5. This is known as the standard normal distribution.

{ figure 6.5 The Standard Normal

Distribution 1Mean = 0 and Standard Deviation = 12

0.50.5

= 1

f (x)

The zscore follows a normal distribution with m = 0 and s = 1, which is known as the standard normal distribution.

At this point, you might be wondering: So what does the standard normal distribution have to do with a random variable that follows the normal probability distribution? Essen- tially, we are converting the value of the random variable to a z-score that follows the stan- dard normal distribution (like converting feet to inches). Once our normal random variable is converted to the z-score, its mean will equal zero, as Figure 6.5 shows, and the standard deviation will equal 1. We can then use the standard normal probability tables shown in Appendix A at the back of the book to calculate the probabilities for our normal distribution. (Looking up the probabilities in the tables will be much easier than using Equation 6.1, which requires an advanced math degree.) Keep in mind that this works for any normal distribution. Any normally distributed values can be standardized with z-scores.

Now, let us return to our original question. We can use the standard normal distribu- tion tables to determine the probability that a person will spend 14 minutes or less on the phone with customer service. Table 6.2 shows a portion of Table 4 in Appendix A. This table

{ table 6.2 | Cumulative Standardized Normal Table Excerpted from Appendix A for Positive z-Scores

first digit of z second digit of z

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359

0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753

0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141

0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224

0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549

0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852

0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133

0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441

1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545

1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633

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254 C h a p t e r 6 | Continuous Probability Distributions

provides the cumulative area under a standard normal distribution curve that lies to the left of the z-score. Because the z-score that corresponds to 14 minutes is 0.67, we use the 0.6 row, which corresponds to the first digit in our z-score, and the 0.07 column, which corresponds to the second digit in our z-score. At the intersection of the two columns, we find the value 0.7486, which is highlighted in yellow. The value 0.7486 (74.86%) represents the probability that a person will be on the phone 14 minutes or less. The shaded area in Figure 6.6 shows this value graphically.

{ figure 6.8 Finding the Value (x) That

Corresponds to 95% of the Area Under a Normal Curve

0.95

0 µ

1.645 x = ?x scale

z scale

{ figure 6.7 The Area Under a Normal Curve

When x 7 14

0.7486

0.2514

14 0 µ

0.67z scale x scale

{ figure 6.6 The Area Under a Normal Curve

When x … 14

0.7486

14 0 µ

0.67 x scale z scale

There is a 74.86% chance that the next person will be on the phone 14 minutes or less, which can be expressed as

P1x … 142 = 0.7486 Remember, this is the same probability as less than 14 minutes because P1x = 142 = 0. What about the probability that the next customer will be on the phone for more than 14

minutes? Because we know the area under the normal curve equals 1.0,

P1x 7 142 = P1z 7 0.672 = 1.0 - 0.7486 = 0.2514 In other words, there is a 25.14% chance that the customer will be on the phone for

more than 14 minutes. The shaded region in Figure 6.7 shows this probability.

You only need to calculate the z-score out to two decimal places because this is the level of precision that Table 6.2 provides.

Suppose that management at a cable company has a goal that 95% of its customers will require less than 18 minutes per call on the phone with customer service. Is this goal being met with the current probability distribution, which has a mean of 12 and standard deviation of 3? To answer this question, we need to determine how many standard deviations to the right of the mean are needed to include 95% of the area under the curve. We do so by using Figure 6.8 and taking the following steps:

• First, we recognize that the area to the left of our point of interest includes 95% of the total area under the curve.

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C h a p t e r 6 | 6.2 Normal Probability Distributions 255

• Second, we search Table 6.2 to find the value in the body of the table closest to 0.95. This value is half-way between 0.9495 and 0.9505, which I have shown in the red box. These values are found in the 1.6 row and the 0.04 and 0.05 columns. This means our point of interest is halfway between these columns at 1.6 + 0.045, or 1.645 standard deviations to the right of the mean. In other words, z = 1.645.

• Third, we need to determine the value of x that corresponds to z = 1.645 by rear- ranging Equation 6.2 in terms of x, as follows:

z = x - m s

x = m + zs x = 12 + 11.6452132 = 12 + 4.94 = 16.94 minutes

• Finally, we can conclude that 95% of the people calling customer service will spend less than 16.94 minutes on the phone. Because this time is less than 18 minutes, manage- ment’s goal appears to have been achieved. The value 16.94 can also be described as the 95th percentile, because 95% of the values in the distribution fall below it.

Now it’s time to move over to the left side of the distribution’s mean. Let’s examine the probability that a customer will be on the phone for 8.5 minutes or less. First, we calculate the z-score for x = 8.5 using Equation 6.2:

z8.5 = x - m s

= 8.5 - 12

3 = - 1.17

Table 6.3 shows an excerpt from Table 3 in Appendix A. The values in this table are the cu- mulative normal probabilities for negative z-scores, which are always on the left side of the standard normal distribution (remember, the mean of this distribution is zero).

{ table 6.3 | Cumulative Standardized Normal Table Excerpted from Appendix A for Negative z-Scores

first digit of z second digit of z

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

- 3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010 - 2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 - 2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 - 2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 - 2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 - 2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 - 2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 - 2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 - 2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 - 2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 - 2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 - 1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 - 1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 - 1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 - 1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 - 1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 - 1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 - 1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 - 1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 -1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 - 1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379

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256 C h a p t e r 6 | Continuous Probability Distributions

This table provides the cumulative area under a standard normal distribution curve that lies to the left of the negative z-score. Because the z-score that corresponds to 8.5 minutes is - 1.17, we look at the - 1.1 row, which corresponds to the first digit in our z-score. Then we look at the 0.07 column, which corresponds to the second digit in our z-score. At the in- tersection of the two columns, we find the value 0.1210, which is highlighted in yellow. The value 0.1210 (12.1%) represents the probability that a person will be on the phone 8.5 min- utes or less. The shaded area in Figure 6.9 shows this value graphically.

{ figure 6.9 The Area Under a Normal Curve

When x … 8.5

128.5

0.1210

0 µ

–1.17 x scale z scale

{ figure 6.10 The Area Under a Normal Curve

When x 7 8.5

128.5

0.1210 0.8790

0 µ

–1.17 x scale z scale

Let’s wrap up this section with one last probability. What is the probability that a person will be on the phone for more than 8.5 minutes? The answer can be found in the shaded area in Figure 6.10.

Using Figure 6.10, we can see that this probability will be

P1x 7 8.52 = P1z 7 - 1.172 = 1.0 - 0.1210 = 0.8790 Thus, there is an 87.9% probability that the next person will be on the phone with cus-

tomer service for more than 8.5 minutes.

Remember, the total area under the normal curve equals 1.0.

When solving these problems, I strongly encourage you to take the time to draw the distribution curves as I have shown you in this chapter. This will help you identify the proper area of interest.

A local gas station waits 4 days to receive a delivery of regular gasoline to replenish its inventory. The wait period to receive inventory is known as the lead time. The demand during the lead-time period for regular gasoline, as measured in gallons, follows the normal distribution with a mean of 930 gallons and standard deviation of 140 gallons.

a. What is the probability that, during the next lead time, the demand for regular gasoline will be less than 800 gallons?

b. What is the probability that, during the next lead time, the demand for regular gasoline will be less than 1,000 gallons?

c. What is the probability that, during the next lead time, the demand for regular gasoline will be equal to 900 gallons?

d. The station manager places the next order for regular gasoline when the inventory is 1,200 gallons (known as the reorder point). What is the probability that the station will not run out of gasoline before the order arrives?

e. Determine the reorder point that will ensure the station a 90% probability that it will have enough inventory of regular gasoline until the next order arrives.

Y o U r t U r N # 1N

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C h a p t e r 6 | 6.2 Normal Probability Distributions 257

revisiting the empirical rule Back in Chapter 3, we introduced the empirical rule, which states if a distribution is sym- metrical and bell-shaped, approximately 68%, 95%, and 99.7% of its data values will fall within one, two, and three standard deviations above and below the mean, respectively. We now have the tools to verify these percentages using Tables 3 and 4 in Appendix A.

Figure 6.11 illustrates the empirical rule using our previous example. Recall that people spend an average of 12 minutes on the phone with customer service, with a standard devia- tion of 3 minutes. Each vertical line in Figure 6.11 represents an increment of one standard deviation from the mean. Because one standard deviation equals 3 minutes, the time interval between vertical lines represents 3 minutes on the phone.

According to the empirical rule, if a distribution follows a bellshape, symmetrical curve centered around the mean, approximately 68%, 95%, and 99.7% of its values will fall within one, two, and three standard deviations above and below the mean, respectively.

{ figure 6.11 The Empirical Rule for

Customer Service Call Length

12 15 18 213 6 9

0.3413 0.3413

68% 95%

99.7%

0 1 2 3 µ

– 3 – 2 – 1 x scale z scale

Let’s examine the shaded region in Figure 6.11, which represents { one standard devia- tion around the mean. According to Table 4 in Appendix A, the area to the left of z = 1.0 is 0.8413. Subtracting 0.5 (the area to the left of the mean) from this value gives us 0.3413 (the area of the shaded region to the right of the mean). Adding the two shaded areas together provides us with the 68% (approximately) stated in the empirical rule. We would expect that 68% of the callers spend between 9 and 15 minutes on the phone.

We’ll use the same logic for { two standard deviations around the mean. According to Table 4 in Appendix A, the area to the left of z = 2.0 is 0.9772. Subtracting 0.5 from this value gives us 0.4772 (the area to the right of the mean). Adding the two areas on either side of the mean together 10.4772 + 0.47722 provides us with the 95% (approximately) stated in the empirical rule. We would expect that 95% of the callers spend between 6 and 18 min- utes on the phone.

Finally, Table 4 in Appendix A shows that the area to the left of z = 3.0 is 0.9987. Subtracting 0.5 from this value gives us 0.4987. Adding the two areas on either side of the mean together 10.4987 + 0.49872 provides us with the 99.7% (approximately) stated in the empirical rule. We would expect that practically all of the callers spend between 3 and 21 minutes on the phone with customer service.

The following Your Turn section will give you the chance to see if you’ve mastered the empirical rule.

Suppose the average fuel economy for the 2013 Ford C Max hybrid car is reported to be 39.2 miles per gallon (mpg) with a standard deviation of 3.7 mpg. Assume the fuel economy for each tankful of gas for this car follows the normal distribution. Determine the interval of miles per gallon around the mean that include the following:

a. Approximately 68% of the reported fuel economies for this car b. Approximately 95% of the reported fuel economies for this car c. Approximately 99.7% of the reported fuel economies for this car

Y o U r t U r N # 2N

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258 C h a p t e r 6 | Continuous Probability Distributions

other Normal probability Intervals We can examine other types of probability intervals using the normal distribution. Let’s look at our snowfall example from the beginning of the chapter. I researched many cities in the United States and found that the annual snowfall amounts in Minneapolis, Minnesota, from 1884 until 2012 tended to follow a normal distribution with a mean of 46.3 in. and a stan- dard deviation of 18.4 in. The histogram for these data are shown in Figure 6.12. As you can see, the data are slightly right-skewed, but they are close enough to a normal distribution for our purposes. (In Chapter 12, I will show you a procedure to test a distribution to see if its data values allow it to be categorized as a normal distribution.)

{ figure 6.12 A Histogram Showing the

Annual Snowfall in Minneapolis

{ figure 6.13 The Probability of Between

30 and 70 in. of Snow Falling in Minneapolis

46.330 70 1.29–0.89 0

0.9015

0.1867 0.7148

z scale x scale

0.9015 – 0.1867 = 0.7148

If the historical data in the figure are a good indicator of future events, what is the prob- ability that between 30 and 70 in. of snow will fall in Minneapolis next year? The area of interest straddles the distribution’s mean of 46.3 inches. For this, we need to calculate the z-scores for x = 30 and x = 70 using Equation 6.2. (Remember, m = 46.3 and s = 18.4.)

z30 = x - m s

= 30 - 46.3

18.4 = - 0.89

z70 = x - m s

= 70 - 46.3

18.4 = 1.29

According to Figure 6.13, the probability of less than 70 in. of snow falling in Minneapolis is

P1x … 702 = P1z … 1.292 = 0.9015

You can find this value in the 1.2 row and 0.09 column in Table 6.2.

However, this probability includes the event that less than 30 inches of snow will fall, so we need to subtract this out to get our final result, which is the shaded region in Figure 6.13.

P1x … 302 = P1z … - 0.892 = 0.1867 P130 … x … 702 = 0.9015 - 0.1867 = 0.7148

You can find this value in the - 0.8 row and 0.09 column in Table 3 in Appendix A.

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C h a p t e r 6 | 6.2 Normal Probability Distributions 259

In other words, there is a greater than 71% chance that between 30 and 70 in. will fall. That’s a lot of snow.

The next example has caused some of my students to accuse me of trickery when they see it on an exam. (OK, it’s a little different from the examples we have looked at so far, but trickery? Decide for yourself.) What is the probability that between 60 and 75 in. of snow will fall in Minneapolis next year? When you examine Figure 6.14, you’ll see the entire shaded area of interest falls to the right of the distribution’s mean of 46.3 in. As with the last example, we need to calculate the z-scores for x = 60 and x = 75 using Equation 6.2.

z60 = x - m s

= 60 - 46.3

18.4 = 0.74

z75 = x - m s

= 75 - 46.3

18.4 = 1.56

According to Figure 6.14, the probability of less than 75 in. of snow falling in Minneapolis is

P1x … 752 = P1z … 1.562 = 0.9406 However, this probability includes the event that less than 60 in. of snow will fall, so we

need to subtract this out to get our final result, which is the shaded region in Figure 6.14.

P1x … 602 = P1z … 0.742 = 0.7704 P160 … x … 752 = 0.9406 - 0.7704 = 0.1702

The probability of less than 75 in. falling includes the probability of less than 60 in. falling. Therefore, we need to remove this probability (0.7704) to obtain our final result.

{ figure 6.14 The Probability of Between 60

and 75 in. of Snow Falling in Minneapolis

46.3 60 1.560 0.74

0.9406

0.17020.7704

0.9406 – 0.7704 = 0.1702

x scale z scale

In other words, the probability of this much snow falling is considerable less—only a little over a 17% chance. Again, so that you don’t get confused (or tricked), draw the distri- bution to show the area of probability more clearly.

Let’s wrap up this section by calculating a shaded area to the left of the mean. (I don’t want to discriminate against you left-handers). Figure 6.15 shows the probability that between 12 and 35 inches of snow will fall in Minneapolis next year. First, we show the z-scores for x = 12 and x = 35 using Equation 6.2:

z12 = x - m s

= 12 - 46.3

18.4 = - 1.86

z35 = x - m s

= 35 - 46.3

18.4 = - 0.61

{ figure 6.15 The Probability of Between

12 and 35 in. of Snow Falling in Minneapolis

46.335 –0.61

12 –1.86 0

0.2709

0.0314

0.2395

0.2709 – 0.0314 = 0.2395

x scale z scale

This is the probability that less than 35 in. of snow will fall next year.

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260 C h a p t e r 6 | Continuous Probability Distributions

Following the logic from the two previous examples and using Table 3 in Appendix A, we have

P1x … 352 = P1z … - 0.612 = 0.2709 P1x … 122 = P1z … - 1.862 = 0.0314 P112 … x … 352 = 0.2709 - 0.0314 = 0.2395

At this point, I have tortured you with every possible type of interval under a normal distribution. Make me proud and nail this type of question in this next example by solving the following Your Turn problem.

The Wall Street Journal studied four National Football League games during a recent season to deter- mine the number of minutes of actual playing time that occurred during a 3-hour game. Surprisingly (at least to me), the average playing time was only 10.7 minutes per game! Assume the actual playing time per game follows the normal distribution with a standard deviation of 0.8 minutes. Determine the probability that the actual playing time for the next game will be between

a. 9.8 and 11.9 minutes b. 11.0 and 12.1 minutes c. 9.7 and 10.3 minutes

Y o U r t U r N # 3N

Answers can be found on u page 286

Calculating Normal probabilities Using excel Excel is a great tool for calculating normal probabilities. To demonstrate its use, I’m going to rely on a research study from the University of California that reported Americans con- sume an average of 34 gigabytes of data and information per day. Let’s assume that the daily consumption of data per person follows a normal distribution with a standard deviation of 10 gigabytes. If so, what is the probability that a randomly selected American will consume less than 46 gigabytes of data today?

First, we’ll calculate this probability using Table 6.2 and Figure 6.16:

z46 = x - m s

= 46 - 34

10 = 1.20

P1x … 462 = P1z … 1.202 = 0.8849 The probability is 88.49%.

{ figure 6.16 The Probability of an

American Consuming Less Than 46 Gigabytes of Data a Day

34 46 1.200

0.8849

x scale z scale

Now we’ll confirm this result with NORM.DIST, Excel’s built-in function, which has the following characteristics:

= NORM.DIST1x, mean, standard_dev, cumulative2 where

cumulative = FALSE if you want the probability density function (we don’t) cumulative = TRUE if you want the cumulative probability (we do)

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C h a p t e r 6 | 6.2 Normal Probability Distributions 261

We’ll choose the cumulative probability option (TRUE) because we want to include the probability of 46 gigabytes or less, which is shown in the shaded region in Figure 6.16.

= NORM.DIST146, 34, 10, TRUE2 = 0.8849 We can also use Excel’s NORM.S.DIST function to determine the probability that a

randomly selected American will consume 27 gigabytes or less today. We’ll start by using Table 3 in Appendix A and Figure 6.17 to find this probability:

z27 = x - m s

= 27 - 34

10 = - 0.70

P1x … 272 = P1z … - 0.702 = 0.2420 The probability is only 24.2%.

{ figure 6.17 The Probability of an

American Consuming Less Than 27 Gigabytes of Data a Day

3427 –0.7 0

0.2420

x scale z scale

We can confirm this result with Excel’s NORM.S.DIST function which provides a prob- ability for a specific z-score and has the following characteristics:

= NORM.S.DIST1z, cumulative2 We’ll set cumulative = TRUE because we want to determine the probability of 27 giga-

bytes or less as shown here:

= NORM.S.DIST1 - 0.70, TRUE2 = 0.2420 This result is consistent with the probability found using Table 3 in Appendix A. Occa-

sionally, there will be a slight discrepancy between the probabilities determined from Tables 3 and 4 in Appendix A and those from Excel’s NORM.DIST and NORM.S.DIST functions. This is because Tables 3 and 4 use only two decimal places for z-scores, whereas Excel uses many more. The rounding difference between the two methods is small enough to be ignored.

PHStat has the capability of providing not only the cumulative probabilities just illus- trated but also the probabilities of intervals. For example, I’ll use PHStat for Windows to determine the probability that a randomly selected American will consume between 27 and 46 gigabytes of data today. Using the probabilities shown in Figure 6.16 and Figure 6.17, we can construct the probability of interest, which is shown in Figure 6.18.

{ figure 6.18 The Probability of an

American Using Between 27 and 46 Gigabytes of Data a Day

3427 46 1.20–0.7 0

0.8849

0.6429

0.2420

x scale z scale

0.8849 – 0.2420 = 0.6429

According to Figure 6.18,

P1x … 462 = P1z … 1.202 = 0.8849 P1x … 272 = P1z … - 0.702 = 0.2420 P127 … x … 462 = 0.8849 - 0.2420 = 0.6429

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262 C h a p t e r 6 | Continuous Probability Distributions

We can use PHStat to calculate this probability directly by taking the following two steps:

1. Go to Add-Ins + PHStat + Probability & Prob. Distributions + Normal, as shown in Figure 6.19A.

{ figure 6.19a Calculating a Normal

Probability Using PHStat (Step 1)

{ figure 6.19b Calculating a Normal

Probability Using PHStat (Step 2)

2. Fill in the Normal Probability Distribution dialog box as shown in Figure 6.19B. Click OK.

Figure 6.19C shows the normal probability we found in Figure 6.18. The slight discrep- ancy is due to rounding.

{ figure 6.19C Finished Normal Probabilities

Calculated Using PHStat

Excel for Mac users can go the textbook’s Web site, www. pearsonhighered.com/ donnelly, for these instructions.

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C h a p t e r 6 | 6.2 Normal Probability Distributions 263

The following Your Turn section will help you hone your skills when it comes to calcu- lating normal probabilities.

A weight-loss program claims that the average weight people lose during the first two weeks of the program is 5.0 lbs. Assuming the weight loss follows the normal distribution with a standard deviation of 3.0 lbs, use Excel and PHStat to answer the following questions.

a. What is the probability that a person on the program will lose less than 7.0 lbs after two weeks? b. What is the probability that a person on the program will gain weight after two weeks? (Hint: For

example, losing 2 lbs means x = 2. Gaining 3 lbs results in x = - 3. Continuous random vari- ables can take on negative values.)

c. What is the probability that a person will lose between 1.0 and 4.0 lbs after two weeks?

Y o U r t U r N # 4N

Answers can be found on u page 286

Using the Normal Distribution to approximate the Binomial Distribution Under the proper conditions, we can use the normal distribution to approximate the binomial probability distribution. Recall from Chapter 5 that a binomial probability distribution is a discrete probability distribution in which only two outcomes—either a success or a fail- ure (hence the term “binomial”). Recall also that the binomial equation will calculate the probability of r successes in n trials, with p = probability of a success for each trial and q = probability of a failure.

The problem is that using the binomial formula from Chapter 5 to calculate probabilities gets very tedious if you have a large number of trials. This is where the normal distribution becomes a convenient tool to use to approximate the binominal distribution. If np Ú 5 and nq Ú 5, we can use the normal distribution to approximate the binomial probability, even though the normal distribution is a continuous distribution and the binomial distribution is not. In other words, if a decent-sized number of successes and number of failures are ob- served (such that np Ú 5 and nq Ú 5), the calculations for a normal distribution—with a few modifications—will work for the binomial distribution.

Binominal approximations are often used in the business world in the area of quality control. For example, Carl’s Jr., a fast-food chain with over 1,000 locations, recently expe- rienced a 10% error rate in its drive-through orders. Carl’s has defined a success as an order that was processed incorrectly, and a failure as an order that was processed correctly. Re- member, the term success used with the binomial distribution is just a label for an outcome and is not necessarily associated with a positive result.

In the following year, the error rate fell to 6%. The reduction was attributed to the chain being able to hire employees with better language skills following a rise in unemployment rates during the period. Essentially, Carl’s Jr. was able to choose applicants from a larger labor pool.

Assuming the chain has a 6% error rate, what is the probability that exactly 7 out of its next 120 drive-through customers will experience mistakes with their orders?

Recall from Chapter 5 that the BINOM.DIST function has the following characteristics:

= BINOM.DIST1x, n, p, cumulative2 where

cumulative = FALSE, if you want to determine the probability of exactly x successes occurring cumulative = TRUE, if you want to determine the probability of x or fewer successes occurring

We can use the normal distribution to approximate the binomial distribution when np Ú 5 and nq Ú 5

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264 C h a p t e r 6 | Continuous Probability Distributions

Because we are counting individual customers, we are using discrete data rather than continuous data. We can use Excel to calculate this binomial probability as follows:

x = 7 n = 120 p = 0.06 = BINOM.DIST17, 120, 0.06, FALSE2 = 0.1531

We use FALSE in the BINOM.DIST function because we are interested in the probabil- ity of exactly seven orders being incorrect.

Also from Chapter 5, the mean and standard deviation of this distribution are as follows:

m = np = 1120210.062 = 7.2 s = 1npq = 11120210.06211 - 0.062 = 16.768 = 2.60

The normal approximation can be used here to calculate binomial probabilities because the following conditions have been met:

np = 1120210.062 = 7.2 Ú 5 nq = 1120211 - 0.062 = 112.8 Ú 5

We can approximate the probability of exactly 7 orders from the next 120 customers be- ing incorrect using the normal distribution. However, we know from earlier in this chapter that P1x = 72 = 0 because the normal distribution is a continuous distribution.

For the normal distribution, we can calculate only the probability for an interval, so we need to make a modification. We create an interval by subtracting 0.5 from 7.0 and adding 0.5 to 7.0, which is known as the continuity correction factor.

We then calculate the z-scores for x = 6.5 and x = 7.5.

z6.5 = x - m s

= 6.5 - 7.2

2.6 = - 0.27

z7.5 = x - m s

= 7.5 - 7.2

2.6 = 0.12

Using Tables 3 and 4 in Appendix A, Figure 6.20, and the normal approximation, we can calculate the probability that exactly seven of the orders will be incorrect as follows:

P1x = 72 = P16.5 … x … 7.52 P1x … 7.52 = P1z … 0.122 = 0.5478 P1x … 6.52 = P1z … - 0.272 = 0.3936 P16.5 … x … 7.52 = 0.5478 - 0.3936 = 0.1542

x represents the number of customers from the 120 whose order was not filled correctly.

The continuity correction factor allows us to approximate the binomial distribution with the normal distribution by adding and subtracting the value 0.5 to create the interval of interest.

{ figure 6.20 The Probability of Exactly

7 Incorrect Orders Occurring Out of 120 Orders

7.2 7.56.5 –0.27 0.120

0.5478

0.5478 – 0.3936 = 0.1542

0.3936

x scale z scale

Notice that this is very close to the results of Excel’s BINOM.DIST function, 0.1531. If we wanted to expand our example to calculate the probability of exactly 7, 8, or 9 incorrect orders, we would examine the area between 6.5 and 9.5 under the normal curve. I’ll be sure to give you a chance to try this type of question in the next Your Turn section.

Let’s explore the probability of 4 or fewer orders of the next 120 being incorrect. Using Excel, we can determine the binomial probability:

= BINOM.DIST14, 120, 0.06, TRUE2 = 0.1473

This is the probability that less than 7.5 of the 120 customers will have an incorrect order.

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C h a p t e r 6 | 6.2 Normal Probability Distributions 265

We use TRUE in the BINOM.DIST function because we are interested in not just the probability of four orders being correct, but the probability of 0, 1, 2, 3, or 4 orders being incorrect.

Using the continuity correction factor to calculate the normal approximation, we add 0.5 to 4, the right boundary of our interval. The left side of the normal distribution is -∞ and therefore does not need a correction factor. We now calculate the z-score for x = 4.5:

z4.5 = x - m s

= 4.5 - 7.2

2.6 = -1.04

Figure 6.21 shows the normal approximation to the probability that 4 or fewer of Carl’s Jr. drive-through customers from the next 120 arrivals will experience a mistake with their order.

P1x … 4.52 = P1z … - 1.042 = 0.1492

{ figure 6.22 The Probability of More Than 8

Incorrect Orders Occurring Out of 120 Orders

7.2 8.5 0.50

0.6915

0.3085

x scale z scale

{ figure 6.21 The Probability of 4 or Fewer

Incorrect Orders Occurring Out of 120 Orders

7.24.5 –1.04 0

0.1492

x scale z scale

Notice that this is very close to the results of Excel’s BINOM.DIST function, 0.1473. Finally, let’s examine the binomial probability that more than 8 of 120 drive-through

customers will receive incorrect orders. Because this question includes the discrete values 9, 10, 11, and 12, on up to the value of 120, we determine this probability as follows using Excel:

P1x Ú 92 = 1 - P1x … 82 = BINOM.DIST18, 120, 0.06, TRUE2 = 0.7061

P1x 7 82 = P1x Ú 92 = 1 - 0.7061 = 0.2939 Now we need to construct the interval for the normal approximation using the continu-

ity correction factor. Because the value 9 is on the left side of the boundary, we subtract 0.5 from 9. The right side of the normal distribution is + ∞ and therefore does not need a correc- tion factor. We now calculate the z-score for x = 8.5:

z8.5 = x - m s

= 8.5 - 7.2

2.6 = 0.5

Figure 6.22 shows the normal approximation for the probability of more than 8 of 120 drive-through customers experiencing mistakes with their orders.

P1x 7 8.52 = P1z 7 0.52 = 1.0 - 0.6915 = 0.3085

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266 C h a p t e r 6 | Continuous Probability Distributions

This probability is fairly close to the results of Excel’s BINOM.DIST function, 0.2939. Sometimes my students have difficulty deciding when to add and when to subtract the

0.5 continuity correction factor. In general, the rules are as follows:

• Subtract 0.5 from the left side of the boundary, which moves it slightly to the left.

• Add 0.5 to the right side of the boundary, which moves it slightly to the right.

• If the left boundary is -∞ or the right boundary is + ∞, no correction factor is needed for that boundary.

Table 6.4 summarizes these rules and lists some examples to help clarify them.

Be careful of the wording of the probability statement so that you don’t confuse phrases like “less than 3” with “3 or less”; they refer to different discrete values.

{ table 6.4 | Rules for Adding and Subtracting the Continuity Correction Factor When Approximating a Binominal Distribution

probability statement discrete values normal probability

less than 3 0, 1, 2 P1x … 2.52 3 or less 0, 1, 2, 3 P1x … 3.52 exactly 5 5 P14.5 … x … 5.52 exactly 7, 8, or 9 7, 8, 9 P16.5 … x … 9.52 6 or more 6, 7, 8, c, n P1x Ú 5.52 more than 6 7, 8, c, n P1x Ú 6.52

As promised, the next Your Turn section will help you test your skills on this procedure.

A March 13, 2010, Wall Street Journal article reported that, to the chagrin of Microsoft management, approximately 10% of Microsoft employees were iPhone users. Of course, iPhones are a product of Apple, Inc., Microsoft’s biggest competitor. The following is an excerpt from the article, titled “Forbidden Fruit: Microsoft Workers Hide Their iPhones” which highlighted the risk of being an iPhone user at a Microsoft function in September, 2009.

“At an all-company meeting in a Seattle sports stadium, one hapless employee used his iPhone to snap photos of Microsoft Chief Executive Steve Ballmer. Mr. Ballmer snatched the iPhone out of the employee’s hands, placed it on the ground and pretended to stomp on it in front of thousands of Microsoft workers, according to people present.”

Suppose 60 Microsoft employees are randomly selected. Use the normal approximation to the binomial distribution to determine the probability that

a. less than five employees are iPhone users. b. exactly six, seven, eight, or nine employees are iPhone users. c. nine or more employees are iPhone users.

Based on: Nick Wingfield, “Forbidden Fruit: Microsoft Workers Hide Their iPhones,” Wall Street Journal, March 12, 2010.

Y o U r t U r N # 5N

Answers can be found on u page 287

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C h a p t e r 6 | 6.2 Normal Probability Distributions 267

6.2 Section Problems Basic Skills 6.1 For a standard normal distribution, determine the

following:

a. P1z … 1.502 b. P1z … - 1.222 c. P1 - 0.86 … z … 1.762 d. P10.32 … z … 2.152

6.2 For a standard normal distribution, determine the following:

a. P1z 7 1.352 b. P1z 7 - 0.422 c. P1 - 1.70 … z … - 0.652 d. P1 - 1.69 … z … 0.202

6.3 A random variable follows the normal probability distri- bution with a mean of 80 and a standard deviation of 20. What is the probability that a randomly selected value from this population

a. is less than 90? b. is less than 65? c. is more than 110? d. is more than 40?

6.4 A random variable follows the normal probability dis- tribution with a mean of 124 and a standard deviation of 27. What is the probability that a randomly selected value from this population

a. is between 100 and 140? b. is between 130 and 170? c. is between 65 and 90? d. is between 120 and 180?

6.5 A binomial distribution has p = 0.60 and n = 40.

a. What are the mean and standard deviation for this distribution?

b. What is the probability of exactly 27 successes? c. What is the probability of less than 30 successes? d. What is the probability of more than 20 successes?

6.6 A binomial distribution has p = 0.22 and n = 75.

a. What are the mean and standard deviation for this distribution?

b. What is the probability of exactly 11 successes? c. What is the probability of 10 to 18 successes? d. What is the probability of 8 to 15 successes?

applications 6.7 The average number of miles driven on a full tank of gas

in a Hyundai Veracruz before its low-fuel light comes on is 320. Assume this mileage follows the normal distribu- tion with a standard deviation of 30 miles. What is the probability that, before the low-fuel light comes on, the car will travel

a. less than 330 miles on the next tank of gas? b. more than 308 miles on the next tank of gas? c. between 305 and 325 miles on the next tank of gas? d. exactly 340 miles on the next tank of gas?

6.8 According to a recent survey by Smith Travel Research, the average daily rate for a luxury hotel in the United States is $237.22. Assume the daily rate follows a normal probabil- ity distribution with a standard deviation of $21.45.

a. What is the probability that a randomly selected luxury hotel’s daily rate will be 1. less than $250? 2. more than $260? 3. between $210 and $240?

b. The managers of a local luxury hotel would like to set the hotel’s average daily rate at the 80th percentile, which is the rate below which 80% of hotels’ rates are set. What rate should they choose for their hotel?

6.9 Major League Baseball teams have become concerned about the length of games. During a recent season, games averaged 2 hours and 52 minutes (172 minutes) to com- plete. Assume the length of games follows the normal distribution with a standard deviation of 16 minutes.

a. What is the probability that a randomly selected game will be completed in 1. 200 minutes or less? 2. more than 200 minutes? 3. 150 minutes or less? 4. more than 150 minutes? 5. exactly 150 minutes?

b. What is the completion time in which 90% of the games will be finished?

6.10 A study conducted by Hershey’s discovered that Americans consumed an average of 11.4 pounds of chocolate per year (seems low to me). Let’s assume that the annual chocolate consumption follows the normal distribution with a standard deviation of 3.6 pounds.

a. What is the probability that I will consume 1. less than 7 lbs of chocolate next year? 2. more than 9 lbs of chocolate next year? 3. between 8 and 12 lbs of chocolate next year? 4. exactly 10 lbs of chocolate next year?

b. What is the annual consumption of chocolate that rep- resents the 60th percentile?

6.11 According to the Internal Revenue Service, the average income tax refund for the 2011 tax year was $2,913. As- sume the refund per person follows the normal probability distribution with a standard deviation of $950.

a. What is the probability that a randomly selected tax return refund from the 2011 tax year will be 1. more than $2,000? 2. between $1,600 and $2,500? 3. between $3,200 and $4,000?

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268 C h a p t e r 6 | Continuous Probability Distributions

b. Confirm the answers to part a using Excel or PHStat. c. What refund amount represents the 35th percentile of

tax returns?

6.12 Assume the cost of an extended 100,000 mile warranty for a particular SUV follows the normal distribution with a mean of $1,600 and a standard deviation of $75.

a. Determine the interval of warranty costs from various companies that are 1. one standard deviation around the mean. 2. two standard deviations around the mean. 3. three standard deviations around the mean.

b. You see an ad for an extended warranty for this type of vehicle for $1,900. Based on the previous results, what conclusions can you make?

6.13 The average price of a 42-in. television on Best Buy’s Web site is $790. Assume the price of these televisions follows the normal distribution with a standard deviation of $160.

a. What is the probability that a randomly selected televi- sion from the site sells for 1. less than $700? 2. between $400 and $500? 3. between $900 and $1,000

b. Confirm the answers to part a using Excel or PHStat. c. The price intervals in parts a)2 and a)3 are equal

($100). Why are the probabilities so different? d. Suppose I am shopping on Best Buy’s Web site for a

new 42-in. television and my wife (also known as she- who-has-veto-power-over-all-electronic-purchases) suggests we set a budget of $750 for the purchase. There are 15 42-inch televisions on the site. How many televisions are within my budget?

6.14 A credit score measures a person’s creditworthiness. According to the credit-scoring company Experian, the average credit score for Americans in March 2013 was 736. Assume the scores are normally distributed with a standard deviation of 40. Determine the interval of credit scores that are

a. one standard deviation around the mean. b. two standard deviations around the mean. c. three standard deviations around the mean.

6.15 According to eMarketer, adults in the United States aver- age 82 minutes per day on mobile devices in 2012. Assume that minutes per day on mobile devices follow the normal distribution and has a standard deviation of 27 minutes.

a. What is the probability that the amount of time spent today on mobile devices by a particular U.S. adult is 1. less than 70 minutes? 2. more than 65 minutes? 3. between 85 and 110 minutes?

b. Confirm the answers to part a using Excel or PHStat. c. Which number of minutes per day represents the 75th

percentile?

6.16 Curling is a Winter Olympic sport in which people slide a stone along a sheet of ice toward a target area. The game is similar to shuffleboard. Two sweepers with brooms lead the stone along the ice, attempting to influence its path toward the target. According to the Journal of Sports Science and Medicine, the average heart rate of a person who sweeps is 185 beats per minute, as compared with 80 beats per minute for an adult at rest. Assume the heart rate for a sweeper follows the normal distribution with a stan- dard deviation of 5 beats per minute.

a. What is the probability that a sweeper’s heart rate is 1. more than 187 beats per minute? 2. less than 181 beats per minute? 3. between 180 and 183 beats per minute? 4. between 188 and 193 beats per minute?

b. Confirm the answers to part a using Excel or PHStat.

6.17 Census data show that 67% of U.S. students went to col- lege or a trade school within a year of graduating from high school. A random sample of 20 high school gradu- ates is selected. Use the normal approximation to the bi- nomial distribution to answer the following:

a. What are the mean and standard deviation for this distribution?

b. What is the probability that less than 16 of the 20 students will go to college?

c. What is the probability that exactly 16 of the 20 students will go to college?

d. What is the probability that more than 14 of the 20 students will go to college?

e. What is the probability that 10, 11, 12, 13, or 14 of the 20 students will go to college?

6.18 According to the Pew Research Center, 33% of U.S. adults owned a tablet in 2012. Use the normal approxima- tion to the binomial distribution to calculate the probabil- ity that, from a sample of 24 U.S. adults,

a. less than 10 adults own a tablet. b. 6, 7, 8, 9, or 10 adults own a tablet. c. 11, 12, 13, or 14 adults own a tablet. d. exactly seven adults own a tablet.

6.19 According to FlightStats.com, 82.8% of American Airline flights were on time during 2012. Assume this percentage still holds true for American Airlines. Using the normal approximation to the binomial distribution, determine the probability that, of the next 30 American Airlines flights,

a. less than 22 flights will arrive on time. b. exactly 26 flights will arrive on time. c. 21, 22, 23, or 24 will arrive on time. d. 24, 25, 26, 27, or 28 flights will arrive on time.

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C h a p t e r 6 | 6.3 Exponential Probability Distributions 269

where

e = 2.71828 l = The mean number of occurrences over the interval x = Any continuous number of interest You were first introduced to the symbol l (lambda) back in Chapter 5 when the

Poisson probability distribution was discussed. In the example used in Chapter 5, l represented the average number of customers who arrived per hour at a drive-through window for a bank. If l = 12 customers per hour, then the average number of minutes between customers, or m, is

m = 1

l =

1 hour

12 customers *

60 minutes

1 hour = 5 minutes per customer

When the number of arrivals per hour follows the Poisson distribution, the average time between arrivals will follow the exponential distribution. In other words, a discrete ran- dom variable that follows the Poisson distribution with a mean equal to l has a counterpart continuous random variable that follows the exponential distribution with a mean equal to m = 1>l.

The shapes of exponential distributions are shown in Figure 6.23 on the next page. As you can see, the shape of each curve is right-skewed and depends on its mean, m. Recall from Chapter 3, a right-skewed distribution is one where the data are concentrated on the right end of the distribution and as we “ski down the slope” of the distribution, we are moving toward the right.

A discrete random variable that follows the Poisson distribution with a mean equal to l has a counterpart continuous random variable that follows the exponential distribution with a mean equal to μ = 1>l.

6.3 Exponential Probability Distributions

The exponential probability distribution is a continuous distribution that is commonly used to measure the time between events of interest, such as the time between customer arrivals or the time between failures in a business process. The mathematical expression that describes the shape of the curve of such a distribution is known as the exponential probability density function. The function for the curve is shown in Equation 6.3.

Formula 6.3 for the Exponential Probability Density Function

f1x2 = le - lx

6.20 According to Gartner, Inc., 39.2% of mobile phones worldwide use the Android operating system in 2013. A random sample of 30 mobile phone users was selected in 2013. Assuming this estimate is true, use the normal approximation to the binomial distribution to answer the following:

a. Calculate the mean and standard deviation for this distribution.

b. What is the probability that 13 or more of the 30 mo- bile phone users have an Android operating system?

c. What is the probability that exactly 14 of the 30 mobile phone users have an Android operating system?

d. What is the probability that 7, 8, or 9 of the 30 mobile phone users have an Android operating system?

e. What is the probability that 14, 15, or 16 of the 30 mo- bile phone users have an Android operating system?

The exponential probability distribution is a continuous distribution that is commonly used to measure the time between events of interest such as the time between customer arrivals or the time between failures in a business process.

Exponential probability density function the mathematical expression that describes the shape of the exponential probability distribution.

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270 C h a p t e r 6 | Continuous Probability Distributions

There are three major differences between exponential and normal distributions:

1. The exponential distribution is right-skewed, whereas the normal distribution is bell- shaped and symmetrical.

2. The shape of the exponential distribution is completely described by only one param- eter, m. On the other hand, the normal distribution requires two parameters, m and s, to describe its shape.

3. The values for an exponential random variable cannot be negative. For example, it doesn’t make sense for the next customer to arrive - 2 minutes after the previous cus- tomer. There is no such restriction for normal random variables.

As I mentioned earlier, the exponential distribution is particularly good at describing random variables, such as time between customer arrivals, where the majority of the data values tend to be on the left side, or lower end, of the distribution.

Calculating exponential probabilities Let’s put the exponential distribution to work with the following example. Suppose the average time between customer arrivals at the deli counter of a Piggly Wiggly grocery store (awesome name for a store) is 4 minutes. What is the probability that the next customer will arrive at the counter within the next 2 minutes? Figure 6.24 shows the exponential distribution for the deli counter. We are looking for the shaded area under the curve labeled P1x … 22.

{ figure 6.23 Three Exponential Probability

Distributions When l = 3.0, l = 2.0, and l = 1.0

0 0

1.0

2.0

3.0

f (x)

= 3.0 ( µ = 0.33)

= 2.0 ( µ = 0.5)

= 1.0 ( µ = 1.0)

{ figure 6.24 An Exponential Distribution for

Customer Arrivals When P(x ≤ 2)

0.25

0 2 4 6 8 100 12

Time Between Customer Arrivals

P (x ≤ 2)

f (x)

To find that area, we use Equation 6.4, which shows the formula for the exponential cumulative distribution function.

Formula 6.4 for the Exponential Cumulative Distribution Function

P1x … a2 = 1 - e - al

Exponential cumulative distribution function used to calculate the probability that a random variable that follows the exponential distribution is less than a specific value.

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C h a p t e r 6 | 6.3 Exponential Probability Distributions 271

where

e = 2.71828 l = The mean number of occurrences over the interval a = Any number of interest First, we need to decide on the value for l, which, quite honestly, is a decision with

which some of my students struggle. One way to help you remember is to realize that l is always a countable rate, such as number of arriving customers per hour. In contrast, m is al- ways a measurable interval, such as the number of minutes between customer arrivals. In our example, we are told that m = 4 minutes per customer, so we need to find l as follows:

l = 1 m

= 1 customer

4 minutes = 0.25 customers per minute

It’s important to remember that l and m are expressed in the same units (that is., minutes in this example). Applying Equation 6.4 to our example with a = 2 minutes, we have the following:

P1x … a2 = 1 - e - al P1x … 22 = 1 - e - 12210.252 = 1 - 12.718282 - 0.5 P1x … 22 = 1 - 0.6065 = 0.3935

There is about a 39% chance that the next customer will arrive within the next 2 minutes.

What about the probability that the next customer will arrive between 4 and 8 minutes from now? We can still use Equation 6.4 by recognizing that

P14 … x … 82 = P1x … 82 - P1x … 42 P1x … 82 = 1 - e - 18210.252 P1x … 42 = 1 - e - 14210.252

= 1 - 0.1353 = 1 - 0.3679 = 0.8647 = 0.6321

P14 … x … 82 = 0.8647 - 0.6321 = 0.2326 Figure 6.25 also shows this shaded area, which is labeled P14 … x … 82.

The parameter l is always a countable rate, such as number of arriving customers per hour. In contrast, m is always a measurable interval, such as the number of minutes between customer arrivals.

{ figure 6.25 An Exponential Distribution for

Customer Arrivals When P14 … x … 82

0.25

0 2 4 6 1080 12

Time Between Customer Arrivals 0.8647

0.6321

0.2326

f (x)

P(x ≤ 4)

P(4 ≤ x ≤ 8)

Therefore, there is a 23.26% chance that the next person will arrive at the Piggly Wiggly deli counter within the next 4 to 8 minutes.

The standard deviation of the exponential distribution is the same as the mean, as Equa- tion 6.5 shows.

Formula 6.5 for the Standard Deviation of the Exponential Distribution

s = m = 1

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272 C h a p t e r 6 | Continuous Probability Distributions

s = m = 4 minutes

As with our previous distributions, the standard deviation of the exponential distribution measures the average variability around the mean of the distribution.

= EXPON.DIST1x, l, cumulative2 where

m = 4 minutes per customer

l = 1 m

= 1 customer

4 minutes = 0.25 customers per minute

An average of 0.25 customers each minute is equivalent to

a 0.25 customers minute

b a 60 minutes hour

b = 15 customers per hour

We’re now ready to use the EXPON.DIST function to calculate P1x … 22 with l = 0.25. = EXPON.DIST12, 0.25, TRUE2 = 0.3935

As you can see, our result, P1x … 22 = 0.3935 matches what we obtained using Equation 6.4.

Take a shot (no pun intended) on the next Your Turn problem that involves using the exponential distribution in the National Hockey League.

a. Calculate the following probabilities using Equations 6.4 and 6.5 and verify them using Excel.

1. The probability that the next shot on goal will occur during the next minute.

2. The probability that the next shot on goal will occur between 2 and 3 minutes from now. b. Verify your answers from part a using Excel. c. What is the standard deviation for this distribution?

(Because I’m such a nice guy, I’ll give you the following hint: You need to convert 32.2 shots per hour into minutes between shots.)

Y o U r t U r n # 6N

Answers can be found on u page 287

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C h a p t e r 6 | 6.4 Uniform Probability Distributions 273

6.3 Section Problems Basic Skills 6.21 An exponential probability distribution has a mean equal

to 10 minutes per customer. Calculate the following prob- abilities for the distribution:

a. P1x … 122 b. P1x … 22 c. P1x … 102 d. P1x … 52

6.22 An exponential probability distribution has a mean equal to 7 minutes per customer. Calculate the following prob- abilities for the distribution:

a. P1x 7 122 b. P1x 7 32 c. P17 … x … 102 d. P11 … x … 52

6.23 An exponential probability distribution has lambda equal to 24 customers per hour. What is the probability that the next customer will arrive

a. within the next minute? b. within the next 30 seconds? c. within the next 5 minutes? d. within the next 10 minutes?

6.24 An exponential probability distribution has lambda equal to 18 customers per hour. What is the probability that the next customer will arrive

a. within the next 45 seconds? b. within the next 1 to 3 minutes? c. within the next 2 to 5 minutes? d. within the next 4 to 10 minutes?

applications 6.25 Suppose that the average time a fully charged 6-volt lap-

top battery will operate a computer is 3.2 hours and fol- lows the exponential probability distribution.

a. Determine the probability that the next charge will last 1. less than 2 hours. 2. between 2.5 and 4.0 hours. 3. more than 3.5 hours.

b. Confirm the answers to part a using Excel.

6.26 Assume the average time between students walking up to my desk during a statistics exam to ask a question in a desperate attempt to gain some insight into how to solve a problem is 8 minutes and follows the exponential prob- ability distribution.

a. What is the probability that the next desperate student will approach my desk 1. within the next 2 minutes? 2. within the next 4 minutes? 3. between the next 3 to 5 minutes? 4. in more than 10 minutes?

b. Confirm the answers to part a using Excel.

6.27 Customers arrive at a local ATM at an average rate of 15 per hour. Assume the time between arrivals follows the exponential probability distribution.

a. What is the probability that the next customer will arrive 1. within the next 3 minutes? 2. in more than 7 minutes? 3. between 4 and 8 minutes?

b. Confirm the answers to part a using Excel.

6.28 During lunch hour, customers arrive at a fast-food drive-through window, on average, every 2.5 minutes. Assume arrival time follows the exponential probability distribution.

a. What is the probability that the next customer will arrive 1. within the next 2 minutes? 2. in more than 5 minutes? 3. between 1 and 4 minutes?

b. Confirm the answers to part a using Excel.

6.4 Uniform Probability Distributions Our third and final continuous distribution is the continuous uniform probability distribution, where the probability of any interval in the distribution is equal to any other interval with the same width. To illustrate this distribution, I’ll use the following example. I have always con- sidered myself an incredibly lucky man because I have a wife who loves to play golf almost as much as I do. As a matter of fact, Deb’s last birthday present was a golf club. (I’ll take shopping for golf equipment over jewelry anytime!)

The pace of play is a very important statistic for golf course management. Courses that have a reputation for slow play risk losing golf clientele who do not wish to spend their

With the continuous uniform probability distribution, the probability of any interval in the distribution is equal to any other interval with the same width.

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274 C h a p t e r 6 | Continuous Probability Distributions

entire round waiting for the group in front of them to finish. Suppose the time that it takes to play a round of golf at Deerfield Golf Club follows the uniform probability distribution with a minimum time of 4 hours (240 minutes) and a maximum time of 4 hours and 50 minutes (290 minutes). This means that any 1-minute interval within this range of time has the same probability of occurring. Let’s explore how to determine the probability that our next round of golf at Deerfield will take less than 4 hours and 12 minutes (252 minutes).

The mathematical expression that describes the shape of the uniform distribution is known as the continuous uniform probability density function. The formula for the function is shown in Equation 6.6.

For example, the probability of finishing a round between 240 and 250 minutes is the same as the probability of finishing between 260 and 270 minutes. Both intervals are 10 minutes long.

Formula 6.6 for the Continuous Uniform Probability Density Function

f1x2 = 1 b - a

if a … x … b

f1x2 = 0 otherwise

where

a = Smallest allowable continuous random variable b = Largest allowable continuous random variable For our Deerfield example,

a = 240 14 hours2 b = 290 14 hours 50 minutes2 f1x2 = 1

b - a =

290 - 240 =

50 = 0.02

The shape of the uniform probability distribution, which is shown in Figure 6.26, is rect- angular. This particular uniform distribution has a height of f1x2 = 0.02 between a = 240 and b = 290 minutes.

{ figure 6.26 A Continuous Uniform

Distribution

a b 240 290

0.02

Time to Complete Round of Golf (Minutes)

area = 1.0

f (x)

The value f (x) in Figure 6.26 refers to the probability of a one-unit interval within the distribution occurring. For example, the probability that the next round of golf will be com- pleted between the 240th and 241st minutes equals 2%. The probability that the next round of golf will be completed between the 241st and 242nd minute also equals 2%, and so on. Because there are 50 of these 1-minute intervals in this distribution, the total probability, or area, for our distribution equals

10.0221502 = 1.0 To find the probability that a continuous uniform random variable will fall between two

particular values, which we will refer to as x1 and x2, we use Equation 6.7, which shows the uniform cumulative distribution function.

Continuous uniform probability density function the mathematical expression that describes the shape of the continuous uniform probability distribution.

Uniform cumulative distribution function used to calculate the probability that a random variable that follows the uniform distribution is between two specific values.

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C h a p t e r 6 | 6.4 Uniform Probability Distributions 275

This equation essentially calculates the area of the distribution between the values x1 and x2. To determine the probability that the next round of golf at Deerfield will take less than 252 minutes, we use Equation 6.7, as follows:

x1 = 240 x2 = 252

P1x1 … x … x22 = x2 - x1 b - a

P1240 … x … 2522 = 252 - 240 290 - 240

= 12

50 = 0.24

There is a 24% chance that the next round of golf at Deerfield will be completed in less than 4 hours and 12 minutes (252 minutes). The shaded area in Figure 6.27 illustrates this probability.

Formula 6.7 for the Uniform Cumulative Distribution Function

P1x1 … x … x22 = x2 - x1 b - a

{ figure 6.27 A Continuous Uniform Distribu-

tion When P(240 ≤ x ≤ 252)

a b 240 252 290

0.760.24

0.02

P(240 ≤ x ≤ 252)

Time to Complete Round of Golf (Minutes)

f (x)

Next, we’ll look at the probability that the next round of golf at Deerfield will be completed between 4 hours and 10 minutes (250 minutes) and four hours and 45 minutes (285 minutes) using Equation 6.7.

x1 = 250 x2 = 285

P1x1 … x … x22 = x2 - x1 b - a

P1250 … x … 2852 = 285 - 250 290 - 240

= 35

50 = 0.70

There is a 70% chance that the next round of golf at Deerfield will be completed within this time interval. The shaded area in Figure 6.28 illustrates this probability.

{ figure 6.28 A Continuous Uniform

Distribution When P(250 ≤ x ≤ 285)

a b 240 250 285 290

0.70

0.02

P(250 ≤ x ≤ 285)

Time to Complete Round of Golf (Minutes)

f (x)

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276 C h a p t e r 6 | Continuous Probability Distributions

What happens if we try to calculate the probability that the next round will take exactly 250 minutes? For this example, both x1 and x2 equal 250. Using Equation 6.7, we have

P1x1 … x … x22 = x2 - x1 b - a

P1250 … x … 2502 = 250 - 250 290 - 240

= 0

50 = 0

No, I didn’t make a mistake here. This result is consistent with the statement made ear- lier in the chapter that we can discuss only the probability of a particular interval occurring for continuous distributions. The probability of arriving at an exact value for these types of distributions will always be equal to zero.

The managers of the Deerfield course have a goal that 80% of the rounds played on the course will be completed within 4 hours and 30 minutes (270 minutes). Is this goal being met under the current uniform distribution? To answer this question, we need to find the value of x2 in Equation 6.7 that covers 80% of the area under the distribution. This is also known as the 80th percentile of the distribution. Figure 6.29 shows this graphically.

The probability that a continuous random variable will equal an exact value, such as 250 in this example, will always be equal to zero.

Recall that percentiles were introduced in Chapter 3 and represent the approximate percentage of values that are below the value of interest.

{ figure 6.29 The 80th Percentile of a

Continuous Uniform Distribution

a b 290

0.80

0.02

Time to Complete Round of Golf (Minutes)

f (x)

x1 = 240 x2 = ?

I’ll rearrange Equation 6.7 in terms of x2, knowing that x1 = 240 and b - a = 290 - 240 = 50.

P1x1 … x … x22 = x2 - x1 b - a

= 0.80

x2 - 240

50 = 0.80

x2 = 10.8021502 + 240 = 280 minutes Eighty percent of the rounds will be completed within 4 hours and 40 minutes

(280 minutes). Because this value exceeds 270 minutes, the goal set by Deerfield’s manag- ers is not being met. Perhaps they need to consider making the course easier to speed up the pace of play. (Sounds like a great idea to me.)

The formulas for calculating the mean and standard deviation of the uniform continuous probability distribution are shown in Equations 6.8 and 6.9.

Formula 6.8 for the Mean of the Continuous Uniform Distribution

m = a + b

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C h a p t e r 6 | 6.4 Uniform Probability Distributions 277

The mean and standard deviation for our Deerfield example are as follows:

m = a + b

2 =

240 + 290 2

= 265

s = b - a112 = 290 - 240112 = 503.46 = 14.45

According to our distribution, the average time to complete a round of golf at Deerfield is 265 minutes with a standard deviation of 14.45 minutes.

The next Your Turn section will give you the opportunity to test your uniform probabil- ity distribution skills.

Formula 6.9 for the Standard Deviation of the Continuous Uniform Distribution

s = b - a112

Suppose the actual flight times for American Airlines nonstop flights between Philadelphia and Orlando follows the uniform probability distribution. Assume the minimum flight time between these two cities is 140 minutes and the maximum time is 170 minutes.

a. Determine the probability that the next flight will take less than 145 minutes. b. Determine the probability that the next flight will take between 148 and 165 minutes. c. Suppose American Airlines has a goal that 75% of the flights will require less than 165 minutes to

complete. Is this goal being met under the current probability distribution? d. Calculate the mean and standard deviation for this distribution.

Y o u r t u r n # 7N

Answers can be found on u page 287

6.4 Section Problems Basic Skills 6.29 A random variable follows the continuous uniform distri-

bution between 20 and 50.

a. Calculate the following probabilities for the distribution: 1. P1x … 252 2. P1x … 302 3. P1x … 452 4. P1x = 282

b. What are the mean and standard deviation of this distribution?

6.30 A random variable follows the continuous uniform distri- bution between 60 and 95.

a. Calculate the following probabilities for the distribution: 1. P1x 7 632 2. P1x 7 702 3. P1x 7 882 4. P1x = 752

b. What are the mean and standard deviation of this distribution?

6.31 A random variable follows the continuous uniform distri- bution between 20 and 120.

a. Calculate the following probabilities for the distribution: 1. P140 … x … 602 2. P155 … x … 1002 3. P180 … x … 1102

b. What are the mean and standard deviation of this distribution?

6.32 A random variable follows the continuous uniform distri- bution between 130 and 300.

a. Calculate the following probabilities for the distribution: 1. P1180 … x … 2602 2. P1130 … x … 2002 3. P1x 7 1502

b. What are the mean and standard deviation of this distribution?

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278 C h a p t e r 6 | Continuous Probability Distributions

applications 6.33 The time required for housekeeping to clean a hotel room

for the next customer varies between 25 and 45 minutes and follows the continuous uniform distribution.

a. Calculate the value of f1x2. b. What are the mean and standard deviation for this

distribution? c. What is the probability that the next room will require

exactly 30 minutes to clean? d. What is the probability that the next room will require

less than 32 minutes to clean? e. What is the probability that the next room will require

more than 36 minutes to clean? f. What is the probability that the next room will require

between 28 and 34 minutes to clean? g. What time represents the 70th percentile of this

distribution?

6.34 Assume the time required to pass through security at a particular airport follows the continuous uniform distribu- tion with a minimum time of 5 minutes and maximum time of 30 minutes.

a. Calculate the value of f1x2. b. What are the mean and standard deviation for this

distribution? c. What is the probability that the next passenger will re-

quire less than 25 minutes to pass through security? d. What is the probability that the next passenger will re-

quire more than 20 minutes to pass through security? e. What is the probability that the next passenger will

require between 8 and 15 minutes to pass through security?

f. What time represents the 75th percentile of this distribution?

6.35 The commute time to work for a particular employee fol- lows a continuous uniform distribution with a minimum time of 10 minutes and maximum time of 22 minutes.

a. Calculate the value of f1x2. b. What are the mean and standard deviation for this

distribution? c. What is the probability that the employee’s next com-

mute to work will require less than 12.5 minutes? d. What is the probability that the employee’s next com-

mute to work will require more than 14.5 minutes? e. What is the probability that the employee’s next com-

mute to work will require between 11 and 20 minutes? f. What commute time represents the 40th percentile of

this distribution?

6.36 A vending machine located in a local office dispenses cof- fee that varies from 7.4 to 8.2 ounces. The volume of cof- fee dispensed follows a continuous uniform distribution.

a. Calculate the value of f1x2. b. What are the mean and standard deviation for this

distribution? c. What is the probability that the next cup of coffee will

contain more than 8.0 ounces of coffee? d. What is the probability that the next cup of coffee will

contain exactly 7.5 ounces of coffee? e. What is the probability that the next cup of coffee will

contain less than 8.1 ounces of coffee? f. What is the probability that the next cup of coffee will

contain between 7.5 and 8.0 ounces of coffee? g. What volume represents the 15th percentile of this

distribution?

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■ Continuous random variables are outcomes that take on any numerical value in an interval, includ- ing numbers with decimal points.

• The probability that a continuous random variable equals a specific value is always equal to zero.

■ A normal probability distribution has the following characteristics:

• The distribution is bell-shaped and symmetrical around the mean.

• Because the shape of the distribution is symmetrical, the mean and median are the same value.

• Random variables around the mean, where the curve is the tallest, have a higher likelihood of oc- curring than values toward the ends of the distribution, where the curve is lower.

• The total area under the curve is always equal to 1.0.

• Because the distribution is symmetrical around the mean, the area to the left of the mean equals 0.5 as does the area to the right of the mean.

• The left and right ends of the distribution extend indefinitely, never quite touching the horizontal axis.

■ A smaller standard deviation results in a skinnier curve that’s tighter and taller around the mean.

■ A larger s (standard deviation) makes for a fatter curve that’s more spread out and not as tall.

■ The normal distribution’s z-score determines the number of standard deviations that a particular value, x, is from the mean.

■ In general, z-scores follow a normal distribution with m = 0 and s = 1.

■ The z-scores for a normal distribution

• are negative for values of x that are less than the distribution mean.

• are positive for values of x that are more than the distribution mean.

• are zero at the mean of the distribution.

■ According to the empirical rule, if a distribution follows a bell-shape, symmetrical curve centered around the mean, approximately 68%, 95%, and 99.7% of the values will fall within one, two, and three standard deviations above and below the mean respectively.

■ We can use the normal probability distribution to approximate a binomial probability distribution if np Ú 5 and nq Ú 5. • The continuity correction factor allows us to approximate the binomial distribution with a normal distribution by adding and subtracting the value 0.5 to the interval of interest.

■ The exponential probability distribution is a continuous distribution that is commonly used to measure the time between events of interest such as the time between customer arrivals or the time between failures in a business process.

■ The exponential distribution is the continuous counterpart of the discrete Poisson distribution from Chapter 5.

• For example, if a random variable follows the exponential distribution with an average time be- tween arrivals of 5 minutes 1m = 52, then the same random variable also follows the Poisson distribution with a mean of 0.2 customers per minute 1l = 0.22.

■ With a continuous uniform distribution, the probability of any interval in the distribution is equal to any other interval with the same width.

Chapter 6 Summary

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280

Continuity correction factor. A factor that allows us to approxi- mate a binomial distribution using the normal distribution by adding and subtracting the value 0.5 to the interval of interest.

Continuous random variables. Outcomes that take on any numerical value in an interval as a result of conducting an experiment.

Continuous uniform probability density function. The math- ematical expression that describes the shape of the continuous uniform probability distribution.

Continuous uniform probability distribution. A distribution in which the probability of any interval occurring is equal to any other interval with the same width.

Empirical rule. A rule that states that approximately 68%, 95%, and 99.7% of a distribution’s data values will fall within one, two, and three standard deviations above and below the mean, respectively, if the distribution is symmetrical and bell shaped.

Exponential cumulative distribution function. Used to calcu- late the probability that a random variable that follows the exponential distribution is less than a specific value.

Exponential probability density function. The mathematical expression that describes the shape of the exponential prob- ability distribution.

Exponential probability distribution. A continuous distribution that is commonly used to measure the time between events of interest, such as the time between customer arrivals or the time between failures in a business process.

Normal probability density function. The mathematical ex- pression that describes the shape of the normal probability distribution.

Normal probability distribution. A continuous distribution that is bell shaped and symmetrical around its mean.

Standard normal distribution. A normal distribution with m = 0 and s = 1.

Uniform cumulative distribution function. Used to calculate the probability that a random variable that follows the uniform distribution is between two specific values.

z-score. Determines the number of standard deviations a particu- lar value, x, is from the mean of its distribution.

Chapter 6 Key Terms

Chapter 6 Equations

P1x … a2 = 1 - e - alFormula 6.4 for the Exponential Cumulative Distribution Function

f1x2 = le - lx 6.3 Exponential Probability Distributions (pp 269–273)

Formula 6.2 for the z-score z = x - m s

f1x2 = 1 s12p e - 11>2231x - m2>s426.2 Normal Probability Distributions (pp 250–269)

f1x2 = 1 b - a

if a … x … b

f1x2 = 0 otherwise

6.4 Uniform Probability Distributions (pp 273–278)

Formula 6.5 for the Standard Deviation of the Exponential Distribution s = m = 1

Formula 6.1 for the Normal Probability Density Function

Formula 6.3 for the Exponential Probability Density Function

Formula 6.6 for the Continuous Uniform Probability Density Function

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s = b - a112Formula 6.9 for the Standard Deviation of the Uniform Distribution

m = a + b

Formula 6.7 for the Uniform Cumulative Distribution Function P1x1 … x … x22 = x2 - x1 b - a

Formula 6.8 for the Mean of the Continuous Uniform Distribution

Chapter 6 Problems 6.37 According to the National Association of Childcare

Resources and Referral Agencies, the average cost of care for an infant in a child-care center in Delaware was $8,769 a year. Assume this cost follows a normal distribution with a standard deviation of $925.

a. What is the probability that a randomly selected child- care center in Delaware costs 1. more than $8,000 per year for infant care? 2. more than $9,000 per year for infant care? 3. between $8,500 and $9,500 per year for infant care?

b. Use Excel or PHStat to confirm the answers in part a. c. Determine the rate a child-care center that wishes to be

in the 60th percentile in the state of Delaware should charge for an infant.

6.38 The average U.S. monthly cable bill in 2012 has been re- ported to be $86. Assume monthly cable bills follow a nor- mal distribution with a standard deviation of $9.50.

a. What is the probability that a randomly selected bill will be 1. less than $80? 2. less than $90? 3. exactly $85? 4. between $75 and $95?

b. Use Excel or PHStat to confirm the answers in part a. c. Which monthly cable bill represents the 75th percentile?

6.39 Assume that the number of miles a particular brand of tire lasts before it needs to be replaced follows a normal distribution with a mean of 45,600 miles and a standard deviation of 5,800 miles.

a. What is the probability that the next tire of this brand sold will last 1. more than 36,000 miles before it needs to be replaced? 2. less than 42,000 miles before it needs to be replaced? 3. between 45,000 and 55,000 miles before it needs to

be replaced? b. Use Excel or PHStat to confirm the answers in part a. c. What is the mileage that 90% of the tires of this brand

will survive before needing to be replaced?

6.40 Assume the room rate for a three-star hotel in Paris follows a normal distribution with a mean of 175 euros and a stan- dard deviation of 16 euros.

a. What is the probability that a randomly selected three- star hotel has a room rate of 1. less than 195 euros? 2. more than 180 euros? 3. exactly 170 euros? 4. between 150 and 182 euros?

b. Use Excel or PHStat to confirm the answers in part a.

c. A particular hotel would like to set its rate so that only 20% of the hotels in Paris will have lower rates. What room rate should the hotel set?

6.41 On weeknights, the average number of hours of sleep per night adults 55–84 years of age get is 7.0. Assume this ran- dom variable follows a normal distribution with a standard deviation of 1.1 hours.

a. What is the probability that a randomly selected adult from this age group will get 1. more than 6.0 hours of sleep tonight? 2. less than 5.0 hours of sleep tonight? 3. between 6.3 and 8.0 hours of sleep tonight? 4. between 5.6 and 6.5 hours of sleep tonight?

b. Use Excel or PHStat to confirm the answers in part a. c. How many hours of sleep will an adult in this age group

get if they are in the 35th percentile?

6.42 Assume the selling prices of homes in a particular com- munity follow a normal distribution with a mean of $285,700 and a standard deviation of $46,100. Deter- mine the range of selling prices in this community that includes

a. one standard deviation around the mean. b. two standard deviations around the mean. c. three standard deviations around the mean.

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6.43 TomTom, the producer of GPS devices, reported that the fastest drivers in the nation are in Mississippi. In Mississippi, interstate drivers average 70.1 miles per hour. Assume that the speed of cars traveling on Mississippi’s interstates follows the normal distribution with a standard deviation of 7.2 miles per hour.

a. What is the probability that a randomly selected car on a Mississippi interstate is traveling 1. faster than 65 miles per hour? 2. slower than 72 miles per hour? 3. between 60 and 66 miles per hour? 4. between 78 and 85 miles per hour?

b. Confirm the answers to part a with Excel or PHStat. c. What average speed should I drive if I want to travel

faster than 70% of the cars on the interstate?

6.44 According to the Organisation for Economic Co-operation and Development (OECD), South Koreans worked an average of 2,090 hours per year in 2011 (compared with 1,787 for work- ers in the United States). Assume this random variable follows a normal distribution with a standard deviation of 420 hours.

a. What is the probability that a randomly selected South Korean worked 1. less than 1,800 hours? 2. between 2,000 and 2,200 hours? 3. between 1,200 and 1,600 hours? 4. between 2,300 and 2,500 hours?

b. Confirm the answers to part a using Excel or PHStat. c. How many hours will a South Korean work if he or she

is in the 90th percentile in terms of the hours an average Korean works?

6.45 In a recent, USA Today article titled Consumers Are Caught in a Sugar Battle, Americans consume an average of 22.2 teaspoons of sugar per day. Assume daily sugar consump- tion follows the normal distribution with a standard devia- tion of 6.5 teaspoons.

a. What is the probability that a randomly selected American will consume 1. more than 25 teaspoons of sugar today? 2. between 20 and 27 teaspoons of sugar today? 3. between 10 and 18 teaspoons of sugar today? 4. between 28 and 35 teaspoons of sugar today?

b. Confirm the answers to part a with Excel or PHStat. c. How many teaspoons of sugar will a person consume

today if he or she is in the 65th percentile?

6.46 According to the U.S. Labor Department, average weekly wages in February 2013 were $822. Assume weekly wages in the United States follow the normal distribution with a standard deviation of $190.

a. What is the probability that a randomly selected worker in the United States earned 1. less than $900 per week? 2. between $800 and $920 per week?

3. between $550 and $700 per week? 4. between $750 and $1,100 per week?

b. Confirm the answers to part a with Excel or PHStat. c. What weekly wage will a person earn if he or she is in

the 45th percentile?

6.47 The RAD Construction Company has been awarded a con- tract with the state of Delaware to replace a bridge on Pike Creek Road. Based on experience with previous projects, RAD feels the project completion time follows a normal distribution with a mean of 92 days and a standard deviation of 9 days. RAD anticipates the net profit for completing the project is $125,000. If the company completes the project in less than 84 days, it will receive an additional $20,000 bonus from the state. If it takes longer than 106 days to com- plete the project, the state of Delaware will penalize RAD $15,000. What is RAD’s expected profit for the project?

6.48 Pew Research Center surveyed adults who were 18–64 years old and had their own children under age 18 living in their household in 2011 on the amount of time spent on various activities. The results indicated that dads spent an average of 10 hours per week doing housework (compared to 18 hours per week for moms). Assume the hours per week doing housework follows the normal probability dis- tribution with a standard deviation of 3.2 hours per week and that these statistics still hold true.

a. What is the probability that a randomly selected dad from this group spent 1. less than 11 hours doing housework this week? 2. more than 12 hours doing housework this week? 3. between 7 and 9 hours doing housework this week? 4. between 8 and 15 hours doing housework this week?

b. Use Excel or PHStat to confirm the answers in part a. c. How many of hours of housework will a dad in this

group perform if he is in the 70th percentile?

6.49 Pew Research Center surveyed adults that were 18–64 years old and had their own children under age 18 living in their household in 2011 on the amount of time spent on various activities. The results indicated that moms spent an average of 21 hours per week doing paid work (compared to 37 hours per week for dads). Assume the hours per week doing paid work follows the normal probability distribution with a standard deviation of 6.7 hours per week and that these statistics still hold true.

a. What is the probability that a randomly selected mom from this group spent 1. less than 18 hours doing paid work this week? 2. more than 18 hours doing paid work this week? 3. between 23 and 30 hours doing paid work this week? 4. between 11 and 19 hours doing paid work this week?

b. Use Excel or PHStat to confirm the answers in part a. c. How many of hours of paid work will a mom in this

group perform if she is in the 20th percentile?

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6.50 The recent economic recession caused a plunge in new cars sales from 2008 to 2010. This trend resulted in a significant shortage of used cars during the next few years which, in turn, pushed up used car prices. Suppose that the average price of a 5-year-old used car is $16,230 with a standard deviation of $4,740. Assume that the price of a 5-year-old used car follows the normal probability distribution.

a. What is the probability that a randomly selected 5-year- old car costs 1. less than $18,500? 2. more than $11,300? 3. between $10,000 and $14,000? 4. between $12,500 and $17,000?

b. Use Excel or PHStat to confirm the answers in part a. c. What is the cost of a 5-year-old car in the 90th

percentile?

6.51 In an effort to hold colleges accountable for the benefits of a degree, legislation has been recently proposed to require institutions to provide prospective students with salary infor- mation for graduates. The state of Virginia has traditionally published starting salary data for their graduating students. For example, the average salary of James Madison Univer- sity (JMU) students who have recently earned a bachelor’s degree was reported to be $35,224. Assume the starting sal- ary for a recent JMU graduate follows the normal probability distribution with a standard deviation of $5,070.

a. What is the probability that a randomly selected JMU graduate from this class had a starting salary 1. less than $40,000? 2. more than $38,000? 3. between $24,000 and $32,000? 4. between $30,000 and $40,000?

b. Use Excel or PHStat to confirm the answers in part a. c. What starting salary of a JMU graduating student from

this class is in the 75th percentile?

6.52 Automotive dealerships have made recent efforts to reduce the amount of time required to purchase a car. One strategy is to use the Internet to streamline the process by provid- ing prospective buyers with car information. According to J.D. Powers, the average number of minutes to negotiate a deal to purchase a car at the dealership was 60.3 minutes in 2012. Assume the number of minutes to negotiate a deal follows the normal probability distribution with a standard deviation of 17.9 minutes and that these statistics still hold true.

a. What is the probability that a random customer spent 1. less than 30 minutes negotiating a deal? 2. more than 45 minutes negotiating a deal? 3. between 50 and 90 minutes negotiating a deal? 4. between 65 and 85 minutes negotiating a deal?

b. Use Excel or PHStat to confirm the answers in part a. c. How many minutes of negotiating are in the 35th

percentile?

6.53 In spite an unemployment rate of 26% and an economy in crisis, 70% of Spaniards responded that they were in fa- vor of keeping the euro rather than returning to the peseta and the national currency in 2012. A random sample of 32 Spaniards was selected. Use the normal approximation to the binomial distribution to answer the following:

a. Calculate the mean and standard deviation for this distribution.

b. What is the probability that 20 or more Spaniards from this sample are in favor of retaining the euro?

c. What is the probability that exactly 22 Spaniards from this sample are in favor of retaining the euro?

d. What is the probability that 15, 16, 17, 18, or 19 Spaniards from this sample are in favor of retaining the euro?

e. What is the probability that less than 24 Spaniards from this sample are in favor of retaining the euro?

6.54 According to Leichtman Research Group, 69% of U.S. households had at least one HDTV in 2012. A random sam- ple of 20 U.S. homes was selected. Use the normal approxi- mation to the binomial distribution to answer the following:

a. Calculate the mean and standard deviation for this distribution.

b. What is the probability that 12 or more households own at least one HDTV?

c. What is the probability that exactly 16 households own at least one HDTV?

d. What is the probability that 11, 12, or 13 households own at least one HDTV?

e. What is the probability that 18 or 19 households own at least one HDTV?

6.55 A process produces strings of Christmas tree lights that historically have experienced a defective rate of 4%. A customer has placed an order for 150 strings of lights. Use the normal approximation to the binomial distribution to answer the following:

a. Calculate the mean and standard deviation for this distribution.

b. What is the probability that less than 2 strings in this or- der will be defective?

c. What is the probability that exactly 7 strings in this order will be defective?

d. What is the probability that 3, 4, or 5 strings in this order will be defective?

e. What is the probability that 8, 9, or 10 strings in this or- der will be defective?

6.56 A recent survey conducted by the Allstate Foundation reported that 68% of teens admitted to texting while driving. A random sample of 40 teens is selected. Use the normal approximation to the binomial distribution to answer the following:

a. Calculate the mean and standard deviation for this distribution.

b. What is the probability that less than 29 teens of the 40 admit to texting while driving?

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c. What is the probability that exactly 24 teens of the 40 admit to texting while driving?

d. What is the probability that more than 32 teens of the 40 admit to texting while driving?

e. What is the probability that 26, 27, or 28 teens of the 40 admit to texting while driving?

6.57 Much of the NCAA men’s basketball tournament, oth- erwise known as March Madness, takes place during the workweek. According to CouponCabin.com, 34% of em- ployed adults who plan to watch at least one game plan to watch tournament games from work in 2013. A random sample of 50 employed adults who plan to watch at least one tournament game was selected. Use the normal ap- proximation to the binomial distribution to answer the following:

a. Calculate the mean and standard deviation for this distribution.

b. What is the probability that more than 19 adults from this sample will watch March Madness at work?

c. What is the probability that exactly 15 adults from this sample will watch March Madness at work?

d. What is the probability that less than 23 adults from this sample will watch March Madness at work?

e. What is the probability that 12, 13, 14, 15, 16, or 17 adults from this sample will watch March Madness at work?

6.58 According to the U.S. Census Bureau, 47.5% of the U.S. population lives in the Eastern Time Zone. A random sample of 50 U.S. residents is selected. Use the normal approximation to the binomial distribution to answer the following:

a. Calculate the mean and standard deviation for this distribution.

b. What is the probability that less than 20 people from the sample live in the Eastern Time Zone?

c. What is the probability that exactly 27 people from the sample live in the Eastern Time Zone?

d. What is the probability that more than 22 people from the sample live in the Eastern Time Zone?

e. What is the probability that 26, 27, or 28 people from the sample live in the Eastern Time Zone?

6.59 The average number of days that a house was on the market in Austin, Texas, in January 2013 was 71 days. Assume the time on the market follows an exponential probability distribution.

a. What is the probability that a randomly selected house will be on the market less than 50 days?

b. What is the probability that a randomly selected house will be on the market between 70 and 100 days?

c. What is the probability that a randomly selected house will be on the market exactly 80 days?

d. Use Excel to verify the results from parts a and b. e. What is the standard deviation for this distribution?

6.60 During a recent men’s college basketball season, the Villanova Wildcats averaged 22.1 personal fouls per 40-minute game. Assume the time between personal fouls follows an expo- nential distribution.

a. What is the probability that the next personal foul com- mitted by Villanova will occur within the first minute of a game?

b. What is the probability that the next personal foul com- mitted by Villanova will occur between the first 2 and 4 minutes of a game?

c. What is the probability that the next personal foul com- mitted by Villanova will occur more than 3 minutes after the start of a game?

d. Use Excel to verify the results from parts a and b. e. What is the standard deviation for this distribution?

6.61 Patients arrive at the emergency room for a local hospital at an average rate of 6.5 per hour. Assume the time between patient arrivals follows an exponential distribution.

a. What is the probability that the next patient will arrive in the emergency room within the next 3 minutes?

b. What is the probability that the next patient will arrive in the emergency room between the next 10 and 15 minutes?

c. What is the probability that the next patient will arrive in the emergency room after the next 12 minutes?

d. What is the probability that the next patient will arrive in the emergency room in exactly 12 minutes?

e. Use Excel to verify the results from parts a and b. f. What is the standard deviation for this distribution?

6.62 A busy intersection in the state of Delaware averages a traf- fic accident every 32 days. Assume the time between acci- dents follows an exponential distribution.

a. What is the probability that the next accident will occur within the next 10 days?

b. What is the probability that the next accident will occur between the next 15 and 25 days?

c. What is the probability that the next accident will occur after the next 40 days?

d. Use Excel to verify the results from parts a and b. e. What is the standard deviation for this distribution?

6.63 Suppose the time between hits on an e-commerce Web site follows the exponential distribution with a mean time of 2.7 minutes.

a. What is the probability that next hit on the Web site will occur within the next 5 minutes?

b. What is the probability that next hit on the Web site will occur within the next 2 to 6 minutes?

c. What is the probability that next hit on the Web site will occur after the next 3 minutes?

d. Use Excel to verify the results from parts a, b, and c. e. What is the standard deviation for this distribution?

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6.64 The optical department at Costco serves customers on a first-come-first-serve basis. Suppose the time between cus- tomer arrivals follows the exponential distribution. Histori- cal data indicates that customers arrive at an average rate of 10.8 per hour.

a. What is the probability that next customer will arrive within the next 3 minutes?

b. What is the probability that next customer will arrive within the next 6 to 12 minutes?

c. What is the probability that next customer will arrive after the next 10 minutes?

d. Use Excel to verify the results from parts a, b, and c. e. What is the standard deviation for this distribution?

6.65 Suppose a number of months before your car battery needs to be replaced follow the exponential distribution with a mean of 50.0 months.

a. What is the probability that your car battery will need to be replaced within 24 months?

b. What is the probability that your car battery will need to be replaced within 30 to 60 months?

c. What is the probability that your car battery will need to be replaced after 84 months?

d. Use Excel to verify the results from parts a, b, and c. e. What is the standard deviation for this distribution?

6.66 Orange Lake Resort is a major vacation destination near Orlando, Florida, adjacent to the Disney theme parks. Be- cause the property consists of 1,450 acres of land, Orange Lake provides shuttle buses for visitors who need to travel within the resort. Suppose the wait time for a shuttle bus follows the uniform distribution with a minimum time of 30 seconds and a maximum time of 9.0 minutes.

a. What is the probability that a visitor will need to wait more than 3 minutes for the next shuttle?

b. What is the probability that a visitor will need to wait less than 5.5 minutes for the next shuttle?

c. What is the probability that a visitor will need to wait between 4 and 8 minutes for the next shuttle?

d. Calculate the mean and standard deviation for this distribution.

e. Orange Lake has a goal that 80% of the time, the wait for the shuttle will be less than 6 minutes. Is this goal being achieved?

6.67 Precision Doors is an organization that installs new garage doors on residential homes. Suppose the installation time for a residence follows the uniform distribution with a minimum time of 190 minutes and a maximum time of 370 minutes.

a. What is the probability that an installation will require less than 4 hours to complete?

b. What is the probability that an installation will require more than 5 hours to complete?

c. What is the probability that an installation will require between 250 and 350 minutes to complete?

d. Calculate the mean and standard deviation for this distribution.

e. Precision Door has a goal that 80% of the time, the in- stallation time for a residence will be less than 6 hours. Is this goal being achieved?

6.68 Suppose the time to deliver a pizza after an order for one is placed on a local restaurant’s Web site follows a uniform distribution with a minimum time of 4 minutes and a maxi- mum time of 18 minutes.

a. Determine the probability that the next delivery will require less than 12 minutes.

b. Determine the probability that the next delivery will require between 6 and 10 minutes.

c. Determine the probability that the next delivery will require more than 7 minutes.

d. Determine the probability that the next delivery will require exactly 15 minutes.

e. Calculate the mean and standard deviation for this distribution.

6.69 Aircraft turnaround time is defined as the time required to prepare an aircraft for its next flight after arriving at a terminal. Assume that United Airlines’ turnaround time follows a uniform distribution with a minimum time of 24 minutes and a maximum time of 56 minutes.

a. Determine the probability that the next United Airlines aircraft will require less than 30 minutes to turn around.

b. Determine the probability that the next United Airlines aircraft will require between 30 and 40 minutes to turn around.

c. Determine the probability that the next United Airlines aircraft will require more than 40 minutes to turn around.

d. Calculate the mean and standard deviation for this distribution.

e. United Airlines wants aircraft turnaround times to be less than 50 minutes 90% of the time. Is this goal being achieved?

6.70 While waiting for an order to arrive from its supplier (known as the lead time), the golf-department manager at a local Dick’s Sporting Goods store will sell an average of 60 boxes of Titleist golf balls. Assume the demand during the lead time for the balls follows a normal distribution with a standard deviation of 12 boxes. Determine the reor- der point that will ensure the store a 95% probability that it will have enough inventory of golf balls until the next order arrives.

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a. NORM.DIST17, 5, 3, TRUE2 = 0.7475 b. NORM.DIST10, 5, 3, TRUE2 = 0.0478 c. Using PHStat 1see at right2 = 0.2782

Y o U r T U r N # 4

a. z11.9 = x - m s

= 11.9 - 10.7

0.8 = 1.50

z9.8 = x - m s

= 9.8 - 10.7

0.8 = -1.13

P19.8 … x … 11.92 = 0.9332 - 0.1292 = 0.8040 b. z12.1 =

x - m s

= 12.1 - 10.7

0.8 = 1.75

z11 = x - m s

= 11 - 10.7

0.8 = 0.38

P111 … x … 12.12 = 0.9599 - 0.6480 = 0.3119

c. z10.3 = x - m s

= 10.3 - 10.7

0.8 = -0.50

z9.7 = x - m s

= 9.7 - 10.7

0.8 = -1.25

P19.7 … x … 10.32 = 0.3085 - 0.1056 = 0.2029

Y o U r T U r N # 3

a. x = m { zs x = 39.2 + 11213.72 = 42.9 x = 39.2 - 11213.72 = 35.5 b. x = m { zs x = 39.2 + 12213.72 = 46.6 x = 39.2 - 12213.72 = 31.8

c. x = m { zs x = 39.2 + 13213.72 = 50.3 x = 39.2 - 13213.72 = 28.1

Y o U r T U r N # 2

a. z800 = x - m s

= 800 - 930

140 = -0.93

P1x … 8002 = P1z … -0.932 = 0.1762 b. z1,000 =

x - m s

= 1,000 - 930

140 = 0.50

P1x … 1,0002 = P1z … 0.502 = 0.6915 c. P1x = 9002 = 0 because x is a continuous random variable. d. The station will not run out of inventory if demand during the lead time is less

than or equal to 1,200 gallons. We need to calculate P1x … 1,2002. z1,200 =

x - m s

= 1,200 - 930

140 = 1.93

P1x … 1,2002 = P1z … 1.932 = 0.9732

e. Find the closest value to 0.90 in the body of Table 6.2. This will be 0.8997 in row 1.2 and column 0.08.

z = x - m s

x = m + zs x = 930 + 11.28211402 = 930 + 179.2 = 1,109.2 gallons

Y o U r T U r N # 1

Chapter 6 Solutions to Your Turn

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Y o U r T U r N # 6

l = 32.2 shots per game

60 minutes per game = 0.537 shots per minute

m = 1

l =

0.537 = 1.86 minutes per shot

a1. P1x … a2 = 1 - e - al P1x … 12 = 1 - e - 11210.5372 P1x … 12 = 1 - e - 0.537 = 1 - 0.5845 = 0.4155

a2. P12 … x … 32 = P1x … 32 - P1x … 22 P1x … 32 = 1 - e - 13210.5372 = 1 - e - 1.611

= 1 - 0.1997 = 0.8003 P1x … 22 = 1 - e - 12210.5372 = 1 - e - 1.074

= 1 - 0.3416 = 0.6584 P12 … x … 32 = 0.8003 - 0.6584 = 0.1419 b. P1x … 12 = EXPON.DIST11, 0.537, TRUE2 = 0.4155 P1x … 32 = EXPON.DIST13, 0.537, TRUE2 = 0.8003 P1x … 22 = EXPON.DIST12, 0.537, TRUE2 = 0.6584 c. s = m = 1.86 minutes per shot

n = 60 p = 0.10 q = 0.90

m = np = 160210.102 = 6.0 s = 1npq = 1160210.1211 - 0.12 = 15.4 = 2.32

a. Less than 5 includes 0, 1, 2, 3, and 4. Add 0.5 to 4 for continuity factor.

z4.5 = x - m s

= 4.5 - 6.0

2.32 = -0.65

P1x 6 52 = P1z 6 -0.652 = 0.2578 b. Subtract 0.5 from 6 and add 0.5 to 9 for continuity factor.

z5.5 = x - m s

= 5.5 - 6.0

2.32 = -0.22

z9.5 = x - m s

= 9.5 - 6.0

2.32 = 1.51

P16 … x … 92 = P15.5 … x … 9.52 = P1 -0.22 … z … 1.512 P1 -0.22 … z … 1.512 = 0.9345 - 0.4129 = 0.5216 c. Nine or more includes 9, 10, 11, c,60. Subtract 0.5 from 9 for continuity

factor.

z8.5 = x - m s

= 8.5 - 6.0

2.32 = 1.08

P1x Ú 92 = P1z Ú 1.082 = 1.0 - 0.8599 = 0.1401

Y o U r T U r N # 5

x1 = 140 x2 = 170

P1x1 … x … x22 = x2 - x1 b - a

a. P1140 … x … 1452 = 145 - 140 170 - 140

= 5

30 = 0.167

b. P1148 … x … 1652 = 165 - 148 170 - 140

= 17

30 = 0.567

c. P1x1 … x … x22 = x2 - x1 b - a

= 0.75

x2 - 140 170 - 140

= 0.75

x2 = 10.7521302 + 140 = 162.5 minutes

Because 162.5 minutes is less than 165 minutes, the goal of management is being met.

d. m = a + b

2 =

140 + 170 2

= 155 minutes

s = b - a112 = 170 - 140112 = 303.46 = 8.67 minutes

Y o U r T U r N # 7

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I n t h I s c h a p t e r , y o u w I l l l e a r n t o :

■ Distinguish between the different types of probability sampling.

■ Identify the limitations of nonprobability sampling.

■ Calculate sampling errors.

■ Identify the sources of nonsampling errors.

■ Use the sampling distribution of the mean.

■ Understand the importance of the Central Limit Theorem.

■ Use the sampling distribution of the proportion.

■ Use the finite population correction factor.

Sampling and Sampling Distributions

C h a p t e r 7 M a p

7.1 Why Sample? | 290

7.2 Types of Sampling | 291

7.3 Sampling and Nonsampling Errors | 298

7.4 The Central Limit Theorem | 301

7.5 The Sampling Distribution of the Proportion | 317

7 C h a p t e r

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290 C h a p t e r 7 | Sampling and Sampling Distributions

Stats in practice: Super Bowl XLVII takes the trophy: Most Watched telecast ever!

That’s right. At the time of this printing, the Baltimore Ravens’ 34-31 victory over the San Francisco 49ers for the NFL Championship on February 3, 2013, in New Orleans, Louisiana was the most watched broadcast in television history with a total audience of 164.1 million viewers. The previous record holder was the 2011 Super Bowl, a game in which the Green Bay Pack- ers defeated the Pittsburgh Steelers in front of 162.9 million total viewers. In fact, the top five most-watched broadcasts are the last five Super Bowls which attests to the popularity of the NFL.

When considering these facts one might ask, “How do they know that 164.1 million people watched the 2013 Super Bowl? I watched the game and nobody counted me.” To answer this important question, we need to explore the topic of statistical sampling. Back in Chapter 1, we defined a population as repre- senting all possible outcomes or measurements of interest and a sample as a subset of a population. Nielson Media Research, a company in the sampling business, provides TV networks and advertisers with information about the sizes of television view- ing audiences across the nation by surveying 5,000 of the 129 million households there are in the United States. Par- ticipating households are selected very carefully so that they are representative of the entire U.S. population. Each person in these households is provided with a “people meter,” which

is a small box the individual uses to indicate when he or she starts and stops watching a TV program. Based on the results of this relatively small sample, Nielsen can develop an accurate estimate of the viewing habits of the entire country.

The information Nielsen collects plays a critical role in the rates that TV networks charge advertisers to run commercials during programs. Generally, the larger a program’s audience is, the more the commercials aired during the show cost advertisers. For example, a 30-second commercial for the top-rated 2013 Super Bowl cost $3.8 million.

In addition to knowing how many viewers are watching a program, advertisers also want to know who is watching. The people meters keep track of both the ages and genders of the viewers. For example, Nielsen’s data shows that the number of male viewers of Super Bowls have traditionally exceeded the number of female viewers. This helps explain why you see so many more beer commercials during the Super Bowl, than say, advertisements for perfume. The opposite holds true with the Olympics TV audience where the majority of viewers are typi- cally female. Figure 7.1 shows the gender breakdown for Super Bowl XLV and the 2010 Winter Olympics.

{ figure 7.1 Percentage of Super Bowl

XLV and 2010 Winter Olympic Viewers by Gender

44% 56%46% 54%

Male Female Male Female

Super Bowl XLV Viewers

2010 Olympic Viewers

7.1 Why Sample? Back in Chapter 1, we introduced the concept of a population and a sample. A population refers to all possible subjects that are of interest to us in a particular study. The term sample refers to a portion of the population that is representative of the population from which it was selected. In the previous Stats in Practice example, population is defined as every television viewer and the sample consists of all the households that possess a Nielsen people meter.

Why not just measure the whole population rather than rely on only a sample? That’s a very good question! Depending on the circumstances, measuring an entire population could be very expensive or just plain impossible. Let’s use a quality control example in business. Suppose the target for salt content set by Kellogg’s for my Cheez-It crackers is 1%. Higher or lower salt content would have a negative impact on the taste of the product.

A population represents all possible subjects of interest. A sample is a subset of a population.

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C h a p t e r 7 | 7.2 Types of Sampling 291

How does Kellogg’s know that its target is being achieved for all of the Cheez-Its produced today? Well, it’s not practical to check the salt content of every cracker coming off the production line because any cracker tested would not be fit for sale to the consumer. A better solution would be to routinely select a small, but representative, sample and test its salt content. These sample results would be used to determine if the process was producing satisfactory crackers. More specifically, we can use the salt content from the sample to draw conclusions about the salt content of the entire production process, which, in this case, is the population. By gathering enough representative data, we can infer what is happening overall. That’s the whole concept of inferential statistics in one paragraph!

Even if we could feasibly measure the entire population, to do so would often be a waste of time and money—resources businesses consider very valuable these days. If a sample is selected properly and the analysis performed correctly, this information can be used to make an accurate assessment of the entire population. There is very little added benefit to continue beyond the sample and measure everything.

One example where a sampling decision was recently made occurred at Goldey- Beacom College, where I currently teach. I am a member of the Academic Honor Code Committee, which launched a project to gather information about the attitudes our college students have toward academic integrity. It would have been possible to ask every student at our college to respond to the survey, but it was actually unnecessary with the availability of inferential statistics. The committee eventually made the intelligent decision to sample only a portion of the students to infer the attitudes of the larger population. Based on our sample, we concluded that the majority of students at our college took academic integrity very seriously.

There are, however, risks involved with making decisions based on sampling. By sampling, we expose ourselves to errors that could lead to poor decisions. As we will see later in the chapter, we can quantify the probability of these errors occurring. As you will see later in this textbook, these probabilities are often small enough to still make sampling worthwhile.

7.2 Types of Sampling There are many options available for gathering samples from a population. The two basic types that we will discuss in this chapter are probability and nonprobability sampling.

probability Sampling A probability sample is a sample in which each member of the population has a known, non- zero, chance of being selected for the sample. However, this does not necessarily mean that every member has the same chance of being selected. With probability sampling, we have

Stats in practice: age Does Matter

Attracting viewers from the coveted 18–49 age group will also boost advertising revenue. For example, commercial time for American Idol during the 2012 season cost $340,825 per 30-second spot. Dancing with the Stars, which has about the same size viewing audience as Idol, commands only $160,466 per commercial. The difference is explained by the average age of the viewers, which is under 50 for Idol and over 50 for Dancing. The reasoning is that we over-50-year-old consumers are more set in our ways and are not as easily influenced by advertising as you “young-uns.”

A probability sample is a sample in which each member of the population has a known, nonzero, chance of being selected for the sample.

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292 C h a p t e r 7 | Sampling and Sampling Distributions

the ability to perform a variety of inferential statistical tests that will allow us to draw reli- able conclusions about the population. These tests are the focus of many of the remaining chapters in this textbook.

There are five main types of probability sampling techniques statisticians use: simple random, systematic, stratified, cluster, and resampling. I will discuss each of the techniques in the following sections.

Simple Random Sampling A simple random sample is a sample in which every member of the population has an equal chance of being chosen. One way to ensure this occurs is to use the sampling tool in Excel’s Data Analysis ToolPak (Windows version). Unfortunately, this sampling procedure is not available on Excel for Mac. To demonstrate, let’s go back to the academic integrity survey at my school. Suppose I want to draw a simple random sample of 10 students out of 1,800 total students to participate in the survey. In Figure 7.2, Column A of the Excel file titled GBC Students.xlsx shows the student ID numbers for each of the 1,800 students. The numbers are sorted in ascending order. (I have hidden Rows 13 through 1,796 to conserve space.) To select 10 students at random for the sample, I take the following steps:

1. Open the Excel file GBC Students.xlsx.

2. Select Data 7 Data Analysis. 3. In the Data Analysis dialog box, select Sampling and click OK.

4. In the Sampling dialog box, click on the text box for Input Range: and select Cells A1:A1800 as shown in Figure 7.2.

5. Select Random under Sampling Method.

6. In the Number of Samples: text box, type 10.

7. Click on the text box for Output Range: select Cell C1, and click OK.

A simple random sample is a sample in which every member of the population has an equal chance of being chosen.

Sampling without replacement means that once a member of the population is selected to be part of the sample, it is not returned to the population. Therefore that member cannot be chosen again in the sample.

{ figure 7.2 Using Excel to Generate a

Simple Random Sample

Column C in Figure 7.3 (on the next page) provides us with the simple random sample of 10 students from the population.

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C h a p t e r 7 | 7.2 Types of Sampling 293

{ figure 7.3 Finished Simple Random

Sample

Formula 7.1 for the systematic sampling constant

k = N n

where

N = Size of the population n = Size of the sample Using my academic integrity survey, with a population of 1,800 students and a sample

of 10,

k = N n

= 1,800

10 = 180

From a listing of the entire population, I would choose every 180th student to be in- cluded in the sample. Excel can also do this for you with the sampling tool I just introduced:

1. Using the GBC Student.xlsx file with a blank Column C, select Periodic in the Sampling dialog box, as shown in Figure 7.4 on the next page.

2. In the Period: text box, type in the value for k, which is 180.

3. Click on the text box for Output Range:, and select Cell C1, and click OK.

The student with ID Number 55516 would be the first member in my sample, followed by ID Number 63282, and so forth. This technique ensures that all 1,800 students have the same chance of being selected for the survey. Also, be advised that Excel’s random sampling tool uses sampling with replacement. This means that after a member of the population has been selected for the sample, he or she is placed back into the population and can be chosen again for the same sample. Therefore, it is possible that a student could be selected more than once for the sample. If this is undesirable, accommodations can be made to replace the duplicated member in the sample.

Systematic Sampling Another way to ensure that samples are selected randomly is to use systematic sampling. This technique results in selecting every kth member of the population being in your sample. Every second member could be selected, or every third member, and so on. (Remember in your physical education classes when players were counted off in a line and placed in teams? Systematic sampling is similar.)

But how do you know which value of k to choose? The value of k, the systematic sam- pling constant, is determined using Equation 7.1.

In systematic sampling, every kth member of the population is chosen for the sample. The value of k is determined by dividing the size of the population (N ) by the size of the sample (n).

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294 C h a p t e r 7 | Sampling and Sampling Distributions

Column C in Figure 7.5 provides us with the systematic sample of 10 students from the population. This list includes the 180th student in our list, followed by the 360th student, and so forth.

{ figure 7.4 Using Excel to Generate a

Systematic Sample

{ figure 7.5 Finished Systematic Sample

Generating a systematic sample is easier to do manually than generating a simple ran- dom sample. For example, if 500 students walk into a classroom building during the first hour of school, and I need a sample of 100 students, using Equation 7.1, I can easily figure out what k is: k = 500>100, or 5. I could then select every fifth student arriving in the morn- ing to participate in the survey. Systematic sampling is better than just randomly selecting students walking by because I could have a bias toward selecting some students rather than others (for example, students I know from my classes).

One concern about systematic sampling is periodicity, which is a pattern in the popula- tion that is consistent with the value of k. For instance, let’s say I’m conducting a survey on campus asking students how many hours they study during the week and I select every fourth week to collect my data. Because we are on an eight-week semester schedule at Goldey-Beacom, every fourth week could end up being either a midterm week or a finals week, which would result in a higher number of study hours than normal (or at least I would hope so). The presence of periodicity can result in a sample that is not representative of the entire population. When using systematic sampling, care must be taken to ensure that period- icity does not impact the collection of your data.

Finally, the starting point for systematic sampling does not need to be the first data value. We can randomly choose the starting position and proceed by counting down k values from this point to choose our sample.

One concern about systematic sampling is periodicity, which is a pattern in the population that is consistent with the value of k.

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C h a p t e r 7 | 7.2 Types of Sampling 295

Stratified Sampling In stratified sampling, we divide the population into mutually exclusive groups, or strata, and randomly sample from each of those groups. For example, suppose we want our aca- demic integrity sample to mirror the proportion of freshmen, sophomore, junior, and senior students in the overall college population and that we would like to select a sample size of 100 students. Because freshmen comprise 30% of the student population our sample of 100 students should contain 30 freshmen. Likewise, because sophomores comprise 22% of the student population our sample of 100 students should contain 22 sophomores. The same logic applies to juniors and seniors. Table 7.1 shows the number of freshmen, sophomores, juniors, and seniors required and calculates the size of each stratum in our sample based on its percentage relative to the population. We can then apply the simple random or systematic sampling methods to select the students from each group.

Recall that you learned in Chapter 4 that two events are considered to be mutually exclusive if they cannot occur at the same time during an experiment.

{ table 7.1 | Stratified Sampling for the Academic Integrity Survey

strata

size (percent) of population sample

Freshmen 540 (30%) 0.30 * 100 = 30

Sophomore 396 (22%) 0.22 * 100 = 22

Junior 504 (28%) 0.28 * 100 = 28

Senior 360 (20%) 0.20 * 100 = 20

Total 1,800 (100%) 100

Why would we go through the trouble of using stratified sampling versus simple ran- dom or systematic sampling? Suppose seniors, in general, take academic integrity much more seriously than freshmen do. Also, suppose that we used systematic sampling, and, by chance, 40% of our sample consisted of seniors. Remember, seniors account for only 20% of the student body. If we use our sample to draw conclusions about the attitudes of the popu- lation, we may be led to believe our students take academic integrity much more seriously than they really do.

We used stratified sampling because we felt that the class the students belonged to— whether they were freshman, sophomores, juniors, or seniors—was an important factor in how they would respond to our survey. This helps ensure that the sample is representative of the overall population.

Cluster Sampling Clusters are defined as mutually exclusive groups, each which is representative of the population. Cluster sampling involves randomly selecting clusters to be part of the final sample. These clusters are often selected based on geography to help simplify the sam- pling process. Using the academic integrity survey as an example, the clusters could be defined as classrooms in use at certain times of the day. We would randomly choose dif- ferent classrooms to participate in the survey. We could then include every student in the classroom as a participant in the survey or use simple random sampling within the class- room to select students.

My students often confuse strata and clusters. The members of a specific stratum all have something in common, such as each student being a freshman from my previ- ous example. As a result, strata tend to be homogeneous collections, each with a cer- tain characteristic of interest. Clusters, on the other hand, are “mini-subsets” of the larger population and therefore tend to be a melting pot of various characteristics. For example, a particular classroom (cluster) could have a mixture of freshman, sopho- mores, juniors, or seniors.

Cluster sampling is often used to test-market new products. Test-market cities are identi- fied, and individual customers within those locations are sampled for their opinions about these products. Cluster sampling can be a cost-effective way to collect a probability sample if performed carefully to ensure that the sample is representative of the population.

Cluster sampling involves dividing the population into mutually exclusive groups, or clusters, that are each representative of the population and then randomly selecting clusters to form the final sample. These clusters are often selected based on geography to help simplify the sampling process.

When choosing a stratified sample, we divide the population into mutually exclusive groups, or strata, and randomly sample from each of those groups. The strata are based on important variables that can have an impact on the data collected and the results that are achieved.

The purpose of cluster sampling is to make the sampling process more economical. The purpose of stratified sampling is to make the sampling process more accurate.

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296 C h a p t e r 7 | Sampling and Sampling Distributions

Resampling* Resampling is a statistical technique where many samples are repeatedly drawn from an available population. A specific type of resampling technique is known as the bootstrap method, developed by Bradley Efron, a faculty member in Stanford University’s Department of Statistics. Bootstrapping involves using computer software to extract many samples with replacement in order to estimate a certain parameter of the population, such as a mean or proportion.

To demonstrate the bootstrap method, consider the following simple example. Suppose the manager at a particular Walmart location would like to estimate the proportion of female shoppers at her store. She randomly selects 100 transactions and discovers that 58 of them were transacted with female customers. We now know that the sample proportion is 58% female, but the population proportion of female shoppers is still unknown.

In an Excel file, we can populate cells A1 to A58 with the number “1” to represent the female shoppers. Cells A59 through A100 would contain the number “0” for the men. Using

Resampling is a statistical technique where many samples are repeatedly drawn from a population.

This Statistics in Practice demonstrates the use of cluster sampling in the marketplace.

Companies that are looking to test their product or services in the American marketplace would like to choose a location that best represents the typical U.S. consumer. To help identify these locales, Acxiom Corporation has ranked the populations in the top 150 Metropolitan Statistical Areas (MSAs) based on how closely they mirror the American consumer. At the top of this list is Albany, New York.

“Albany has almost the same proportion of consumers as is found in the nation as a whole—from people just leaving home or school, buying a home, raising kids, approaching retirement, launching the kids out of the house, to being fully retired,” said Acxiom’s Tiffany Weatherly, who supervised the research for this project.

The following MSAs made the top 10 as far as being the most representative of the U.S. consumer:

1. Albany, N.Y. 2. Rochester, N.Y. 3. Greensboro, N.C. 4. Birmingham, Ala. 5. Syracuse, N.Y. 6. Charlotte, N.C. 7. Nashville, Tenn. 8. Eugene, Ore. 9. Wichita, Kans. 10. Richmond, Va.

Companies would also be interested in knowing which locations to avoid conducting test markets due to the lack of suitability of the local population. The following list is the bottom 10 MSAs according to Acxiom.

1. El Paso, Tex. 2. Columbia, Mo. 3. Tallahassee, Fla. 4. Brownsville, Tex. 5. Provo, Utah 6. Ocala, Fla. 7. McAllen, Tex. 8. Honolulu, Hawaii 9. San Francisco, Calif. 10. New York City, N.Y.

Based on: Jonathan Portis, “Which American City Provides The Best Consumer Test Market?” Acxiom Corporation News Release, May 24, 2004, http://www.businesswire .com/news/home/20040524005650/en/American-City-Consumer-Test-Market-Mirror-America.

Stats in practice: Which City Best represents the american Consumer? the Winner is albany, N.Y.

*This content is optional. Special thanks to Dr. Sunil Sapra from California State University for recom- mending the inclusion of this topic.

The bootstrap method involves using computer software to extract many samples with replacement in order to estimate a parameter of the population, such as a mean or proportion.

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C h a p t e r 7 | 7.2 Types of Sampling 297

Excel’s Sampling tool in Data Analysis, shown in Figure 7.2, we can generate a random sample of size 100 and calculate the proportion of female shoppers for this sample. Using bootstrapping software to assist us, we can repeat this sampling process, say 10,000 times or more (otherwise known as resampling). Some of these sample proportions of female shop- pers will be less than 58%, whereas others will be more than 58%. We now have a range of proportions that will provide us an idea of where the true population proportion lies based on our original sample of 100 shoppers. This concept is related to confidence intervals, which will be discussed at great length in Chapter 8. Remember, Excel’s random sampling tool uses sampling with replacement. This means that after a member of the population has been selected for the sample, he or she is placed back into the population and can be chosen again as part of the same sample. This is an important feature of bootstrapping.

As you can see in this example, the bootstrap method can be applied to estimate param- eters such as the population proportion, mean, median, or variance. Bootstrapping can also be used to estimate the standard deviation of a sample mean or proportion and is not limited by underlying assumptions that are necessary to conduct many statistical analyses.

Nonprobability Sampling A nonprobability sample is a sample in which the probability of a population member being selected for the sample is not known. A common type of nonprobability sample is a conve- nience sample.

A convenience sample is used when members of the population are chosen to become part of the sample simply because they are easily accessible. This type of sampling is useful when you are simply trying to gather some general information about the population. For ex- ample, suppose you would like some quick feedback from your customers about a new bak- ery product you’ve recently developed. Sending out free samples and having customers who happen to enter your store rate the product would be an example of a convenience sample. In most cases, it would not be appropriate to use these data to perform inferential statistics like we would with probability sampling.

For example, if I would like some student feedback on this textbook, I could select my current statistics classes as my sample. This would be a convenience sample because it is most likely not representative of all of the students in the nation who read statistics text- books; therefore, I need to be very careful with interpreting the results.

One of the most common types of convenience samples is an Internet poll. Most in- dividuals who participate in Internet polls have very little statistical training. They are im- pressed when they see thousands of responses recorded in real time and tend to believe the results. What they fail to see, however, is that this is a convenience sample, and that often the conductor of the poll has very little control over who participates in it. Generally, anyone with an Internet connection is eligible to participate, and people can respond to the survey multiple times by re-accessing the poll on a different day or from a different browser or computer. Of course, this reduces the credibility of the results. Suffice it to say, a sampling technique that gives you no control over who participates in your survey does not qualify as probability sampling.

A nonprobability sample is a sample in which the probability of a population member being selected for the sample is not known.

This is especially true because the students in my classes might be telling me what I wanted to hear in an effort to pass my course!

Stats in practice: Sampling in presidential races

Back in Chapter 1, I discussed one of the most famous conve- nience samples in history that caused much embarrassment for the poll takers. During the 1936 U.S. presidential race, the Literary Digest, which had correctly predicted the previous five presidential winners, predicted that Alf Landon would win the election over the actual winner, Franklin D. Roosevelt. The problem was that the Digest conveniently drew its sample from phone books and automobile registrations. People with phones and cars in 1936 tended to be wealthier Republicans and were

not representative of the entire voting population. Even though the Digest randomly selected names from these sources, they did not choose them from the proper population. In comparison, a 1968 Gallup poll used a probability sample to predict that Richard Nixon would receive 43% of the votes for the 1968 presidential election when in fact he won 42.9%. This Gallup poll was based on a sample size of only 2,000 people, whereas the disastrous 1936 Literary Digest poll sampled 2,000,000 people. Probability samples rule when conducted properly!

According to my father-in-law, Lindsay Greenplate, who was 12 years old during this election, there was a rumor that Roosevelt did not fly while campaigning because he was afraid of “Landon”. I did not make this up.

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298 C h a p t e r 7 | Sampling and Sampling Distributions

Try your hand with this next Your Turn section to make sure you understand the differ- ence between the sampling techniques we have discussed.

7.3 Sampling and Nonsampling Errors So far in this chapter, I realize I’ve been ranting about the marvels of sampling and how life on this planet could not exist as we know it today without this wonderful tool statisticians (sort of) invented. Maybe I’m exaggerating slightly, but sampling is a pretty cool idea that has made our lives a little better. However, I also tell my students that there is no free lunch in statistics because the benefits of sampling come with a cost. When we rely on relatively small samples to draw conclusions about larger populations, we expose ourselves to sam- pling errors (cue the scary music).

Before discussing the details of sampling errors, I need to introduce two new terms: parameters and statistics. Parameters are values that describe some characteristic of a popu- lation, such as its mean or median. For the most part, parameters are unknown because we rarely observe or measure an entire population. Statistics are values calculated from a sam- ple, such as the mean or the median. Statistics are used to estimate the value of a parameter. It is common practice, for example, to use the sample mean (a statistic) to estimate the value of the population mean (a parameter).

Because a statistic is based on just a portion of the population, it would be unreason- able to expect the sample mean and population mean to be the same. The difference between these two values is known as the sampling error of the sample mean. The formula for the sampling error is shown in Equation 7.2.

Parameters are values that describe some characteristic of a population, such as its mean or median.

Statistics are values calculated from a sample, such as the sample’s mean or median.

Beware of convenience sampling. Because no effort is made to ensure the sample is representative of the population, the results based on this type of sample are unreliable.

Identify the type of sampling technique for each of the following:

a. The first Monday of each month, I ask my customers who come to my store to fill out a satisfaction survey.

b. I randomly select four stores in a mall and ask each customer in those stores about his or her opinion of the latest health care legislation.

c. I position myself on a busy intersection of a city street and ask people what their opinions are of a local sports team.

d. Sixty percent of the students attending my college are female. I construct a random sample that consists of 60% females from the student population to ask what their opinions are of the college’s food service.

e. Using computer software, I randomly select 20 employees to participate in a job satisfaction survey.

Y o u r t u r N # 1N

Answers can be found on u page 329

Formula 7.2 for the sampling error of the sample Mean

Sampling Error = x - m

where

x = The sample mean m = The population mean

To demonstrate sampling errors and how they typically behave, consider the following example. Suppose I’m teaching a small, graduate-level statistics class composed of 10 stu- dents with the following ages.

24 33 28 41 38 30 24 33 37 32

Sampling error of the sample mean is defined as the difference between the sample statistic and the population parameter.

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C h a p t e r 7 | 7.3 Sampling and Nonsampling Errors 299

If I consider this class my population, the population mean can be found using Equation 3.2 from Chapter 3.

m = a N

i = 1 xi

N =

24 + 33 + 28 + g + 37 + 32 10

= 320

10 = 32.0

The average age of my class is 32.0 years old, which is my population parameter. Next, I would like to draw a random sample consisting of two students to estimate this parameter. I realize we already know the value of the parameter, but I ask you to play along and humor me. The method to my madness will become obvious soon.

Let’s say the random sample consists of the first two students 1n = 22 in our previous list. They are ages 24 and 33. The sample mean can be found using Equation 3.1 from Chapter 3:

x = a

i = 1 xi

n =

24 + 33 2

= 28.5

I use Equation 7.2 to calculate my sampling error:

Sampling Error = x - m = 28.5 - 32.0 = - 3.5

My sample statistic is 3.5 years below the population parameter due to the negative sampling error. Positive sampling errors would be the result of sample means greater than the population mean. Now, suppose I gather another random sample of two students, which happen to be the last two students in my original list of 10. They are ages 37 and 32. This sample mean is found to be

x = a

i = 1 xi

n =

37 + 32 2

= 34.5

I use Equation 7.2 again to calculate my latest sampling error.

Sampling Error = x - m = 34.5 - 32.0 = 2.5

This sample statistic is 2.5 years above the population parameter. This illustrates the first lesson about sampling errors: They can differ from one sample to the next. Sampling errors can also be positive or negative, depending on whether they are above or below the population mean.

In the previous examples, I used a sample size of two students. What happens when I increase the sample size to, say, n = 5? If this random sample consists of the first five stu- dents in my original list, my new sample mean becomes

x = a

i = 1 xi

n =

24 + 33 + 28 + 41 + 38 5

= 32.8

This results in a sampling error given as follows:

Sampling Error = x - m = 32.8 - 32.0 = 0.8

Obviously, this latest sample provides a much better estimate of the actual average age of my statistics class. In general, we expect larger sample sizes to provide smaller corre- sponding sampling errors. It’s important to note, though, that a larger sample size will not guarantee you a smaller sampling error for every sample selected. For example, if my ran- dom sample of five happens to include the five oldest students in the class, the sample aver- age and corresponding sampling error would be

x = a

i = 1 xi

n =

41 + 38 + 37 + 33 + 33 5

= 36.4

Sampling Error = x - m = 36.4 - 32.0 = 4.4

Sampling errors can differ from one sample to the next. Sampling errors can also be positive or negative.

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300 C h a p t e r 7 | Sampling and Sampling Distributions

The error from this sample is larger than both errors from my examples with a sample size of two students. Larger samples simply increase the likelihood that the sampling error will be smaller, but with statistical sampling, there are few guarantees.

Now, in most statistical settings, the population parameter is not known because it is of- ten not feasible to observe or measure the entire population. Therefore, we will rarely know what the actual sampling error will be for a particular sample that we select. However, in upcoming chapters, I will show you how to determine what the probability is of sampling er- rors occurring. Stay tuned!

Before we wrap up this section, I feel compelled to mention that sampling errors are not the only pitfall you can encounter with statistics. Nonsampling errors occur for a variety of reasons. A common source is survey questions that are ambiguous in the eyes of the respon- dents, which results in their not answering the questions in a reliable manner. Other survey questions can be written in a way that leads the respondent to certain “correct” answers, which will influence the conclusions.

Data collection errors also fall into the category of nonsampling errors. A data collection error occurs when a person responsible for recording the information being collected as part of a survey makes an input error. Or a measurement device could be faulty, such as a scale that weighs a box of cereal. Unlike sampling errors, nonsampling errors cannot be reduced by simply increasing the size of the sample. Reducing nonsampling errors requires a thor- ough analysis of the survey’s design and how the data are collected.

The following Your Turn section will test your sampling error skills using the stock market.

Nonsampling errors occur as a result of issues such as ambiguous survey questions, questions that lead respondents to a certain “correct” answer, and data collection errors.

Increasing the sample size tends to decrease the sampling error. However, there is no guarantee that a particular sampling error will be smaller with a larger sample size.

Suppose you have an investment portfolio that consists of 16 stocks that have closed today with the fol- lowing stock prices.

$21 $24 $36 $23 $26 $31 $26 $28

$17 $30 $34 $27 $21 $36 $48 $33

Calculate the sampling error if the sample includes the following stocks:

a. the first two stocks ($21, $24) b. the first four stocks ($21, $24, $36, $23) c. the first six stocks ($21, $24, $36, $23, $26, $31)

What conclusions can be drawn about sample size and sample error for this population?

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7.3 Section Problems applications 7.1 Suppose the average grade point average (GPA) for the

undergraduate student population at Goldey-Beacom College is 3.10. The following list is a random sample of the GPAs of five students from the college. Calculate the sampling error.

2.45 3.76 3.48 2.81 3.34

7.2 Jeff Bollinger of Prudential Realtors has 20 residential listings under contract. The following table shows the number of days each of these 20 houses has been on the market as of today.

20 21 2 17 35 21 40 54 52 24 40 53 46 10 42 11 37 62 21 76

a. Calculate the mean for this population. b. Calculate the sampling error using the first five homes

in the first row as your sample. c. Calculate the sampling error using all 10 homes in the

first row as your sample. d. How does increasing the sample size affect the sam-

pling error? e. Using a sample size of 5, what is the largest sampling

error that can be observed from this population?

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C h a p t e r 7 | 7.4 The Central Limit Theorem 301

7.3 The following table shows the total points scored in the 16 National Football League games played during Week 1 of the 2012 season.

41 62 64 50 47 33 49 72 76 40 52 36 26 50 57 36

a. Calculate the mean for this population. b. Calculate the sampling error using the first four games

in the first row as your sample. c. Calculate the sampling error using all eight games in

the first row as your sample. d. How does increasing the sample size affect the sam-

pling error? e. Using a sample size of 4, what is the largest sampling

error that can be observed from this population?

7.4 Avis rental car in Whittier, Alaska, currently has 15 rental cars on its lot, each with the following mileage:

19,871 11,483 18,519 10,977 8,325 6,514 14,394 6,225 4,702 11,145 4,768 16,329 15,094 13,317 19,537

a. Calculate the mean for this population. b. Calculate the sampling error using the five cars in the

first row as your sample. c. Calculate the sampling error using the 10 cars in the

first two rows as your sample. d. How does increasing the sample size affect the sam-

pling error?

7.5 The Excel file titled states.xlsx contains the median income in 2011 for all 50 states and the District of Columbia.

a. Use Excel to draw a simple random sample consisting of four states, and then calculate the sampling error for the sample.

b. Use Excel to draw a simple random sample consisting of eight states, and then calculate the sampling error for the sample.

c. Use Excel to draw a simple random sample consisting of 12 states, and then calculate the sampling error for the sample.

d. Compare the sampling error for parts a, b, and c and explain the reason for the differences.

7.6 The Excel file titled verizon.xlsx contains daily revenues for the past 350 business days earned from cell phone ac- cessories (Bluetooth headsets, memory cards, and so on) sold at a local Verizon retail store in Delaware.

a. Use Excel to draw a systematic sample consisting of 10 days, and then calculate the sampling error for the sample.

b. Use Excel to draw a systematic sample consisting of 25 days, and then calculate the sampling error for the sample.

c. Use Excel to draw a systematic sample consisting of 50 days, and then calculate the sampling error for the sample.

d. Compare the sampling error for parts a, b, and c and explain the reason for the differences.

e. What problems might be encountered with the sample obtained in part c?

7.4 The Central Limit Theorem When my students encounter the word theorem, I often see that glassy-eyed look that tells me their minds are wandering off to some faraway place. But believe me when I tell you that the Central Limit Theorem (CLT) is the “mother of all theorems” and is worthy of your at- tention. The Central Limit Theorem states that the sample means of large-sized samples will be normally distributed regardless of the shape of their population distributions. Without this gem, the field of statistics would probably have never made the cut as a required course for many college students. So you might as well learn about the real reason you are reading this book today.

The best way to describe the CLT is with the following example. My wife, Deb, and I are going out to dinner at a moderately priced restaurant (I am a teacher, after all) that offers the four types of entrées shown in Table 7.2 on the next page. If an entrée from this list is randomly selected, there is a 25% probability that it will cost $12. The same prob- ability holds true for the price of the three other entrées. The graph of this probability dis- tribution is shown in Figure 7.6, also on the next page.

According to the Central Limit Theorem, sample means from samples of sufficient size, drawn from any population, will be normally distributed.

I can’t overemphasize the importance of the CLT. We will rely on this behavior of sample means over and over again throughout the rest of this book.

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302 C h a p t e r 7 | Sampling and Sampling Distributions

Because Table 7.2 includes all the entrée types that are available to choose from, these data would be considered a population. The mean of this population is found using Equation 5.1:

m = a N

i = 1 xiP1xi2

m = 1$12210.252 + 1$14210.252 + 1$16210.252 + 1$18210.252 m = $3.00 + $3.50 + $4.00 + $4.50 m = $15.00

In other words, the mean price for this population is $15. Next, I want to calculate the standard deviation of this population using Table 7.3 and Equation 5.4.

{ table 7.2 | Entrees, Their Prices, and the Probability of Their Being Randomly Selected

entrée type price ($) probability of being randomly selected

Vegetarian (V) 12 0.25

Chicken (C) 14 0.25

Fish (F) 16 0.25

Steak (S) 18 0.25

{ figure 7.6 Probability Distribution for

Four Randomly Selected Entrée Choices

0.05

$12 $14 $16 $18

0.15

0.25

Entrée Prices

Pr ob

ab ili

ty Probability Distribution for Entrée Choices 0.3

0.2

0.1

{ table 7.3 | Calculating the Standard Deviation for the Population of Entrée Prices

xi xi 2 P1xi2 xi2 P1xi2

12 144 0.25 36

14 196 0.25 49

16 256 0.25 64

18 324 0.25 81

Total = 230

a N

i = 1 x 2i P1xi2 = 230

s2 = ¢ aN i = 1

x 2i P1xi2≤ - m2 s2 = 230 - 11522 = 230 - 225 = 5 s = 2s2 = 15 = $2.24

This calculates the variance of a discrete probability distribution.

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C h a p t e r 7 | 7.4 The Central Limit Theorem 303

To summarize so far, we have described a population with the following parameters:

m = $15.00 s = $2.24

Now I’ll introduce you to the wonders of the Central Limit Theorem. Suppose Deb and I each randomly select our entrées. (In reality, I’d never do this. I’d be so disappointed if I were stuck with the vegetarian entrée. Fish dishes make me nervous, too. Combine the two scenarios, and I would have a 50% chance of walking out hungry.) What are all the possible meal combinations Deb and I could end up with if we both randomly selected our meals? Table 7.4 shows the entrée prices for all possible 16 combinations. The top-left entry of 12, 12 represents the combination of both Deb and me (Bob) selecting a vegetarian entrée (V), each costing $12. The entry directly to the right of this one (12, 14) shows the cost of me choosing the vegetarian entrée (V), which costs $12, and Deb choosing a chicken entrée (C), which costs $14—and so on through the rest of the table. My entrées are highlighted in yel- low and Deb’s are highlighted in blue.

{ table 7.4 | Possible Entrée Combinations (in $)

deb (v) deb (c) deb (f) deb (s)

Bob (V) 12, 12 12, 14 12, 16 12, 18

Bob (C) 14, 12 14, 14 14, 16 14, 18

Bob (F) 16, 12 16, 14 16, 16 16, 18

Bob (S) 18, 12 18, 14 18, 16 18, 18

{ table 7.5 | The Population of Average Prices for Two Entrées Randomly Selected by Deb and Bob (in $)

deb (v) deb (c) deb (f) deb (s)

Bob (V) 12 13 14 15

Bob (C) 13 14 15 16

Bob (F) 14 15 16 17

Bob (S) 15 16 17 18

What we have essentially done here is show all the possibilities of selecting a simple random sample of size 2 1n = 22 with replacement from this population. The term with re- placement refers to the fact that Deb and I can both select the same entrée. Sampling without replacement means that if Deb selected vegetarian, then I would have to choose one of the other three entrees.

Next, I calculate the sample average for each pair of entrées selected in Table 7.4. All 16 averages are shown in Table 7.5. For example, if I choose chicken and Deb chooses vegetar- ian (the second entry in the first column in Table 7.4), then the sample average is

x = $14 + $12

2 = $13

This value is shown as the second entry in the first column of Table 7.5.

Finally, I construct frequency and relative frequency distributions of the sample aver- ages from Table 7.5. They are shown in Table 7.6 on the next page.

Each value in this table represents a sample average for the pair of values shown in Table 7.4. This value is the average of 14 and 12.

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304 C h a p t e r 7 | Sampling and Sampling Distributions

The distribution shown in Table 7.6 has a very special name—the sampling distribution of the mean. This table describes the pattern that sample averages tend to follow when ran- domly drawn from a population. Figure 7.7 shows this distribution graphically.

{ table 7.6 | Frequency and Relative Frequency Distributions for Table 7.5: Average Entrée Prices

sample average xi frequency relative frequency P1xi2 $12 1 0.0625

$13 2 0.125

$14 3 0.1875

$15 4 0.25

$16 3 0.1875

$17 2 0.125

$18 1 0.0625

Total 16 1.00

2/16 = 0.125

The sampling distribution of the mean describes the pattern that sample averages tend to follow when randomly drawn from a population.

{ table 7.7 | Sample Means for Entrée Prices

deb (v) deb (c) deb (f) deb (s)

Bob (V) 12 13 14 15

Bob (C) 13 14 15 16

Bob (F) 14 15 16 17

Bob (S) 15 16 17 18

{ figure 7.7 The Sampling Distribution of

the Mean: Average Entrée Prices

0.05

$12 $13 $14 $15 $16 $17 $18

0.15

0.25

Average Entrée Price

Sampling Distribution of the Mean 0.3

0.2

0.1Pr ob

ab ili

Now compare the shape of the sampling distribution of the mean with the population probability distribution in Figure 7.6. Even though the population distribution was uniform (all four entrées had a 25% chance of being randomly selected), the sample means distribu- tion tends to be bell-shaped and symmetrical around the population’s mean ($15), which re- sembles the normal probability distribution from Chapter 6. This is because the sample mean of $15 has four different ways to occur, as Table 7.7, which replicates Table 7.6, shows.

By contrast, the sample means of $12 and $18 each appear only once in our sampling distribution.

In general, the CLT tells us the following: For any population distribution, with a suffi- ciently large sample, the sample means will be normally distributed. In most cases, sample sizes of 30 or larger will result in sample means being normally distributed, regardless of the shape of the population distribution. If the population happens to follow the normal probability distribu- tion, the sample means will also be normally distributed, regardless of the size of the samples.

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C h a p t e r 7 | 7.4 The Central Limit Theorem 305

We can verify the mean of the population by calculating the mean of the sample means, or mx, using Table 7.6:

mx = a N

i = 1 xiP1xi2

mx = 1$12210.06252 + 1$13210.1252 + 1$14210.18752 + 1$15210.252 + 1$16210.18752 + 1$17210.1252 + 1$18210.06252 mx = $0.75 + $1.625 + $2.625 + $3.75 + $3.00 + $2.125 + $1.125 mx = $15.00

Notice that I distinguish between the two different means.

m = The population mean mx = The mean of the sample means

It is not a coincidence that m = mx = $15.00. The average of many sample means drawn from a particular population should be very close or equal to the population mean, as we have seen with my entrée example.

Bear with me for one last important point. Let’s calculate the standard deviation for the sample means, or sx, using Table 7.6. I will once again rely on Equation 5.4, substituting x for x and mx for m. Table 7.8 shows the necessary calculations.

Because of sampling error, the mean of the sample means 1mx2 may not always be equal to the population mean 1m2. They are equal in this example because we included every possible sample mean from the population using n = 2.

{ table 7.8 | Standard Deviation Calculations for the Sample Means: Average Entrée Prices

xi xi 2 P1xi2 xi2 P1xi2

12 144 0.0625 9

13 169 0.125 21.125

14 196 0.1875 36.75

15 225 0.25 56.25

16 256 0.1875 48

17 289 0.125 36.125

18 324 0.0625 20.25

Total = 227.5

The higher the standard error of the mean, the more variation you will notice from one sample mean to the next as they are drawn from the population.

a N

i = 1 x 2i P1xi2 = 227.5

sx 2 = ¢ aN

i = 1 x 2i P1xi2≤ - mx2

sx 2 = 227.5 - 11522 = 227.5 - 225 = 2.5

sx = 2sx2 = 12.5 = $1.58 In other words, the standard deviation for the sample means is $1.58. Notice that I dis-

tinguish between the two different standard deviations. s = The population standard deviation, which measures the variation around the popu-

lation mean. sx = The standard deviation of the sample means, which measures the average varia-

tion around the mean of the sample means for all samples of size n. This is also known as the standard error of the mean.

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306 C h a p t e r 7 | Sampling and Sampling Distributions

Let’s stop for a minute to summarize what we have found so far.

population sample means

m = $15.00 mx = $15.00 s = $2.24 sx = $1.58

If we know the population standard deviation, s, we can calculate the standard error of the mean, sx, using Equation 7.3.

Formula 7.3 for the Standard Error of the Mean

sx = s1n

where

s = The population standard deviation n = The sample size This is a very important equation that you will see much of in the upcoming chapters.

Let’s put it to work in our entrée example.

sx = s1n = $2.2412 = $1.58

In our example, n = 2 because we randomly selected two entrees way back in Table 7.4 to construct our sampling distribution. Equation 7.3 gives us the same result that we obtained using Table 7.8 and Equation 5.4, but with a lot less work!

Let’s take this time to summarize what we have learned so far:

• The sample means for a population that follows the normal distribution will also be normally distributed, regardless of the size of the samples.

• If the population does not follow the normal probability distribution, the CLT tells us that the sample means will be normally distributed with a sufficiently large sample size. In most cases, sample sizes of 30 or larger will result in sample means being normally distributed, regardless of the shape of the population distribution.

m = mx

sx = s1n

These properties unlock many statistical procedures that have made significant contribu- tions to decision making. The following section will demonstrate how we can put the CLT to work in the business world. But first, I encourage you to check your understanding of the theorem by trying to solve the next Your Turn problem.

Y o u r t u r n # 3N

The standard error of the mean is the sample mean standard deviation, which measures the average variation around the mean of the sample means.

Debbie and Katie are planning a visit to the Elegant Day Spa where they will each randomly choose one of the following three services that are offered.

Service Cost

Facial (F) $40

Pedicure (P) $30

Manicure (M) $20

Define the population as the three services and a sample of size two as the service that Debbie chooses and the service that Katie chooses.

a. Identify the frequency distribution for the population. b. Identify the frequency distribution for all combinations of sample means for Debbie and Katie

choosing a service.

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C h a p t e r 7 | 7.4 The Central Limit Theorem 307

putting the Central Limit theorem to Work The purpose of this section is to convince you that the CLT is an extremely useful tool wor- thy of all the praise that I have bestowed on it. To illustrate, I’ll use data from a 2012 Pew Research Center study, which discovered that U.S. teenagers send and receive an average of 60 text messages a day. I know little or nothing about the probability distribution for this population and, therefore, cannot answer the following question: What is the probability that a randomly selected teenager will send and receive more than 69 text messages today?

However, using the CLT, I can answer the following question: What is the probability that a randomly selected sample of 30 teenagers will average more than 69 text messages today (sent and received).

With this sufficiently large sample 1n Ú 302, I can answer the second question because I know that my sample averages will follow a normal probability distribution with the fol- lowing mean:

mx = m = 60

My next step is to calculate the standard error of the mean, sx, using Equation 7.3. However, this equation requires me to know s, the population standard deviation, which is not always available. There are techniques available to estimate s that we will investigate in later chapters. In the meantime, for our text messaging example, we’ll assume that we have evidence to conclude that the population standard deviation equals 36 text messages per day. We can now calculate the standard error of the mean using Equation 7.3.

s = 36 n = 30

sx = s1n = 36130 = 365.48 = 6.57

To calculate our normal probability, I need to rely on the z-score for sample means shown in Equation 7.4.

A common mistake that some of my students make is to use s rather than sx in Equation 7.4. Be careful not to be one of these unfortunate statistics!

Formula 7.4 for the z-Score for the Sample Mean

zx = x - mx sx

where

x = Sample mean mx = Mean of the sample means sx = Standard error of the mean This equation is a variation of the z-score that was discussed in Chapter 6. In this sam-

ple mean version of the z-score:

• x replaces x

• sx replaces s

• mx replaces m

Using Equation 7.4, the z-score for x = 69 is

z 69 = x - mx sx

= 69 - 60

6.57 = 1.37

Answers can be found on u pages 329–330

c. Verify that the population mean and mean of all possible sample means are equal. d. Calculate the standard error of the mean two different ways, first using Equation 5.4 and then with

Equation 7.3.

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308 C h a p t e r 7 | Sampling and Sampling Distributions

In other words, the sample mean of 69 is 1.37 standard deviations to the right of the mean for the sampling distribution, mx, and is shown in Figure 7.8.

{ figure 7.8 The Sampling Distribution

of the Mean: Text Messages Example

0.9147

60 69 1.37

0.0853

µx z scale x scale

Using Table 4 in Appendix A, I find that the area to the left of z = 1.37 equals 0.9147. Therefore,

P1x 7 692 = P1z 7 1.372 = 1.0 - 0.9147 = 0.0853 The shaded area in Figure 7.8 shows us that the probability that our sample of 30 teen-

agers will average more than 69 text messages a day is 8.5%. What I find remarkable is that we know this without any information about the population probability distribution. Call me a nerd, but I think this is so cool. Companies like Verizon and AT&T can use this informa- tion to help them design pricing levels for service plans for their younger customers.

Before moving on, please try your hand at the next Your Turn problem to be sure you understand the CLT.

According to the U.S. Labor Department, the average hourly wage for private-sector production and non- supervisory workers was $20.04 in February 2013. Assume the standard deviation for this population is $6.00 per hour. A random sample of 35 workers from this group was selected. What is the probability that the mean for this sample is less than $19.00?

Y o u r t u r n # 4N

Answer can be found on u page 330

using the Central Limit theorem to test Claims But wait—there’s more. The CLT can also be used to verify the validity of claims made by individuals or organizations. For example, let’s say I make the claim that my average drive of a golf ball while teeing off is equal to a whopping 240 yards in length (picture Deb look- ing at me doubtfully). You see, Deb has proofread all of my statistics chapters and knows how to put my claim to the test. She can do so by randomly sampling 45 of my drives to satisfy the CLT requirement (we have no knowledge of the shape of the population distribu- tion). Suppose the average drive from this sample is 233 yards, and the standard deviation for my driving population is 20 yards. Is there enough evidence to support my claim?

On the surface, you might be led to believe that, because 233 yards is less than 240, my claim is not supported. However, the mean of 233 yards is based on a sample. We know from earlier in the chapter that a sampling error occurs when a sample is used to estimate a popu- lation parameter. The sample mean does not have to equal 240 to support my claim; it can be slightly lower and still suffice. However, to answer this question thoroughly, Deb needs to employ the CLT. Her goal is to determine the probability of observing a sample mean of 233 yards or less, given that the sampling distribution mean (which we assume equals the population mean) is truly 240 yards. Sit back and watch her go to work.

Don’t forget to draw the curve!

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C h a p t e r 7 | 7.4 The Central Limit Theorem 309

First, Deb sets up our sampling distribution by assuming the mean does equal 240 yards. Even though we don’t know if this claim is true at this point, we assume it is in order to test my claim.

mx = m = 240 yards

Next, Deb calculates the standard error of the mean:

s = 20 yards n = 45

sx = s1n = 20145 = 206.71 = 2.98

Now, Deb calculates the z-score for x = 233:

z233 = x - mx sx

= 233 - 240

2.98 = - 2.35

Finally, she calculates the probability that the sample mean will be less than or equal to 233 yards if the actual sampling distribution mean equals 240 yards. I’m calculating the probability that my sample mean is less than 233 yards because 233 yards is less than my claim of 240 yards. Had my sample mean been greater than 240 yards, I would have calcu- lated the probability that my sample mean is greater than 240 yards. Using Table 3 in Appendix A, we have

P1x … 2332 = P1z … - 2.352 = 0.0094 This probability is shown graphically in the shaded region in Figure 7.9.

{ figure 7.9 The Sampling Distribution of

the Mean: Golf Example

240233 –2.35 0

µx

0.0094

x scale z scale

This is bad news for me. If the true population mean for my drives (my average drive) is indeed 240 yards as I claimed, there is less than a 1% chance that Deb would have been able to collect a sample with a mean of 233 or less. It’s more likely that my true population mean is somewhat less than 240 yards. It looks like I hit the ball farther in my mind’s eye than I actually do on the golf course.

So far, we have discovered that my sample mean of 233 yards did not support my claim that my average drive was 240 yards. What sample mean would support my claim? Obvi- ously, it would need to be closer to 240 yards than 233, so sample means slightly above 240 yards 1xU2 and below 240 yards 1xL2 would suffice. The shaded region in Figure 7.10 on the next page shows how we find these two values. Most statisticians would agree that these values can be identified in the symmetrical interval that includes 95% of the sample means if the true population mean was 240 yards.

A sample mean between the values of xL and xu would support my claim!

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310 C h a p t e r 7 | Sampling and Sampling Distributions

To find xU and xL, we need to identify the z-scores that define the shaded region in Figure 7.10. If our 95% region is in the center of the sampling distribution, we have 2.5% of the area to the right of the shaded region and 2.5% of the area to the left.

This means that the area to the left of xU is 0.95 + 0.025 = 0.975, which is shown in Figure 7.10. We then search Table 4 in Appendix A to find the value in the body of the table closest to 0.975, which happens to be z = 1.96. We want to find the sample mean that is 1.96 standard deviations to the right of 240. Because the sampling distribution is sym- metrical and the 95% region is in the middle of the curve, the z-score that corresponds to xL is - 1.96. In other words, we are also looking for the sample mean that is 1.96 standard deviations to the left of 240 yards. We do this by rearranging Equation 7.4 in terms of x, as follows:

z = x - mx sx

xU = mx + zsx xU = 240 + 11.96212.982 = 245.8 yards xL = mx - zsx xL = 240 + 1 - 1.96212.982 = 234.2 yards

To satisfy my claim that I average 240 yards off of the tee, the sample average of 45 drives would have to be between 234.2 and 245.8 yards. If the sample average is below 234.2 yards, I would have evidence that my population mean is less than 240 yards. A sam- ple mean above 245.8 yards would indicate my average drive exceeds 240 yards.

In the next section, we will explore the effect sample size has on the sampling distribu- tion. But before we get there, be sure to test the claim made by Kellogg’s in the following Your Turn problem.

{ figure 7.10 Ninety-Five Percent of the

Sample Means: Golf Example

240

0.025

xU = ?xL = ?

95%

0.9750 µx

z = –1.96 x scale z scalez = 1.96

The CLT can be used in a quality control setting, such as a process that fills boxes with cereal. Suppose Kellogg’s claims that the average weight for a box of Frosted Flakes is 18 ounces. To make sure that the filling process is working properly, a sample of 30 boxes is selected and found to have a mean of 18.2 ounces. Historical records show that the standard deviation of the filling process is 0.32 ounces.

a. Is the Frosted Flakes filling process working as Kellogg’s claims? b. Identify the symmetrical interval that includes 95% of the sample means if the true population

mean is 18 ounces.

Y o u r t u r n # 5N

Answers can be found on u page 330

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C h a p t e r 7 | 7.4 The Central Limit Theorem 311

the effect of the Sample Size on the Sampling Distribution Earlier in this chapter, I mentioned that sample size plays an important role in the CLT. As we increase the size of our samples, the standard error of the mean becomes smaller, which in turn reduces the sampling error. Let’s return to my debunked claim of averaging 240 yards with my drives on the golf course. (Thanks a lot, Deb.) Using s = 20 yards, n = 45 (the sample size), and Equation 7.3, we found the standard error of the mean to be sx = 2.98 yards. Figure 7.11 shows the impact of increasing the sample size from 45 to 100.

With a larger sample size the standard error is reduced, which results in the sample means having more of a tendency to crowd around the mean of the distribution.

{ figure 7.11 The Effect Changing the Sample

Size Has on the Sampling Distribution: Golf Example

= = = 2.0020 45

σ n

σ = 20 yards n = 45

240 0 µx

σ = 20 yards n = 100

σx = = = 2.98 20100 σ n

σx

x scale z scale

Increasing the sample size to 100 changes the shape of the sampling distribution. The standard error of the mean drops from 2.98 to 2.00, causing the distribution to be taller and narrower, as shown in Figure 7.11. With a larger sample size the standard error is reduced, resulting in the sample means having more of a tendency to crowd around the mean of the distribution, which in this case is 240. We can see this by calculating the interval that will contain 95% of the sample means for n = 100.

xU = mx + zsx xU = 240 + 11.96212.002 = 243.9 yards xL = mx - zsx xL = 230 + 1 - 1.96212.002 = 236.1 yards

Table 7.9 compares these results to the interval calculated earlier for n = 45. As you can see, the interval of sample means for n = 100 is narrower than the one for n = 45.

{ table 7.9 | Intervals That Contain 95% of Sample Means When the Sample Size Changes from 45 to 100: Golf Example

n xL xU

45 234.2 245.8

100 236.1 243.9

Increasing the sample size also effectively reduces the sampling error, which we defined as x - m earlier in Equation 7.2. Think of it this way: As you gradually increase the sample size until it eventually becomes the population size, the sample means will approach the population mean. If the entire population is “sampled,” the mean of that sample is the popu- lation mean. This is known as a census. Once the entire population is measured, then x = m, which makes the sampling error equal to zero. I can summarize this as follows:

As n S N Then x S m Eventually x = m Making x - m = 0 1Sampling Error2

A census is the process of recording information from the entire population of interest.

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312 C h a p t e r 7 | Sampling and Sampling Distributions

In most cases, sampling is still the preferred approach over a census (unless you are the U.S. government). When the U.S. government counts the population of the country, it tries to count everyone—even people who are hard to find, such as the homeless. However, for most people and organizations, populations are usually too large to measure or not completely ac- cessible. This concept demonstrates that sample size can be used to control the inherent er- rors that can occur when relying on a sample to describe a larger population.

The shape of the distribution of the population will also affect the shape of the sampling distribution, as will the size of the sample. Figure 7.12 shows the histograms of three popu- lations, along with their corresponding sampling distributions.

{ figure 7.12 Population Distributions and

Their Corresponding Sampling Distributions (for Three Different-Sized Samples)

Population 1 Population 2 Population 3

Sample Means n = 2

Sample Means n = 5

Sample Means n = 30

Right-Skewed Normal U-Shaped

In all three cases, the sampling distribution for n = 30 can be described as having the shape of the normal distribution, which is taller and narrower than the distributions with smaller sample sizes. You can also see that the normally distributed population in the middle column has a normal sampling distribution for all sample sizes. The same can’t be said for Populations 1 and 3, which obviously are not normal. Figure 7.12 visually shows that when samples are large enough, their means follow a normal distribution regardless of the shape of a population’s distribution. This is what the CLT predicts.

I hope I have convinced you of the usefulness of the CLT. If I were your instructor, you’d be seeing all sorts of exam questions on this topic, so give the next Your Turn problem a try.

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C h a p t e r 7 | 7.4 The Central Limit Theorem 313

the Sampling Distribution of the Mean with a Finite population Up to this point in the chapter, we have assumed that all of the samples have been selected from a very large, or infinite, population. In these cases, the proportion of the sample size to the population size, n>N, is very small. By contrast, when the population is smaller, or finite, this proportion can be much larger. When sampling from a smaller population when the pro- portion n>N is greater than 5%, I recommend making an adjustment to the standard error of the mean calculation using Equation 7.5. So, for example, if our population consisted of 100 people and we were sampling more than 5 of them, we would adjust the standard error of the mean using Equation 7.5.

The fuel consumption for the 2013 Toyota Prius was rated by the U.S. Environmental Protection Agency (EPA) to average 48 miles per gallon (MPG) for combined city and highway driving. Assume the popula- tion standard deviation is 7 MPG. Determine the interval that will contain 95% of the sample means for a sample size of

a. 40 cars b. 80 cars

Y o u r t u r n # 6N

Answers can be found on u page 330

Finite population correction factor A factor used to adjust the standard error of the mean or proportion when the proportion of sample size to population size, n >N, is greater than 5%.

When the proportion of sample size to population size, n/N, is greater than 5%, the finite population correction factor is used to adjust the standard error of the mean.

Formula 7.5 for the Standard Error of the Mean (Finite Population)

sx = s1n A N - nN - 1

where

s = Population standard deviation n = Sample size N = Population size

The term A N - nN - 1 in Equation 7.5 is known as the finite population correction factor. To illustrate sampling from a finite population, consider Sum-R-Fun, a Delaware pool ser- vice company that has 100 residential customers. Historically, Sum-R-Fun’s customers have given the company an average rating for customer service of 7.2 on a scale of 1–10 with a standard deviation of 0.7. However, a recent sample of 40 customers led to an average rating of 7.5 for the company. Is there any evidence that the average customer service rating for the entire customer base is different from 7.2? In other words, does the 7.5 rating indicate that if all customers were surveyed, the firm’s average customer rating might be higher than 7.2? That would suggest that the company’s customers were becoming more satisfied. Or, per- haps the higher 7.5 rating is just a result of sampling error. Let’s find out which it is.

Because 40 out of 100 customers were sampled, the proportion n>N = 40>100 = 40%. This value is greater than 5%, so this would be considered sampling from a finite population, requiring the use of Equation 7.5.

sx = s1n A N - nN - 1 = 0.7140 A 100 - 40100 - 1

sx = 0.7

6.32 A 6099 = 10.111210.606 sx = 10.111210.7782 = 0.086

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314 C h a p t e r 7 | Sampling and Sampling Distributions

We now calculate the z-score for x = 7.5:

z7.5 = x - mx sx

= 7.5 - 7.2

0.086 = 3.49

Using Table 4 in Appendix A, we have

P1z … 3.492 = 0.9998 However, we are looking for

P1z 7 3.492 = 1 - 0.9998 = 0.0002 There is only a very small probability (0.02%) that a sample mean of 7.5 or greater

could have occurred if the true population mean were 7.2. Therefore, it appears the popula- tion mean is greater than 7.2. Apparently, the customer service at Sum-R-Fun has improved recently.

I’m sure you’re asking why we need a different standard error formula for finite popula- tions. For an explanation, look at Table 7.10, which shows the standard errors for Sum-R- Fun when samples sizes of 40, 60, 80, and 100 are used. Note that a sample size of 100 is considered a census because the population consists of 100 customers.

The first column in Table 7.10 is the sample size. The second column shows the stan- dard error without the correction factor, which assumes our population is very large (infinite) when compared to the size of the sample. The third column shows the finite population cor- rection factor for each sample size amounts. The last column shows the standard error for each sample size with the correction factor included.

{ table 7.10 | Sampling Errors for Sum-R-Fun’s Sample Sizes When the Finite Population Equals 1001N = 1002

sample size

standard error

finite correction factor

standard error with finite correction factor

n Sx = S1n A N - nN - 1 Sx = S1n A N - nN - 1

40 0.111 0.778 0.086

60 0.090 0.636 0.057

80 0.078 0.449 0.035

100 0.07 0 0

Recall our sampling-error discussion in the previous section. We learned that as the sample size approaches the population size, the sampling error approaches zero. When the sample mean equals the population mean there is no variation between the two. In other words, sx should be equal to zero. However, look at the second column in Table 7.10: As n goes to 100, the standard error only drops to 0.07. This is because the standard error equa- tion in the second column assumes the population is infinite. The finite population correction factor comes to our rescue. When n = N, the correction factor equals zero, resulting in a corrected standard error also equaling zero.

In general, the correction factor reduces the standard error to reflect the fact that the sampling error is lower for a finite population. If you use the infinite population standard er- ror in the second column when the population is finite, you’ll be exaggerating your sampling error and adversely affecting your probability calculations.

What happens to the finite correction factor when the population size becomes much larger when compared to the sample size? Let’s plug in a population size of 10,000 with a sample size of 30 to find out:

N = 10,000 n = 30A N - nN - 1 = A 10,000 - 3010,000 - 1 = A 9,9709,999 = 10.997 = 0.998 ≈ 1.0 We use the correction factor to reduce the standard error because the sampling error is lower for a finite population.