814 CHAPTER 12 Statistics To change the area under the normal curve to a percent, multiply the area by 100%. In Example 2(a), we determined the area to the left of = − z 1.00 to be 0.1587. To change this area to a percent, multiply 0.1587 by 100% = × = 0.1587 0.1587 100% 15.87% Therefore 15.87% of the normal curve is less than a score that is one standard deviation below the mean. Below, we summarize the procedure to determine the percent of data for any interval under the normal curve. This answer agrees with our answer obtained by subtracting the area to the left of = z 1.19 from 1. d) To determine the area between two z-scores, we subtract the smaller area from the larger area (Fig. 12.29). Using Table 12.8(b), we see that the area to the left of = z 2.57 is .9949 (Fig. 12.29(a)). Using Table 12.8(a), we see that the area to the left of = − z 1.62 is .0526 (Fig. 12.29(b)). Thus, the area between = − z 1.62 and = − z 2.57 is 0.9949 0.0526, or 0.9423 (Fig. 12.29(c)). 0 (a) 2.57 – = 2.57 0 (b) –1.62 –1.62 0.0526 0 (c) z-scores z-scores z-scores 0.9949 0.9423 Figure 12.29 7 Now try Exercise 29 DETERMINING THE PERCENT OF DATA BETWEEN ANY TWO VALUES IN A NORMAL DISTRIBUTION 1. Draw a diagram of the normal curve, indicating the area or percent to be determined. 2. Use the formula = − z value of the piece of data mean standard deviation to convert the given values to z-scores. Indicate these z-scores on the diagram. 3. Look up the areas that correspond to the specified z-scores in Table 12.8. a) When determining the area to the left of a negative z-score, use Table 12.8(a). 0 z b) When determining the area to the left of a positive z-score, use Table 12.8(b). 0 z PROCEDURE
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