Question

A survey found that women's heights are normally distributed with a mean 63.6 in and standard devation 2.5 in. A branch of the military requires women's heights 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 tallest 2%, what are the new height requirements?

Answer 1

Mean = 63.6

S.D = 2.5

to find P( 58 < x < 80)

a)

Z score at X = 58

Z = (X - μ) / σ
Z = (58 - 63.6) / 2.5
Z = -2.24

Z score at X = 80

Z = (X - μ) / σ
Z = (80 - 63.6) / 2.5
Z = 6.56

From Z score table

P( 58 < x < 80) = P(-2.24 < Z < 6.56) = P( Z < 6.56)-P(Z < -2.24)= 1 - 0.0125 = 0.9875

98.75% woman's are meeting the rquirement, 1.25% were denied

b)

For the shortest 1%

P Value = 0.01

Z score from Z score table

Z = -2.32635

X value for 1%

Z = (X - μ) / σ
-2.32635= (X - 63.6) / 2.5
X =57.784

For the Tallest 2%

P Value = 0.98

Z score from Z score table

Z = 2.05375

X value for 1%

Z = (X - μ) / σ
2.05375= (X - 63.6) / 2.5
X =68.734

New height requirement will be between 57.784 to 68.734