team assignment
Problem 1
| The box plots below compare the base yearly salaries of employees in marketing and research for a large company. | |||
| Identify the five number summaries for each profession. Interpret each in everyday language. | |||
| A. Five Number Summary | Lind et al. (2021) LO4-3 Construct and analyze a box plot | ||
| Example Interpretation | |||
| Marketing | Research | Approximate the values | Their body weight was observed as more than 99 and under 234 pounds. One-half of the subjects' weight was 149 pounds or greater. Twenty-five percent of the weight was under 125 pounds and over 169 pounds. There were no observed outliers. |
| Min | |||
| Max | |||
| First Quartile | |||
| Second Quartile | |||
| Third Quartile | |||
| B. Interpretations | Example | ||
| Min | 99 | ||
| Interpretation Marketing | Max | 234 | |
| Median | 149 | ||
| 1st Quartile | 125 | ||
| 3rd Quartile | 169 | ||
| Interpretation Research | |||
Problem 2
| An insurance company determines that in every 100 claims, 3 are fraudulent. | |
| What is the probability that the next claim the company processes will be | |
| fraudulent? | |
| Please state the probability as a decimal value. | Lind et al. (2021) LO5-2 Classical Probability |
| (show your calculation) |
Problem 3
| Probability Distribution | Lind et al. (2021) LO6-1 What Is a Probability Distribution? | |
| A. A review of emergency room records at rural William Tell Hospital was performed to determine the probability | ||
| distribution of the number of patients entering the emergency room during a 1-hour period. The following table | ||
| lists this probability distribution. | ||
| Patients per hour | Probability | |
| 0.00 | 0.2440 | |
| 1.00 | 0.3543 | |
| 2.00 | 0.2001 | |
| 3.00 | 0.0998 | |
| 4.00 | 0.0101 | |
| 5.00 | 0.0084 | |
| 6.00 | 0.0833 | |
| Determine the probability that the number of patients entering the emergency room | ||
| during a randomly selected 1-hour period is | ||
| a) 2 or more | Show which values were used. | |
| b) exactly 5 | Show which values were used. | |
| c) fewer than 3 | Show which values were used. | |
| d) at most 1 | Show which values were used. | |
| B. Determine whether the distribution given below are valid probability distributions or not. | ||
| x | P(x) | |
| 0 | 0.300 | |
| 1 | 0.240 | (Choose one) |
| 2 | 0.250 | |
| 3 | 0.160 | |
| 4 | 0.050 | |
| C. Find the missing probability in the following probability distribution. | ||
| x | P(x) | |
| 0 | 0.070 | |
| 1 | 0.180 | |
| 2 | 0.220 | |
| 3 | ||
| 4 | 0.130 |
Problem 4
| Binomial Distribution | Lind et al. (2021) LO6-4 Binomial Probability Distribution |
| A recent survey by the American Accounting Association revealed 52% of accountants have the Certified Public Accounting (CPA) certificate. | |
| Suppose we select a sample of 15 accountants. | |
| a. What is the probability that five of the 15 accountants are CPAs? | |
| b. What is the probability that seven or eight of the 15 accountants are CPAs? |
Problem 5
| Determine whether the random variable x is discrete or continuous | Lind et al. (2021) LO6-2 Random Variables |
| a. Let x represent your body height (inches) | (Choose one) |
| b. Let x represent the number of students in class | (Choose one) |
Problem 6
| The weekly incomes of shift foremen in the glass industry follow the normal probability distribution with a population mean of $1,500 and a population standard deviation of $100. | ||
| Lind et al. (2021) LO7-3 The Standard Normal Probability Distribution | ||
| a. What is the z value for the income of a foreman who earns x = $1,600 per week? | z = | Show your work. |
| b. What is the z value for the income for a foreman who earns x = $1300 per week? | z = | |
| What is the probability associated with a., and b. above? | ||
| Look-up Lind et al. (2021) Appendix B.3 Areas under the Normal Curve | Or use the Lane Probability Calculator | Lind et al. (2021) LO7-3 Finding Areas under the Normal Curve |
| a. | p = | |
| b. | p = | |
| What is the likelihood of selecting a foreman whose weekly income is between $1300 and $1600? | p = | |
| References | ||
| Lane, D. (n.d.). Normal distribution . http://onlinestatbook.com/2/calculators/normal_dist.html | Probability Calculator | |
| Lind, D., Marchal, W., & Wathen, S.A. (2021). Statistical techniques in business and economics (18th ed.). McGraw-Hill. | Course Textbook |
Lane Binomial Calculator
| Lane Binomial Calculator | http://onlinestatbook.com/2/calculators/binomial_dist.html | ||||
| Recent audit of computer chips found 15% produced in a particular factory are defective. If 25 chips are randomly selected, | |||||
| a. What is the probability that none of the chips are defective? | |||||
| b. What is the probability of less than 2 chips are defective? | |||||
| Excel Function Method | Lane Online Calculator | ||||
| =BINOM.DIST(Number,Trials.Probability,Cumulative) | Enter Data | ||||
| N | 25 | 25 | |||
| a. | =BINOM.DIST(0,25,0.15,FALSE) | 0.0172 | π | 0.15 | 0.15 |
| x | 0, 0 | Less than 2 | |||
| Explanation of a. | Calculated | ||||
| FALSE is used for a single value, 0 | p = | 0.0172 | 0.0931 | ||
| b. | =BINOM.DIST(1.99999,25,0.15,TRUE) | 0.0931 | |||
| Explanation of b. | |||||
| Number 1.99999 is used to indicate less than 2 | |||||
| TRUE is used for multiple values, 0 to 1.99999 | |||||
Lane Probability Calculator
| Lane Probability Calculator | http://onlinestatbook.com/2/calculators/normal_dist.html | |||||
| Using Lind et al. (20210 Appendix B.3 is the manual means for calculating probability. It is difficult because | ||||||
| until the learning curve is achieved. Lane, who is a professor at Rice University, Houston, TX., | ||||||
| created this web calculator, which provides a visual. Both methods will be illustrated. | ||||||
| Manual Method | Lane Online Calculator | |||||
| µ = 100 | σ = 10 | x = 75 | x = 110 | |||
| Enter Data | ||||||
| z = (x - µ) / σ | z = (75-100) / 10 | z = (110-100) / 10 | Mean | 100 | 100 | 100 |
| z = | -2.50 | 1.00 | SD | 10 | 10 | 10 |
| x | 75, 100 | Left of mean | 100, 110 | Right of mean | 75, 110 | Between |
| a. What is the probability : | Calculated | |||||
| Appendix B.3 | p | 0.4938 | 0.3413 | 0.8351 | ||
| z = -2.50 | z | p | ||||
| z = 1.0 | ||||||
| b. What is the probability between z = -2.5 and 1.0 = | 0.4938 | |||||
| 0.3413 | ||||||
| 0.8351 | ||||||
References
| References |
| Lane, D. (n.d.). Binomial distribution. http://onlinestatbook.com/2/calculators/binomial_dist.html |
| Lane, D. (n.d.). Normal distribution. http://onlinestatbook.com/2/calculators/normal_dist.html |
| Lind, D., Marchal, W., & Wathen, S.A. (2021). Statistical techniques in business and economics (18th ed.). McGraw-Hill. |
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