12.5 The Normal Curve 817 Example 4 Horseback Rides Assume that the length of time for a horseback ride on the trail at Triple R Ranch is normally distributed with a mean of 3.2 hours and a standard deviation of 0.4 hour. a) What percent of horseback rides last at least 3.2 hours? b) What percent of horseback rides last less than 2.8 hours? c) What percent of horseback rides last at least 3.7 hours? d) What percent of horseback rides last between 2.8 hours and 4.0 hours? e) In a random sample of 500 horseback rides at Triple R Ranch, how many last at least 3.7 hours? Solution a) In a normal distribution, half the data are always above the mean. Since 3.2 hours is the mean, half, or 50%, of the horseback rides last at least 3.2 hours. b) Convert 2.8 hours to a z-score. = − = − z 2.8 3.2 0.4 1.00 2.8 Use Table 12.8(a) to determine the area of the normal curve that lies below a z-score of −1.00. The area to the left of = − z 1.00 is 0.1587. Therefore, the percent of horseback rides that last less than 2.8 hours is 15.87% (Fig. 12.36). c) At least 3.7 hours means greater than or equal to 3.7 hours. Therefore, we are seeking to determine the percent of data to the right of 3.7 hours. Convert 3.7 hours to a z-score. = − = z 3.7 3.2 0.4 1.25 3.7 From Table 12.8(b), we determine that the area to the left of = z 1.25 is .8944. Therefore, 89.44% of the data are below = z 1.25. The percent of data above = z 1.25 (or to the right of = z 1.25) is − 100% 89.44%, or 10.56% (Fig. 12.37). Thus, 10.56% of horseback rides last at least 3.7 hours. d) Convert 4.0 to a z-score. = − = z 4.0 3.2 0.4 2.00 4.0 From Table 12.8(b), we determine that the area to the left of = z 2.00 is .9772 (Fig. 12.38(a)). Therefore the percent of data below a z-score of 2.00 is 97.72%. From part (b), we determined that = − z 1.00 2.8 and that the percent of data below a z-score of −1.00 is 15.87% (Fig. 12.38(b)). To determine the percent of data between a z-score of −1.00 and a z-score of 2.00, we subtract the smaller percent from the larger percent. Thus, the percent of horseback rides that last between 2.8 hours and 4.0 hours is − 97.72% 15.87%, or 81.85% (Fig. 12.38(c)). 3.2 0 2 (a) 4.0 Original values – = z-scores 3.2 0 2.8 –1 (b) Original values z-scores 15.87% 3.2 0 2.8 –1 2 (c) 4.0 Original values z-scores 97.72% 81.85% Figure 12.38 e) In part (c), we determined that 10.56% of all horseback rides last at least 3.7 hours. We now multiply 0.1056 times the number in the random sample, 500, to determine the number of horseback rides that last at least 3.7 hours. There are × = 0.1056 500 52.8, or approximately 53, horseback rides that last at least 3.7 hours. 7 Now try Exercise 57 3.2 0 2.8 –1 Original values z-scores 15.87% Figure 12.36 3.2 0 3.7 1.25 Original values z-scores 89.44% 10.56% Figure 12.37 Timely Tip Following is a summary of some important items presented in this section. • The normal curve is symmetric about the mean. • A negative z-score indicates that the corresponding value in the original distribution is less than the mean. • A positive z-score indicates that the corresponding value in the original distribution is greater than the mean. • A z-score of 0 indicates that the corresponding value in the original distribution is the mean. • Table 12.8 provides the area to the left of a specified z-score. • When using Table 12.8 to determine the area to the left of a specified z-score, locate the units value and tenths value of your specified z-score under the column labeled z. Then move to the column containing the hundredths value of your specified z-score to obtain the area. My Good Images/Shutterstock
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