chap 7 assign 5

# chap 7 assign 5 - Normal Distribution 9 The mean of a...

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Normal Distribution 9.) The mean of a normal probability distribution is 500; the standard deviation is 10. a.) About 68 percent of the observations lie between what two values? 490, 510 500 – 1(10) = 490 500 + 1(10) = 510 b.) About 95 percent of the observations lie between what two values? 480, 520 500 – 2(10) = 480 500 + 2(10) = 520 c.) Practically all of the observations lie between what two values? 470, 530 500 – 3(10) = 470 500 + 3(10) = 530 19.) According to the Internal Revenue Service, the mean tax refund for the year 2007 was \$2,708. Assume the standard deviation is \$650 and that the amounts refunded follow a normal probability distribution. a.) What percent of the refunds are more than \$3,000? .3264 3000 – 2708 / 650 = .45 z = 0.1736 .5000 - .1736 = .3264 b.) What percent of the refunds are more than \$3,000 but less than \$3,500? .2152 3000 – 2708 / 650 = .45 z = .1736 3500 – 2708 / 650 = 1.22 z = .3888 .3888 - .1736 = .2152 c.) What percent of the refunds are more than \$2,500 but less than \$3,500? .5143 2500 – 2708 / 650 = -.32 z= .1255 3500 – 2708 / 650 = 1.22 z=. 3888 .3888 + .1255 = .5143 22.) Among U.S. cities with a population of more than 250,000 the mean one-way commute to work is 24.3 minutes. The longest one-way travel time is New York City, where the mean time is 38.3 minutes. Assume the distribution of travel times in New

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York City follows the normal probability distribution and the standard deviation is 7.5
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