For further practice in using the normal distribution table, look up the prob- ability associated with a z score of 1.0; that is, a score one standard deviation from the mean. For this z score the table displays the now familiar probability of .3413. Now, look up the probability for a z score of 2.0 (two standard deviations from the mean). No surprise here either: Th e probability is the equally familiar .4772. Th e probability for a z score of 3.0 also confi rms our earlier u nderstanding.
Th e table shows that .4986 of the data values lie between the mean and a score three standard deviations above it.
Applications to Public and Nonprofi t
the z score in question and the mean (the total area under the curve is 1.0). In this example, proceed down the fi rst column until you fi nd a z score of 1.9. According to the numbers across the top of the table, the fi rst number next to the 1.9 represents a z score of 1.90; the second number is for a z score of 1.91; the third number represents a z score of 1.92 and is the one that interests us. Th e value found here is .4726. Th is means that 47.26% of the police exam scores fall between the mean (100) and a z score of 1.92 (119.2). Because 50% (or half ) of the scores fall below the mean, a total of 97.26% of the scores fall below a score of 119.2. In probability terms, the probability that a randomly selected individual will score a 119.2 or better on the Jeff erson police exam is .0274 (1.0 2 .9726). Another way of stating this result is that the indi- vidual in question scored at the 97th percentile. In sum, the applicant did very well on the exam.
An important aspect of the normal distribution is its fl exibility. Th e follow- ing examples illustrate some of the numerous applications of the normal distribution.
Suppose the Jeff erson, California, police chief wants to know the percent- age of job applicants who scored between 100 and 106 on the Jeff erson police exam. Because the mean is 100 (and the standard deviation is 10), this ques- tion is equivalent to asking what percentage of the applicants score between 106 and the mean. To answer, follow the two-step procedure discussed earlier and repeated here.
Step 1: Convert the score of interest into a z score. Use the formula for the z score:
z5 X2 m s 5 1062100
10 5.60
Th e z score corresponding to a raw score of 106 is .60. In other words, this score is six-tenths of a standard deviation above the mean.
Step 2: Look up a z score of .60 in the normal distribution table (Table 1) at the end of the book. Th e police chief wants to know the percentage of job applicants who fall between this z score (raw score 5 106) and the mean. We read this probability directly from the table; the table reveals a probability of .2257. Th us, about 22.6% of all applicants score between 100 and 106 on the exam. Th is area is shaded in Figure 8.6.
What percentage score above 106? Th is area, unshaded in Figure 8.6, is equal to .5 2 .2257 5 .2743, so about 27.4% of job applicants surpass this criterion.
What percentage of applicants scores between 88 and 112 on the exam?
Calculating this probability is easy if you split it into two questions with refer- ence to the mean. First, what percentage of applicants fall between the mean of 100 and a score of 112? Calculate the z score in the space provided. Th en, look up the corresponding probability in the normal distribution table.
Step 1: Convert the score of interest into a z score:
z5 1122100 10 512
10 51.20
Step 2: Look up a z score of 1.20 in the normal distribution table. Th e per- centage of job applicants who fall between this z score and the mean is .3849. Th us, 38.49% of all applicants score between 100 and 112 on the police examination.
Now, consider the second part of the question: What percentage of applicants score between 88 and the mean of 100? Note that 88 is the same distance from the mean as 112, but in the opposite (negative) direction. Therefore,
Figure 8.6 Probability for z Score 5 .60
m z 5 .60
it will yield the same absolute z value with a negative sign, 2 1.20. The associated probability is the same, .3849. To fi nd the percentage of applicants who score between 88 and 112, add these probabilities, for an answer of about 77%
(.3849 1 .3849 5 .7698). Th is area of the normal curve is shaded in Figure 8.7.
What is the probability that a randomly selected applicant will score between 117 and 122 on the Jeff erson police examination? Note that the probability of scoring between these two values is equal to the probability of scoring between the mean and 122 minus the probability of scoring between the mean and 117;
this diff erence is highlighted in Figure 8.8. To fi nd this area of the normal curve, follow the two-step procedure presented earlier. First, calculate the z scores for the two scores of interest, 117 and122:
z5 1172100
10 z 51222100 10 5 17
10 5 22 10 51.70 52.20
Next, look up the probabilities associated with z scores of 1.70 and 2.20 in the normal distribution table; .4554 of the area under the curve lies between 1.70 and the mean, and .4861 lies between 2.20 and the mean. As shown in Figure 8.8, the probability of scoring between 117 and 122, therefore, is equal to the diff erence between these two probabilities, or .0307 (.4861 2 .4554).
What is the probability that an applicant will score 125 or more on the Jef- ferson police exam? Because we can calculate the probability that someone will score between the mean and 125, we can solve this problem. Th e probability
Figure 8.7 Probability Between z 5 21.20 and z 5 1.20
m
z 521.20 z 5 1.20
of scoring between 100 and 125 (z score 5 ?) is .4938. Because 50% of all persons score below the mean, the probability of scoring 125 or below is .9938 (.50 1 .4938). This means the probability of scoring 125 or above is .0062 (1.0 2 .9938).
Calculating a probability when given a score or a range of scores is only one way the normal distribution is used. In public and nonprofi t administration, the manager often needs to fi nd the raw score that will cut off a certain percentage of the cases—for example, the top 10% or the bottom 25%. Say the Jeff erson police force has a policy of accepting only those who score in the top 20% of the per- sons taking the police exam. What raw score should be used as the cutoff point?
Th is question can be answered by following these steps:
Step 1: Th e score associated with the top 20% is the same score that is associ- ated with the bottom 80%. Th is problem is easier to solve for the bot- tom 80% than for the top 20%, so we will fi nd the cutoff point for the bottom 80%. We know from the shape of the normal curve that 50% of the bottom 80% fall below the mean of 100. Another 30% fall between the cutoff value and the mean. To fi nd the z score associated with this cutoff point, look up a probability of .3000 in the body of a normal distribution table. Th en read to the far left column and top row of the table to fi nd the corresponding z score. Note that this procedure reverses the one we have been following—that of calculating a z score from raw numbers and then fi nding the associated probability in the body of the table. Here, we know the probability we want to set off (.3000), and we need to fi nd the associated z score. Th erefore, we start in the body of the table.
Figure 8.8 Probability Between Two z Scores
m z 5 1.70 z 5 2.20
Scanning the body of the normal distribution table, we fi nd that the closest probability to .3000, without going under it, is .3023. The z score associated with this probability is .85 and is the one we seek. (If we had chosen a z score with an associated probability less than .3000, we would cut off slightly more than the top 20% of scores on the police examination.)
Step 2: Convert this z score to a raw score. Recall that a z score is the number of standard deviations a score of interest is from the mean. We want the raw score associated with a z score of .85. We multiply this z score times the standard deviation (10) to get 8.5; our score of interest is, thus, 8.5 above the mean. To complete the conversion, add this number to the mean to get 100 1 8.5 5 108.5. If the police academy class is limited to the top 20% of all applicants, only those who score 108.5 or above on the examination should be admitted.
An example from nonprofi t management is in order. Th e Gold Star Mentoring Program selects peer mediators from among Cub Scouts in the area. To be eli- gible, a Cub Scout must score in the top 75% of applicants on a test of social knowledge and interest in helping other people. Because the Gold Star Mentor- ing Program would like to involve as many Cub Scouts as possible, it accepts the top 75% of all those taking the exam. Last year, the mean score on this exam was 80, with a standard deviation of 6. Exam scores were normally distributed. At what value should the minimum acceptable score be set to admit only the top 75% of Cub Scouts who apply?
Because we know 50% of the applicants will score above 80 (mean score), we need to know the score below the mean that will contain 25% of the scores between it and the mean. Looking up a probability of .25 in the normal table, we fi nd a value of .2486 associated with a z score of .67. Because we are interested in scores below the mean, this z score is 2.67. Converting a z score of 2.67 to a raw score, we get a minimum acceptance score of 76 [( 2.67 3 6) 1 80 5 76].
Th us, the Gold Star Mentoring Program should accept Cub Scouts who score 76 or above on the test of social knowledge and interest in helping other people.