BUS 308 Assignment
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Learning Objectives
After reading this chapter, you should be able to:
1. Calculate the probability of the occurrence of an event.
2. Describe the characteristics of normal distributions.
3. Compare and contrast normal distributions to the standard normal distribution.
4. Explain the utility of the standard normal distribution.
5. Calculate and explain z scores.
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Probability, Probability Distributions, and z Scores
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CHAPTER 2Section 2.1 A Primer in Probability
Chapter Outline
2.1 A Primer in Probability Calculating Probability Probability and Data Distributions
2.2 The Normal Distribution Skewness Kurtosis The Characteristics of Samples
2.3 The Standard Normal Distribution z Scores Interpreting the z Score Comparing Measures The Area Under the Curve
2.4 Using Excel to Calculate z Scores
2.5 Summary
2.1 A Primer in Probability
Part of the value of statistical analysis is that besides providing tools for examining events that have already occurred, it provides a way to estimate the likelihood that uncertain outcomes will occur. Business leaders have relatively little interest in analyzing things that never happen, or for that matter, things that always happen. From the stand- point of statistical analysis at least, certainties hold no interest for the analyst because they involve no surprises. It is the uncertain outcome, the things that vary, that hold interest. The manager of the city’s water treatment plant probably spends little time wondering whether there will be a demand for potable water. It is a certainty that the demand will be there; people need drinkable water. There may be questions, however, about whether the manager can supply enough water, particularly if there have been problems with con- tamination at the water source.
When probabilities outcomes are quantified, they take on values that range from 0 to 1.0. Something that never occurs has a probability of 0, symbolized p 5 0. An outcome that occurs every time has a probability of 1.0, or p 5 1.0. Note that these values are theoreti- cal outcomes. The fact that someone watches a roulette wheel for 5 consecutive spins and finds that the result is red every time should not conclude that the probability of a black outcome is p 5 0. For such cases, probabilities are based on a very large number of repeti- tions rather than short-term results. For those who find it attractive, the allure of gambling is precisely that characteristic of course. Over the short term, any outcome is possible, even if some are a good deal less probable than others.
Calculating Probability
The probability that people will need drinkable water is p 5 1.0. We know from experi- ence that everyone requires water. While that outcome is a certainty, the probability that
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CHAPTER 2Section 2.1 A Primer in Probability
the local water supply will be adequate is not. If at least some potable water is available (the water treatment plant has not been completely shut down), the probability that the demand can be met for a specified period will range between 0 and 1.0. The manager’s calculation of that value will be affected by factors such as rainfall, the proper functioning of the equipment in the treatment plant, the availability of needed resources such as elec- tricity, and so on. Because these factors are in themselves variable, calculating the precise probability that demand will be met can be an involved process. In simple terms, how- ever, the probability of an event is the number of times the event actually occurs, divided by the total possible number of outcomes:
p 5 number of times the event occurs ÷ the total number of possible outcomes
Formula 2.1 If e 5 the number of times the event occurs and o 5 the number of times it could possibly occur, the formula reduces to p 5 e/o.
If a student takes a test on probability, and the test is made up of multiple-choice items that each have four choices with just one correct choice, the prob- ability of guessing correctly on a single test item is indicated as follows:
p 5 e/o 5 ¼ 5 .25
If the items are true/false, the number of possible outcomes changes to two, and the prob- ability of guessing correctly changes accordingly:
p 5 e/o 5 ½ 5 .50
When there are only two outcomes, the outcomes are said to be dichotomous. The answers on the true/false questions are dichotomous. Questions about whether manag- ers are female or male provide for a dichotomous response. If each outcome on the true/ false item has an equal probability of occurrence (e.g., there is an equal number of true and false responses on the test as a whole), the probability of guessing correctly on any particular true/false test item is p 5 .5.
Whether the water treatment plant will be able to provide enough water can also be viewed in dichotomous terms. If, rather than calculating the percent of demand man- aged, the plant’s performance is predicted in terms of whether it will or will not meet the demand, the outcome is dichotomous.
Other occurrences also have a discrete number of outcomes but are not dichotomous. Undergraduate students belong to classes that are mutually exclusive. Students can be in their freshman, sophomore, junior, or senior years. With four classes and each under- graduate a member of just one, the probability that a student is a member of the junior class can be determined as follows:
Key Terms: Dichotomous outcomes have only two pos- sibilities. A true/false item calls for a dichotomous outcome.
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CHAPTER 2Section 2.1 A Primer in Probability
p 5 e/o 5 1⁄4 5 .25
The condition for determining probability with Formula 2.1, of course, is that membership in each of the four classes must be equally likely. If a student is transferring to the university, and transfer students are much more likely to be sophomores than they are to be freshmen, juniors, or seniors, the prob- ability that the transfer student will be a junior will be higher than p 5 .25.
Consider this type of problem from another vantage point. What is the probability of randomly selecting a student who belongs to the junior class given the following data?
Freshman: 400
Sophomore: 300
Junior: 200
Senior: 100
Of course, p 5 .25 only if each class includes the same number of students. If the number of students varies from class to class, the probability of selecting a student from a particular class will reflect the proportion of the population of all undergraduate students that the particular class represents. Since junior class students represent 200 out of 1,000 under- graduate students, the proportion of juniors 5 200/1,000 5 .20. Thus, rather than the p 5 .25 of selecting a junior if the juniors made up ¼ of all undergraduates, the probability is p 5 .2; junior students actually constitute 1⁄5 of all undergraduates.
Probability and Data Distributions
In the very simple examples above, the probability that certain events will occur was cal- culated directly. The calculations make it possible to predict the probability, for example, that one undergraduate student selected at random will turn out to be a member of the junior class. This was done according to two scenarios, one where the classes have equal memberships and one where they do not. If data were available for the age and gender of consumers of large flat-screen TVs, it would likewise be possible to predict the probability that a purchaser of a 60-inch plasma screen TV would be a woman in her 30s. Such data could have a great value to an advertiser, who could know to which demographic group an advertising campaign should be directed; the calculated probabilities would indicate which consumers are most likely to purchase which kinds of products.
As attractive as it would be to have access to such information, gathering all the data necessary for calculating all the probabilities of occurrence for every event that might interest someone in business would be a huge task. Happily, whether the events are dichotomous or have more than two outcomes, whether they are mutually exclusive or related, in large numbers the probabilities of all the outcome possibilities tend to reflect
Review Question A: A multiple-choice item on an employment test has five options, A through E. If there is only one correct choice, what is the probability that an applicant will guess the correct answer?
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CHAPTER 2Section 2.1 A Primer in Probability
a predictable pattern. When it is represented visu- ally, the pattern indicates one of the most enduring concepts in statistical analysis, the bell-shaped curve that is the normal distribution.
A normal distribution is based on measures that are continuous, which means that the scores can have any value. But distributions can also be created for dichotomous outcomes, and in fact these distributions can have a great deal in common with a normal distribu- tion based on continuous measures. How can a distribution that has only two possible outcomes be anything like normally distributed? Technically, it cannot, but it turns out to be close enough that many of the principles that hold for a distribution of continuous mea- sures hold also for a distribution of dichotomous outcomes. This can be illustrated with the true/false test item that was discussed earlier. Suppose a student responds to a true/false item question twice on two occasions, or responds to two different true/false questions items on the same test. The entire range of possible responses is the following:
true, true
true, false
false, true
false, false
Each time the student responds, the probability is p 5 .5 that the student is correct. But what is the probability that the student is correct with both responses? The probability of being correct on two, independent, dichotomous outcomes is the product of the prob- abilities of being correct on each outcome. Since the probability that the student will have responded correctly on each item is p 5 .5, the probability that the student will be correct on both of the true/false items is then .5 3 .5 5 .25. So for the four possible combinations of outcomes, and with a probability of p 5 .5 that each response is correct, the probability that the student responded correctly with each pair of items can be determined as follows:
Response Probability of Responding Correctly
true, true .5 3 .5 5 .25
true, false .5 3 .5 5 .25
false, true .5 3 .5 5 .25
false, false .5 3 .5 5 .25
For the sake of illustration, assume that the correct answer to both items is “true.” Given the preceding combinations above, note how many correct responses there were for each combination of two responses:
Response Number of Correct Responses
true, true 2
true, false 1
false, true 1
false, false 0
Key Terms: The normal distribution is symmetrical, unimodal, and mesokurtic.
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CHAPTER 2Section 2.1 A Primer in Probability
Next, note how many different ways there are to get the listed number of correct responses. For example, there is only one way to get two correct responses with one pair of responses, which is to answer “true” to both items. There are two ways that one could get one correct response from a pair of responses, however. The student can answer “true” to the first item and “false” to the second, or do the reverse and answer “false” and then “true.”
Number of Correct Responses Number of Ways to Achieve
2 1
1 2
0 1
Finally, recall that the probability of a correct response when there are independent, dichotomous outcomes is the sum of the probabilities of each, and the probability of being correct on multiple outcomes is the product of the probabilities of being correct with each decision. The probabilities associated with achieving each of the possible outcomes are as follows:
Number of Correct Responses Number of Possibilities Probability of Occurrence
2 1 .5 3 .5 5 .25
1 2 .25 1 .25 5 .5
0 1 .5 3 .5 5 .25
With two true/false test items for which the answer to each is true, and independent responses to each item, there are three correct possible combinations of the number. With random guesses, the student could:
1. Answer both items correctly, for which the probability is the product of the probabilities that the student will select the correct response on each item independently, which is .5 3 .5 5 .25.
2. Answer one of the two items correctly, for which the probability is the sum of the probabilities that each item will be answered correctly, or in other words, .25 1 .25 5 .5.
3. Answer both items incorrectly, for which the probability is the product of the probabilities that the student will select the incorrect response on each of the two items independently, or .5 3 .5 5 .25.
If the three possible outcomes above are graphed, with the horizontal (x) axis indicating the number of correct choices on the two items, and the vertical (y) axis indicating the probability of the outcome, Figure 2.1 is the result.
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CHAPTER 2Section 2.2 The Normal Distribution
Figure 2.1: The probability of 0, 1, or 2 correct answers when guessing on 2 dichotomously-scored test items
Number of Correct Choices
P ro
ba bi
lit y
.5
.4
.3
.2
.1
0
0 1 2 3 4
2.2 The Normal Distribution
Because it is based on just four outcomes, the Figure 2.1 distribution looks crude and lacks the smooth character that normal distributions have, but it begins to suggest a normal distribution even with so few data points. Had the process above been completed for a larger number of outcomes, the similarities between this distribution and a normal distribution would be even more evident. The point is that in very large numbers, even when events are dichotomous, as they are with true/false outcomes, graphing the out- comes results in a frequency distribution that takes on the appearance of a normal curve. If a graph had been developed that was based on continuous data, something such as the various weights of all members who frequent a fitness gym, the similarity to a normal distribution would emerge even more quickly.
Normal distributions are defined by the following three properties:
• normal distributions are unimodal, • normal distributions are symmetrical, and • in a normal distribution the standard deviation will be about one-sixth of the
data’s range
Recall from Chapter 1 that the mode is the most frequently occurring value in a data set. When a distribution is unimodal, it means that there is just one value that occurs with the greatest frequency. By contrast, sometimes data distributions are bimodal; they have two values that occur with greatest frequency, or two modes. Perhaps a shift manager in a manufacturing plant is preparing a report on worker safety and finds that there are two weeks during the year when accident rates tend to be high—the two weeks following a vacation period. During the weeks after the plant has been closed for the winter and sum- mer holiday seasons, accidents tend to spike. Those two periods when accidents occur with greater frequency create a bimodal distribution in the graph indicating accidents at the plant. Distributions can also be “tri-modal,” of course, or indeed have any number of modes, but when the data are normal the distribution has just one mode.
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CHAPTER 2Section 2.2 The Normal Distribution
Skewness
If the other measures of central tendency are also cal- culated in a normal distribution, it would be apparent that the median and mean have the same value that the mode has. That characteristic, Mo 5 Mdn 5 M, indicates that the data distribution is symmetrical. It does not mean that the distribution is normal, necessarily, since normality requires more than just symmetry, but if Mo 5 Mdn 5 M, at least the distribution is symmetrical. When the distribution is not symmetrical the data are said to be skewed. Consider an office staff with the following years of experience: 3, 4, 4, 5, 5, 5, 6, 6, and 7.
With the values presented in an ordered group as they are above, the measures of central tendency are easy to determine:
• The most frequently occurring value is 5, so Mo 5 5. • The middle-most (5th) number in the distribution is also 5, Mdn 5 5. • The mean, M 5 x/n 5 45/9 5 5.0.
The fact that this distribution is symmetrical is also suggested by plotting the years of experience in a frequency distribution (see Figure 2.2):
Figure 2.2: A frequency distribution of employees’ years of experience
Years of Experience
Fr eq
ue nc
y
0 1 2 3 4 5 6 7 8 9
3
2
1
0
A skewed distribution occurs when there is an imbalance of extreme scores. When extreme scores, called outliers, occur on one side of the distribution and are not countered by equally extreme scores on the other side the distribution, the result is a skewed distribution. The direction of the skew is indicated by how the mean is oriented to the other two measures of central tendency. If the employees whose years of experience are plotted in Figure 2.2 are joined by someone who has nine years of experience, note the impact, first on the plotted data distribution (Figure 2.3), and then on the three measures of central tendency.
Key Terms: A non-symmetrical distribution is skewed. Skew is created by an imbalance of extreme scores, or outliers, on one side of the distribution.
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CHAPTER 2Section 2.2 The Normal Distribution
Figure 2.3: The effect of an outlier on a distribution of years of experience
Years of Experience
Fr eq
ue nc
y
0 1 2 3 4 5 6 7 8 9
3
2
1
0
If the three measures of central tendency are recalculated with the addition to the group of the person with nine years of experience, note what happens to each statistic. The new data set is:
3, 4, 4, 5, 5, 5, 6, 6, 7, and 9.
Mo 5 5
Mdn 5 5
M 5 5.4
The additional person’s years of experience have a differential impact, depending upon the particular measure of central tendency. The new score doesn’t affect the mode since the most frequently occurring value is still 5, nor does it affect the median, since the mean of the now 2 middle numbers in the distribution remains 5 ((5 1 5) 2 5 5). The value of the mean changes from 5.0 to 5.4, however. Of the three measures of central tendency, the mean is always the one most affected by outliers. For this reason, salaries or home prices typically are described in terms of medians rather than means. Salaries and home prices usually have a few extremely high values. It is the impact of the company president’s salary averaged in with the salaries of the rank and file workers, or the professional basketball player’s home price averaged in with the prices of more modest homes, that skews a data set. As the amount of skew increases, the value of the mean as an indicator of central tendency dimin- ishes, particularly when it is reported alone.
If there are extreme values in both directions, there is no skew. If, for example, someone with 1 year of experience joins the office staff as well as the person with 9 years, note that the mean becomes 5.0 again. It isn’t extreme values in a distribu- tion that create skew, but rather extreme values in only one tail of the distribution.
Review Question B: Can a distribution be symmetrical and mesokurtic, but not normal?
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CHAPTER 2Section 2.2 The Normal Distribution
There are formulas for calculating skew, but they tend to be quite tedious to complete. For our purposes, examining the value of the mean relative to the other members of central tendency will be a sufficient indicator of whether a distribution is skewed and when it is, in which direction. When there are one or two extremely high values in a distribution, the mean becomes higher than the other measures of central tendency, and the distribution indicates positive skew (sk1). When there are a few extremely low values compared to the balance, the mean is lower than the other measures of central tendency, and the result is a distribution with negative skew (sk2).
Kurtosis
The third characteristic that indicates normality has to do with how closely “bunched” the data are. Kurtosis refers to how a population is distributed. When data are such that the value of the standard deviation is about one-sixth of the range (s 5 1/6 R), the result is a mesokurtic distribution. The “meso” prefix means “middle.” The Greek work “kurtos” means bulging, so mesokurtic describes a middle, or intermediate, bulge or curve in the data distribution.
When a distribution includes the odd outlier but the bulk of the individual measures vary little from the mean of the distribution, it is described as leptokurtic. This is what happens, for example, when job performance data from a group of management trainees include a very low score and a very high score, which results in a substantial range, but the great majority of measures reflect very similar values. The standard deviation value in that case would likely be less than one-sixth of the range (s , 1/6 R).
When there is too much variability for a distribution to be normal, it is said to be platykurtic. The evidence for a platykurtic distribution is that the standard deviation is greater than one-sixth of the value of the range (s . 1/6 R). Samples, especially small samples, are often platykurtic. Bimodal distributions, where much of the data is concen- trated in two extremes, also tend to be platykurtic. This might occur, for example, if sales results from a franchise in a relatively rural location are compared to those from an urban store in the same chain.
Earlier it was noted that a distribution does not qualify as normal simply because it is symmetrical. The same can be said of kurtosis. While a normal distribution is mesokurtic, not all mesokurtic distributions are normal. Normality requires all three characteristics— unimodality, symmetry, and a standard deviation that is an appropriate proportion of the entire range.
The Characteristics of Samples
The examples earlier in the chapter all employed rather small samples to illustrate the characteristics that indicate normality. Normality, however, is a characteristic of popula- tions rather than samples. Except for very large samples that include several hundred
Key Terms: Kurtosis refers to how a population is distrib- uted. Normal distributions are mesokurtic. If s . 1/6 R, the distribution is platykurtic. If s , 1/6 R, the distribution is leptokurtic.
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CHAPTER 2Section 2.3 The Standard Normal Distribution
measures, samples generally involve too few data points to be normal. Furthermore, small samples often provide too few data points to yield a meaningful mode. Modes based on two or three observations are not helpful indicators of distribution characteristics; for that reason, when the data set is comparatively small, it is the orientation of the mean relative to the median that indicates whether a distribution is skewed.
2.3 The Standard Normal Distribution
The characteristics of normality are particularly relevant to the business manager. Many of the things we wish to analyze in business are normally distributed when they are in their population form. When data are normal, the general characteristics of the data distri- bution can be known even before any data are gathered. Because all normal distributions share common characteristics, statisticians have described normal distributions in great detail. In any normal distribution, for example:
• Just over two-thirds of the population occurs between the points that are one standard deviation below the mean and one standard deviation above the mean.
• About 95% of the population will occur between plus and minus 2 standard deviations of the mean.
• Only the most extreme outliers will be beyond plus or minus 3 standard deviations from the mean.
The three bullet points above indicate that if the mean and standard deviation for any normal distribution are known, anyone can know a good deal about how the individual data values will be distributed. All that is required is a little basic calculation:
• If a data distribution has mean 5 22 and a standard deviation of 3, for exam- ple, about 2⁄3 of the population will occur between 19 and 25.
• About 95% of the entire population will occur between 16 and 28. • Very little of the entire population will occur below 13 or above 31.
Consider how probability can become part of the normality discussion. Since 2⁄3 of a normal distribution occurs between 61 standard deviation of the mean, the probability that an individual from a normal distribution selected at random will be within 1 standard devia- tion of the mean is about p 5 .67 (it’s actually p 5 .68, but more on that later).
Assume that one of those normal distributions is the ages of those who purchase new automobiles, for example. If the mean age of purchasers is 32.547 with a standard devia- tion of 3.221, about 2⁄3 of those who will buy new cars next year will be between the ages of 29.326 (32.547 3 3.221) and 35.768 (32.547 1 3.221). That information means that those who work on the advertising know before they begin what the primary age demographic is for those who will be buying most of the new cars, and consequently those to whom advertising must be tailored.
As long as the mean and standard deviation are available, it is possible to work out the percentages occurring in particular ranges for any normally distributed population. The problem is that each normally distributed population has its own mean and standard deviation, and values that are accurate for one distribution will not explain another.
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CHAPTER 2Section 2.3 The Standard Normal Distribution
Furthermore, knowing that about 2⁄3 of a population occurs between 11 and 21 standard deviation from the mean is small comfort if we don’t know what the value of the mean and standard deviation are, and these values can be hard to come by for populations. Although the government conducts an exhaustive census study every 10 years and publishes descrip- tive data for a number of population characteris- tics, those data are not likely to include what busi- ness people wish to know, like the mean age of cell phone users or the standard deviation of the amount spent dining out. This is where the standard normal distribution, or the z distribution, comes in.
z Scores
There are many normal distributions, but there is just one standard normal distribu- tion. It is a distribution for which the mean always has a value of 0, and the standard deviation is 1.0 (see Figure 2.4). For this one distribution, the percentages of the entire population occurring in specified areas have been worked out in sufficient detail that we can answer nearly any question relating to the percentage of the population in any par- ticular area of the distribution. Because that information is available, questions can also be posed about the probability that a particular measure will occur in a specified area.
Figure 2.4: The standard normal distribution
-4 -3 -2 -1 1 2 3 40
Since all normal populations are distributed the same way, if the characteristics of one are established in great detail, what is known about that particular distribution can be gener- alized to any other normal distribution if that other distribution can be related somehow to the standard normal distribution. The task is to create a metric that is common to all normal distributions, that makes them equivalent. This equivalence is created by recal- culating the data from the new distribution so that it too has a mean of 0 and a standard deviation of 1.0.
Key Terms: The standard normal distribution, or z distribution has mean of 0 and standard deviation of 1. Data are conformed to this distribu- tion by using the z transfor- mation to create z scores.
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CHAPTER 2Section 2.3 The Standard Normal Distribution
Where
z 5 the transformation of the raw score into a score fitting the z distribution x 5 the raw score M 5 the sample mean s 5 the sample standard deviation
Because s and M represent the standard deviation and mean of a sample, technically For- mula 2.2 is for the transformation of sample data from a normally distributed population into z scores. If the mean and standard deviation of the population were available, the z transformation would take this form:
Formula 2.2 z 5 x 2 M
s
Turning data from any normal distribution into data that fit the standard normal distribu- tion involves the use of the z transformation. When it is employed, data with means and standard deviations that in their raw form can have any value come to have the same values that the standard normal distribution has, a mean of 0 and a standard deviation of 1.0. Once transformed, the scores are called z scores. The formula for the z transformation is the following:
Formula 2.3 z 5 x 2 m
s
Where
z 5 the transformation of the raw score into a score fitting the z distribution x 5 the raw score m 5 the population mean 5 the population standard deviation
As noted earlier, detailed data about population characteristics are a good deal less avail- able than samples. As a result, the concentration will be almost exclusively Formula 2.2 from this point forward in the chapter.
For the sake of illustration, assume that the amount that couples spend dining out at res- taurants is normally distributed. Assume further that a restaurant manager is reviewing the menu items that are offered at the restaurant in order to change any prices or menu items that may be necessary to accommodate the preferences of the restaurant clientele. For a sample of eight couples drawn from that normally distributed population, the man- ager finds that the amounts couples spend dining are as follows:
45.11, 49.75, 55.19, 58.35, 62.95, 65.74, 69.90, 72.55
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CHAPTER 2Section 2.3 The Standard Normal Distribution
The lowest dining charge in the sample is $45.11. If the man- ager wishes to know what that value is as a z score fitting the standard normal distribution, the following steps will be taken. First, the mean and standard deviation for these amounts spent are determined. Verify that
M 5 59.943 s 5 9.635
Having calculated the descriptive statistics that Formula 2.2 requires, the manager can now determine the z-score equivalent for a restaurant bill of $45.11. All that is necessary to determine the z equivalent of $59.943 is to enter the appropriate values into the formula and complete the calculations. First, the formula:
z 5 x 2 M
s
The “x” value is the value for which z is being calculated. In this case, it is the lowest of the restaurant bills, $45.11. Since M and s have already been determined,
z 5 45.11 2 59.943
9.635 5
214.833 9.635
5 21.539
The result indicates that a “raw score” of 45.11 from a distribution where the mean is 59.943 and the standard deviation is 9.635 is equivalent to a z score of 21.539 in that dis- tribution where the mean 5 0 and the standard deviation 5 1.0. The two distributions can be compared as follows:
Raw Score z Score Equivalent
Mean $59.943 0
Standard deviation $9.635 1.0
Individual score $45.11 21.539
In the Raw Score column are the original values for the mean and standard deviation for the data set, and for the cost of that one meal. Once transformed by using Formula 2.2, the raw score values become z values that fit the standard normal distribution.
The transformation does not alter the shape of the distribution. We assumed that the orig- inal data distribution was normal, but if it happened to have positive skew, for example, the transformed distribution would also have positive skew. Transforming raw scores into z scores does not make them normal if they aren’t part of a normal distribution to begin with.
Review Question C: If a value is z 5 1.25, how many standard deviations from the mean of the popula- tion is the original raw score?
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CHAPTER 2Section 2.3 The Standard Normal Distribution
Interpreting the z Score
Because the numerator in the z calculation is a difference score (x 2 M), and the denomi- nator is a measure of data variability (the standard deviation), the resulting z score is a ratio. It is a ratio of difference to variability, and the ratio indicates the distance between x and M in units of standard deviation. For the z value just calculated, z 5 21.539 indicates that a restaurant bill of $45.11 is 1.539 standard deviations below the mean of all restau- rant bills for couples, since the sample is being used to indicate what occurs in the entire population.
Recall that the mean of the standard normal distribution is zero. The fact that this particu- lar z value is negative is a reminder that the value for which this z was calculated was an x less than the mean of the data set. By extension, any z score that is calculated for a value below the mean, which is half of all possible z scores in the population (the left-hand side of the x-axis under the normal distribution curve; see Figure 2.4), is going to result in a negative z score. By itself, this is no great revelation. At a glance it is evident that $45.11 is less than $59.943, but the calculated z value indicates precisely how much below M this particular x falls, and the units of measurement are standard deviations.
Earlier it was noted that in normal distributions, virtually the entire population occurs between 63 standard deviations. Since x 5 $45.11 is the lowest value in the data set, how can it be only 1.539 standard deviations below the mean? Once again we are reminded that these are sample data. The calculation indicates that in the population from which this sample was drawn, x 5 $45.11 would probably occur about halfway between the mean and the lower extreme of the distribution, a judgment that will be accurate if the sample represents the population well.
The above is meant to suggest that some hand calculation can be a great help to under- standing the logic and reasoning that statistical analysis involves. To accommodate hand calculations, the samples in the chapters are kept quite small. Small samples will help us avoid the calculation tedium that comes with larger samples. The trade-off is that small samples can rarely provide an accurate representation of the characteristics of the popu- lation. Just be reminded that in practice a great deal hinges on the degree to which the sample can mirror the population, something difficult to achieve with very small n values.
Comparing Measures
Calculating z scores creates a common metric for whichever data are transformed. It does not matter that their characteristics might have been different when the measures were originally taken. The z transformation makes the data fit a distribution that has a mean of 0 and a standard deviation of 1.0. In spite of what might be very different distributions with entirely different characteristics, the z transformation allows the two measures to be compared directly. For example, suppose that an insurance company pays sales commis- sions to its agents based on the dollar amounts of the policies they sell, and it asks cus- tomers to rate their satisfaction level with the agents with whom they deal. The data for a group of eight agents are as follows:
Sales commissions in thousands: $12, $15, $21, $21, $23, $23, $25, $27
Average customer satisfaction ratings: 9, 11, 14, 12, 15, 17, 11, 13
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CHAPTER 2Section 2.3 The Standard Normal Distribution
The 5th agent had sales commissions of $23,000 and an average customer satisfaction rat- ing of 15. On which measure did the agent do better: sales commissions or customer satis- faction? The two values are difficult to compare directly because the characteristics of the two distributions are different; they each have different means and standard deviations, for example. If the two data points are both turned into z scores, however, their values will each indicate their respective differences from their particular sample means, and those differences can be compared directly. To complete the comparison, the descriptive statistics for both distributions will need to be calculated, and then the z values for a sales commission of $23,000 and for a customer satisfaction rating of 15. Then the two z values can be compared directly. The descriptive statistics for each distribution are below:
Mean Standard Deviation
Commissions 20.875 ($k) 5.027
Satisfaction 12.750 2.550
Using the z transformation to produce z scores results in the following:
z 5 x 2 M
s For sales commissions, z 5
23 2 20.875 5.027
5 .423
For customer satisfaction, z 5 15 2 12.750
2.550 5 .882
Both z values are positive, which indicates that both raw scores are higher than their respective means. However the sales commission of $23,000 is .423 standard deviations above its mean, while the customer satisfaction level of 15 is .882 standard deviations above its mean. Relative to their respec- tive distributions, the customer satisfaction level is the higher score. In other words, this sales agent’s commissions are closer to the average for sales than the satisfaction score is to its average. In customer satisfaction, the agent is scoring is well above average. Perhaps the agent is spending more time servicing existing accounts, a behavior that may not generate additional sales and com- missions (at least not immediately) but can lead to more satisfied customers. This agent’s manager might want to consider an additional bonus or some other form of reward to rec- ognize the agent’s efforts and the superior service the agent is providing to the customers.
The Area Under the Curve
The point of using the z transformation was to adapt data from any normal distribution to the standard normal distribution. The adaptation allows what is known about the stan- dard normal, or z distribution, to help provide answers to questions that are more detailed than just what lies within 61, 2, or 3 standard deviations of the mean. Using the informa- tion in Table 2.1 is the key to the answers to those more specific questions.
Review Question D: What is the prob- ability that a value selected at random from a normal popu- lation will have a negative z value equivalent?
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CHAPTER 2Section 2.3 The Standard Normal Distribution
Table 2.1: Proportions of the normal distribution occurring between a value of z and the mean
0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359
0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549
0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952
2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964
2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974
2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981
2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986
3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990
Source: Standard Normal Distribution Table. (2012). Retrieved from http://math.tutorvista.com/statistics/z-score-table.html
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CHAPTER 2Section 2.3 The Standard Normal Distribution
Sometimes called the “z score table” or the “area under the normal curve table,” this table indicates how much of the standard normal distribution occurs in specific ranges. Not all versions of the data that make up Table 2.1 are arranged the same way. Some z tables indicate the amount of the distribu- tion below a specified z value. This particular table indicates the amount of the normal distribution that occurs between any value of z and the mean of the distribution.
Some z tables indicate results in terms of percentages. This particular table is in proportions of the entire distribution. Pro-
portion values have the same range as probability values, from 0 to 1.0. A proportion of 0 means none. A proportion of 1.0 indicates the entire distribution. Neither of those values ever occurs in a table of proportions of the distribution because there is no z value that accounts for either none, or all, of the distribution. It is a reminder that there is always at least the theoretical possibility of yet lower or higher measures associated with a particu- lar distribution. Note that the z values represented in the table are all those from z 5 .01 to z 5 3.09. The table is arranged so that:
• The whole numbers and tenths for values of z from 0.0 to 3.0 are in the extreme left column.
• The corresponding hundredths are in the columns across the top of the table beginning with 0 hundredths and continuing to 9 hundredths (.09). So if the proportion of the distribution between the mean and a value of z 5 1.12 was needed:
• Read down the left column to find 1.1 (the whole number and tenth). • Read over in the columns to find .02 (the hundredth), and then move down to
where that column intersects with the row for 1.1.
The intersecting value is .3686, which means that of the whole distribution, a proportion of .3686 occurs between z 5 1.12 and the mean. This is illustrated in Figure 2.5. Note that because it is a positive value of z, the proportion of the distribution between the value of z and the mean is to the right of the mean. If it had been a negative value of z, the proportion of the distribution would have been to the left of the mean.
Review Question E: Since Table 2.1 has the values for half of the distribution, why does no value in that table account for a .5 pro- portion of the normal distribution?
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CHAPTER 2Section 2.3 The Standard Normal Distribution
Figure 2.5: The proportion of the distribution for z 5 1.12
Mean = 0 z = 1.12
Proportion = .3686
Note that there are no values for negative z. Because the standard normal distribution, like all normal distributions, is symmetrical, the values that describe the positive side of the distribution will also be accurate for the same values on the negative side. The proportion of the distribution between z 5 21.0 and the mean of the distribution is exactly the same as the proportion between z 5 11.0 and the mean of the distribution.
Note two other things about the table. First, the table only accommodates z values to the hundredth place. While the general practice in the book is to round to thousandths, we make an exception with this table in order to limit its length. As a result, once z values are calculated they will need to be rounded to two decimals before consulting the table.
Second, there are no z values in the table more extreme than z 5 3.0. A z 5 3.0 accounts for .499 out of .5 of the distribution. Any value for which z 3.0 will be unusual and therefore occur only rarely. There will be little need for it in the table.
In the restaurant bills example, z 5 21.539. Table 2.1 can help answer several related ques- tions—questions that will have greater clarity if we take the time to draw a distribution like Figure 2.5 so that we can be oriented more easily to what is being asked. Now ques- tions such as the following are within reach:
1. What proportion of the population of dining couples will have dining charges that are somewhere between $45.11 and the mean (which was determined to be $59.943)?
To answer this question, • Round the calculated z value to the nearest hundredth (z 5 21.54). • In the left-hand column of the table move down to the row for 1.5. • Then move horizontally to the right to the column headed .04.
The intersecting value for z 5 21.54 is .4382. The proportion of the entire population of couples dining out that will spend between $45.11 and $59.943 on a meal is .4382. As a percentage, a little less than half, 43.82%, will spend somewhere between $45.11 and $59.943.
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CHAPTER 2Section 2.3 The Standard Normal Distribution
2. What proportion of the population of couples dining out will spend less than $45.11 for dinner? • It was already established that for z 5 21.54, the proportion from the table
is .4382. It can be helpful to indicate the z value and its corresponding table value this way: z 5 21.54 (.4382). Since half of the distribution is a proportion of .5, the proportion of the pop- ulation below z 5 21.54 must be must be .5 3 .4382 5 .0618. The proportion of the population of couples dining out who will spend less than $45.11 is .0618.
3. What proportion of the population of couples dining out will spend more than $45.11 for dinner? • Once again, starting from z 5 21.54 (.4382), a proportion of .4382 of the
total population of couples dining out will spend between $45.11 and $59.943 on a meal.
• To include the upper half of the population, the half that spends more than $59.943 on a meal, add .5 to .4382: .5 1 .4382 5 .9382. The proportion of the population of couples dining out who will spend more than $45.11 is .9382.
The three questions above are illustrated in Figure 2.6.
Figure 2.6: The proportions of the distribution for various values of z
z = 1.54
Proportion below z = 1.54 = .5 – .4382 = .0618
Proportion between z = -1.54 and the mean = .4382
Proportion above -1.54 – .4382 + .5 = .9382
From Proportions to Probabilities and Percentages
Table 2.1 provides data in proportions of the entire population. Because probability uses the same metric as proportions, we can extend the application by saying that:
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CHAPTER 2
1. The probability that a couple will spend between $45.11 and $59.943 on a meal is p 5 .4382.
2. The probability that a couple will spend less than $45.11 on a mean is p 5 .0618. 3. The probability that a couple will spend more than $45.11 on a mean is p 5 .9382.
If percentages are required, multiplying either the proportion or probability values by 100 will indicate the relevant percentage of the distribution between a value of z and the mean of the distribution.
1. The percentage of the population of couples dining out who will spend less than $45.11 is 6.18%.
2. The percentage of the population of couples dining out who will spend more than $45.11 is 93.82%.
Consider in the above all the information that such data can provide to a restaurant man- ager who is planning menus and meal prices. In terms of the most likely customers, the data suggest that the emphasis should be placed on meals for two that cost in the $60 range. Meals that are substantially below about $45 represent a missed opportunity since relatively little of the population of dining couples will probably order meals in that price range, and there will likewise be perhaps a limited market for that bacon-wrapped filet and lobster meal for two that is too much beyond $75 or so.
Individual Scores Versus Ranges
One of the questions that cannot be answered is what proportion of the population of cou- ples will order meals that cost $45.11 (or any other discrete value). The horizontal axis of a population distribution is called the abscissa, and the vertical axis is called the ordinate. Theoretically, the points along the abscissa that represent the different values for z, or the scores for any other distribution, for that matter, have no width and so by themselves can represent no proportion of the population. The only way a point indicates a proportion of
the distribution is when it is used with reference to some other point on the abscissa, such as the mean or another score. Table 2.1 can be used to determine what percentage of the population of dining couples will likely order meals that cost between $45.11 and $59.943, but it will not indicate what percentage will likely order meals costing precisely either value.
2.4 Using Excel to Calculate z Scores
Although Excel provides dedicated applications for several statistics procedures, it won’t automatically calculate z scores. However, all business spreadsheets are very good at performing the repetitive calculations that the z transformation involves. By way of illustration, perhaps the personnel specialist at a company that provides temporary office workers collects data on how flexible its employees are about moving from one type of job to another. These “job flexibility” measures for 10 employees are as follows:
Job Flexibility: 43, 49, 55, 59, 62, 63, 68, 73, 77, 82
Key Terms: In a frequency distribution graph, the horizon- tal axis is called the abscissa. The vertical axis is called the ordinate.
Section 2.4 Using Excel to Calculate z Scores
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CHAPTER 2Section 2.4 Using Excel to Calculate z Scores
The higher the value, the more flexible the individual is about adapting to different assign- ments. Because the personnel specialist has determined that the most productive employ- ees have flexibility scores 55 or higher, the question is, what percentage of the entire popu- lation of temporary workers does that include?
In a blank spreadsheet:
• Enter the scores (without the commas) in the first column with one score in each cell from cell A1 to cell A10.
• In cell A11 enter the formula 5average(A1:A10). This will produce the sample mean, 63.1.
• In cell A12 enter the formula 5standev(A1:A10). This will produce the sample standard deviation, 12.28775.
• In cell B1 enter the formula 5(A1-63.1)12.288. This will produce the z equiva- lent for 43, which rounds to 21.64.
• With the cursor on B1 hold the shift key down and use the down arrow to highlight all the cells from B1 to B10.
• Click the Fill icon in the tool bar. It could be on either the left or right side of the tool bar, depending upon the version of Excel. This will repeat the process producing the z values for all of the other raw scores.
Figure 2.7 is a screenshot of the result.
Figure 2.7: Using Excel to calculate z scores
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CHAPTER 2Chapter Summary
Note that the z value that is calculated for a raw score of 55 (z 5 20.65918) rounded to two decimals so that it fits Table 2.1 becomes z 5 2.66. The associated table value for z 5 .66 is a proportion of .2454. That proportion, multiplied by 100, indicates that 24.54% of the distribution occurs between z 5 2.66 and the mean of the distribution. Since the question is what percentage is above a score of 55, adding the 50% that represents the upper half of the distribution to the percentage for z 5 2.66 will provide the answer: 24.54% 1 50% 5 74.54% of the distribution above a raw score of 55. The most productive employees are those who have the highest 74.54% of flexibility scores.
Chapter Summary
This chapter began with an introduction to probability. It is a topic that can become part of a very complicated discussion, but some exposure to the way probabilities are cal- culated and the range of values that probabilities can take are what were needed for this chapter as part of the z score discussion (Objective 1), and for the discussion of statistical significance that will come up in Chapter 3.
Understanding probability is one of the keys to understanding the normal distribution. All normal distributions have certain characteristics in common. They are symmetrical (one piece of evidence to support this is that the measures of central tendency have the same value), they have just one mode, and the standard deviation tends to have a value about one-sixth of the range (Objective 2). Because of those common elements, normal distributions also have consistent characteristics related to probability. Those common probability characteristics mean that the likelihood that the particular measures making up the normal distribution will occur are known even before data are gathered.
Consistent characteristics mean that the proportions that will occur in certain areas of any normal distribution can be calculated. However, as long as the descriptive values such as mean and standard deviation are particular to a specific distribution, the results are difficult to generalize to other distributions. The exception is the standard normal distribu- tion. For this population, in fact, a great deal is known. Its characteristics are established in sufficient detail that tables exist, allowing us to ask about the probability that specific measures will occur in nearly any area of the population. The standard normal distribu- tion can serve as something of a window to any normally distributed population, as long as the data from those other populations are made to conform to this distribution where the mean is 0 and the standard deviation is 1.0. This is the point of the z transformation (Objective 4). By employing the z transformation, data from any normally distributed population take on the characteristics of the standard normal distribution. The resulting values of z indicate how distant any raw score is from its mean in standard deviation units (Objective 5). By relying on Table 2.1, which lists proportions of the population under specified areas of the normal curve, calculating z scores indicates the probability that cer- tain outcomes will occur.
In Chapter 3, z values are extended to samples. There the probability of certain outcomes will extend the discussion to statistical significance and the first statistical test.
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CHAPTER 2Management Application Exercises
Answers to Review Questions
A. If an item has 5 choices, and the respondent selects at random, the probability of a correct guess is p 5 number of times the event occurs ÷ the total number of possible outcomes; p 5 1⁄5 5 .2.
B. In theory at least, a bimodal distribution could be mesokurtic and symmetri- cal, but it certainly would not be normal.
C. If z has a value of 1.25, it indicates that the raw score value for which z was calculated is 1.25 standard deviations from its mean.
D. Since half of a normal distribution will have values below the mean, and since all those values will have z scores that are negative, half of all z values will be negative. The probability of a negative z, therefore, is p 5 .5.
E. Although they continue to come closer as they extend to more extreme val- ues, the tails of the curve in a frequency distribution never actually touch the abscissa. This accommodates the possibility that there could yet be a more ex- treme measure, higher or lower. Consequently, there is no value that accounts for half the entire distribution. We don’t know that there might not yet be a more extreme value.
Chapter Formulas
Formula 2.1 p 5 e/o This is the formula for calculating the probability of an outcome, when all possible outcomes are equally likely.
Formula 2.2 z 5 x 2 M
s This is the z transformation. It alters the value of raw scores
so that they fit a distribution where the mean 5 0 and the standard deviation 5 1.0. This particular formula is based on sample data.
Formula 2.3 z 5 x 2 m
s This is the z transformation for population data.
Management Application Exercises
Unless otherwise stated, use p 5 .05 in all your answers.
1. Human resource records in an organization show the following statistics of employee age: M 5 45, Mdn 5 38, Mo 5 36, R 5 38, s 5 12.
a. In terms of normality, how should the distribution be described? b. In lay terms, how would you characterize the age distribution of this orga-
nization’s workforce?
2. The manager of a sandwich shop gathers data on what people spend on lunch on a particular day of the week. The results are $4.20, $4.22, $2.35, $4.32, $5.25, $6.48, $6.78, $8.59, $6.95, $5.52, $6.83, $7.35, $4.36, $9.39, $6.42. To the degree that this rep- resents the population of all those who eat lunch at sandwich shops, what percent- age of customers spend:
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CHAPTER 2Management Application Exercises
a. Less than $4.34? b. Between $3.50 and $4.80? c. Less than $2.35? d. Why is the answer to “b” not zero, since there isn’t anyone in the group
who spent less than that amount?
3. The Management Ability Test (MAT) has m 5 38.00 and 5 4.50. a. What is the equivalent z score for someone with MAT 5 33? b. What is the z score for someone with MAT 5 39? c. What proportion of scores will occur between scores of 33 and 39? d. What proportion will occur above MAT 5 40?
4. A particular job applicant has scored 41 on a test of analytical ability. For that particular test, the characteristics for the rest of the group of which this trainee is a member are M 5 38.101 and s 5 4.508. The same applicant scored 72 on a test of social intelligence with M 5 64.00 and s 5 7.753. According to these test scores, is this applicant stronger on analytical ability or social intelligence?
5. The ANxiety General Stress Test (ANGST, for short) has been designed to gauge the level of psychological stress management trainees experience when they are under pressure. For a random sample of trainees the scores are as follows: 47, 49, 53, 53, 54, 58, 61, 64, 75, 81.
a. What is the z score equivalent of ANGST 5 81? b. What is the probability that someone selected at random will score 81 or
lower? c. What percentage of all trainees will score between 60 and 75?
6. A human resources manager has data representing a normal distribution of man- ual dexterity for clerical workers. What percentage of manual dexterity scores will fall between the following z scores?
a. z 5 21 and z 5 11 b. z 5 21.96 and z 5 11.96
7. If the mean of all achievement motivation scores among sales representatives is 10 with a standard deviation of 2.0, why is it highly unlikely that someone with an achievement motivation score of 16 will apply for a job?
8. Regarding the previous item, why can an employment manager be assured that people with achievement motivation scores of 8 are just as common as those with achievement motivation scores of 12?
9. Referring to the data in item 7, what’s the probability that someone applying for a position will have an achievement motivation score 14 or higher?
10. If the mean for achievement motivation among sales representatives is 10 with a standard deviation of 2.0, as was noted in item 7, and verbal aptitude for the same population has a mean of 23.5 with a standard deviation of 4.957, a particular candidate who has achievement motivation of 12.3 and verbal aptitude of 28.0 is higher in which characteristic?
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CHAPTER 2Key Terms
Key Terms
• Dichotomous outcomes have only two possibilities. A true/false question calls for a dichotomous outcome.
• The normal distribution is symmetrical, unimodal, and platykurtic. • A nonsymmetrical distribution is skewed. Skew is created by an imbalance of extreme
scores, or outliers, on one side of the distribution. • Kurtosis refers to how a population is distributed. Normal distributions are
mesokurtic. If s . 1/6 R, the distribution is platykurtic. If s , 1/6 R, the distribu- tion is leptokurtic.
• In a frequency distribution graph, the horizontal axis is called the abscissa. The vertical axis is called the ordinate.
• The standard normal distribution is a population in which the mean has a value of 0, and the standard deviation equals 1.0. It is also sometimes called the z distribution.
• The z transformation is a formula which turns “raw” scores into z scores so that they find the standard normal distribution. For that reason, the standard normal distribution is also sometimes called the z distribution.
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