Question

A survey found that​ women's heights are normally distributed with mean 63.5 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

Solution :

Given that,

mean = = 63.5

standard deviation = =2.5

a ) P (58 < x < 80 )

P ( 58 - 63.5 / 2.5 ) < ( x -  / ) < ( 80 - 63.5/ 2.5 )

P ( - 5.5 / 2.5 < z < 16.5 / 2.5 )

P (-2.2 < z < 6.6)

P ( z < 6.6 ) - P ( z < -2.2)

Using z table

=1 - 0.0139

=0.9861

Probability =0.9861

c ) P(Z < z) = 1%

P(Z < z) = 0.01

P(Z < - 2.326) = 0.01

z = -2.33

Using z-score formula,

x = z * +

x = -2.33 * 2.5 + 63.5

x = 57.67

d ) P( Z > z) = 2%

P(Z > z) = 0.02

1 - P( Z < z) = 0.02

P(Z < z) = 1 - 0.02

P(Z < z) = 0.98

z = 2.05

Using z-score formula,

x = z * +

x =2.05 * 2.5 + 63.5

x = 68.62