Question

A survey found that​ women's heights are normally distributed with mean 63.6 in and standard deviation 2.5 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

Mean = 63.6 in

Standard deviation = 2.5 in

P(X < A) = P(Z < (A - mean)/standard deviation)

a) P(meeting the height requirement) = P(58 < X < 80)

= P(X < 80) - P(X < 58)

= P(Z < (80 - 63.6)/2.5) - P(Z < (58 - 63.6)/2.5)

= P(Z < 6.56) - P(Z < -2.24)

= 1 - 0.0125

= 0.9875

= 98.75%

Not many women are denied the opportunity to join this branch of the military because they are too short or too​ tall

b) Let the minimum height be M and maximum height be N

P(X < M) = 0.01

P(Z < (M - 63.6)/2.5) = 0.01

Take Z value corresponding to 0.01 from standard normal distribution table

(M - 63.6)/2.5 = -2.33

M = 57.775 in

P(X > N) = 0.02

P(X < N) = 1 - 0.02 = 0.98

P(Z < (N - 63.6)/2.5) = 0.98

(N - 63.6)/2.5 = 2.05

N = 68.725 in

The new height requirements are minimum 57.775 in and maximum 68.725 in