Question

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

63.963.9

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?

Answer 1

a)
Here, μ = 63.9, σ = 2.3, x1 = 58 and x2 = 80. We need to compute P(58<= X <= 80). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (58 - 63.9)/2.3 = -2.5652
z2 = (80 - 63.9)/2.3 = 7

Therefore, we get
P(58 <= X <= 80) = P((80 - 63.9)/2.3) <= z <= (80 - 63.9)/2.3)
= P(-2.5652 <= z <= 7) = P(z <= 7) - P(z <= -2.5652)
= 1 - 0.0052
= 0.9948

Ans: 99.48%
No.

b)
lower limit = 63.9 - 2.33*2.3 = 58.54

upper limit = 63.9 + 2.05*2.3 = 68.62

new height requirements are 58.54 in to 68.62 in