1. Copyright © Cengage Learning. All rights reserved. Normal Curves and Sampling Distributions 7

2. 2 6.a) 50% b) 95% c)0.15% 8.a) 95% or 950 chicks b) 68% or 680 chicks c) 50% or 500 chicks d) 99.7% or 997 chicks 10.a) 0.16 b) 0.84 c) about 136 cups 1.a) no, skewed b) no, crosses the horizontal line c) it has three peaks d) no, not smooth curve 2.) µ=16, µ+σ = 18, σ=2 3.) 7-9 has larger standard deviation,10,4 4.) e.) no, the values of µ and σ are independent

3. Copyright © Cengage Learning. All rights reserved. Section 7.2 Standard Units and Areas Under the Standard Normal Distribution

4. 4 Focus Points Given  and , convert raw data to z scores. Given  and , convert z scores to raw data.

5. 5 z Scores and Raw Scores

6. 6 Normal distributions vary from one another in two ways: The mean  may be located anywhere on the x axis, and the bell shape may be more or less spread according to the size of the standard deviation . The differences among the normal distributions cause difficulties when we try to compute the area under the curve in a specified interval of x values and, hence, the probability that a measurement will fall into that interval.

7. 7 z Scores and Raw Scores It would be a futile task to try to set up a table of areas under the normal curve for each different  and  combination. We need a way to standardize the distributions so that we can use one table of areas for all normal distributions. We achieve this standardization by considering how many standard deviations a measurement lies from the mean. In this way, we can compare a value in one normal distribution with a value in another, different normal distribution. The next situation shows how this is done.

8. 8 z Scores and Raw Scores Suppose Tina and Jack are in two different sections of the same course. Each section is quite large, and the scores on the midterm exams of each section follow a normal distribution. In Tina’s section, the average (mean) was 64 and her score was 74. In Jack’s section, the mean was 72 and his score was 82. Both Tina and Jack were pleased that their scores were each 10 points above the average of each respective section. However, the fact that each was 10 points above average does not really tell us how each did with respect to the other students in the section.

9. 9 z Scores and Raw Scores In Figure 7-11, we see the normal distribution of grades for each section. Distributions of Midterm Scores Figure 7-11

10. 10 z Scores and Raw Scores Tina’s 74 was higher than most of the other scores in her section, while Jack’s 82 is only an upper-middle score in his section. Tina’s score is far better with respect to her class than Jack’s score is with respect to his class. The preceding situation demonstrates that it is not sufficient to know the difference between a measurement (x value) and the mean of a distribution. We need also to consider the spread of the curve, or the standard deviation. What we really want to know is the number of standard deviations between a measurement and the mean. This “distance” takes both  and  into account.

11. 11 z Scores and Raw Scores We can use a simple formula to compute the number z of standard deviations between a measurement x and the mean  of a normal distribution with standard deviation  :

12. 12 z Scores and Raw Scores The mean is a special value of a distribution. Let’s see what happens when we convert x =  to a z value: The mean of the original distribution is always zero, in standard units. This makes sense because the mean is zero standard variations from itself. An x value in the original distribution that is above the mean  has a corresponding z value that is positive.

13. 13 z Scores and Raw Scores Again, this makes sense because a measurement above the mean would be a positive number of standard deviations from the mean. Likewise, an x value below the mean has a negative z value. (See Table 7-1.) x Values and Corresponding z Values Table 7-1

14. 14 Example 2 – Standard score A pizza parlor franchise specifies that the average (mean) amount of cheese on a large pizza should be 8 ounces and the standard deviation only 0.5 ounce. An inspector picks out a large pizza at random in one of the pizza parlors and finds that it is made with 6.9 ounces of cheese. Assume that the amount of cheese on a pizza follows a normal distribution. If the amount of cheese is below the mean by more than 3 standard deviations, the parlor will be in danger of losing its franchise.

15. 15 Example 2 – Standard score How many standard deviations from the mean is 6.9? Is the pizza parlor in danger of losing its franchise? Solution: Since we want to know the number of standard deviations from the mean, we want to convert 6.9 to standard z units. cont’d

16. 16 z Scores and Raw Scores

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