Question

Women's heights are normally distributed with mean 64 inches and a standard deviation of 2.5 inches.

(a) What percent of women are less than 60 inches tall?

(b) The requirement of all Navy pilots is between 62 inches and 78 inches. What percent of women meet the Navy requirement?

(c) If the Navy changed its requirement to eliminate the "shortest" 5% of all women, what would the requirement be on the lower end of the scale?

Answer 1

Solution :

Given that,

mean = = 64

standard deviation = = 2.5

A ) P( x < 60 )

P ( x -  / ) < ( 60 - 64 / 2.5 )

P ( z < - 4 / 2.5 )

P ( z < - 1.6 )

Using z table

= 0.0548

Probability = 0.0548

B ) P ( 62 < x < 58 )

P ( 62  - 60 / 2.5 ) < ( x -  / ) < ( 78 - 60 / 2.5)

P ( 2 / 2.5 < z < 18 / 2.5 )

P (0.8 < z < 7.2 )

P ( z < 7.2 ) - P ( z < 0.8)

Using z table

= 1 - 0.7881

= 0.2119

Probability = 0.2119

C ) Using standard normal table.

P(Z < z) = 5%

P(Z < z) = 0.05

P(Z < -1.645) = 0.05

z = -1.64

Using z-score formula,

x = z * +

x = -1.64 * 2.5 + 60

= 55.9

x = 55.9