In a survey of women in a certain country (ages 20-29), the mean height was 66.6

Question

In a survey of women in a certain country (ages 20-29), the mean height was 66.6 inches with a standard deviation of 2.78 inches. Answer the following questions about the specified normal distribution (a) What height represents the 95th percentile? (b) What height represents the first quartile? Click to view page 1 of the table. Click to view page 2 of the table (a) The height that represents the 95th percentile is inches Round to two decimal places as needed.) (b) The height that represents the first quartile is inches. Round to two decimal places as needed.) Enter your answer in each of the answer boxes

Explanation / Answer

Mean Height = 66.6 inches

standard deviation = 2.78 inches

(a) If 95th percentile x that means if X is the height of any random person then

Pr(X < x) = NORM(X <x ; 66.6; 2.78) = 0.95

so Z - value for p value = 0.95 is

Z = 1.645

(X - 66.6)/ 2.78 = 1.645

X = 66.6 + 1.645 * 2.78 = 71.17 inches

(b) First quartile means 25% percentile let say it is x so

that means if X is the height of any random person then

Pr(X < x) = NORM(X <x ; 66.6; 2.78) = 0.25

so Z - value for p value = 0.25 is

Z = -0.675

(X - 66.6)/ 2.78 = -0.675

X = 66.6 - 0.675 * 2.78 = 64.72 inches

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