Question

A survey found that​ women's heights are normally distributed with mean

63.763.7

in and standard deviation

2.32.3

in. A branch of the military requires​ women's heights to be between 58 in and 80 in.

a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall?

b. If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%, what are the new height​ requirements?

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a. The percentage of women who meet the height requirement is

nothing​%.

​(Round to two decimal places as​ needed.)

Answer 1

This is a normal distribution question with


a)
P(58.0 < x < 80.0)=?

This implies that

PS: you have to refer z score table to find the final probabilities.

Since, 99.34% of teh women are selected so, we can say that most of women are selected and very least are denied.

b)
If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%.
We will need z score for left of 0.01 = -2.3263
and right of 0.02 = -2.0537

New height requirements:
since,

z score for left of 0.01 = -2.3263
x = 63.7+(-2.3263)*2.3 = 58.3495

z score for right of 0.02 = -2.0537
x = 63.7+(2.0537)*2.3 = 68.4235

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