Question

A-C, assume the following:
Men’s heights are normally distributed with mean 71.3 inches and standard deviation 2.8 inches.
Women’s heights are normally distributed with mean 65.7 inches and standard deviation 2.6 inches.
Most of the live characters at Disney World have height requirements with a minimum of 58 inches and a maximum of 77 inches.

A. Find the percentage of women meeting the height requirement.

B. Find the percentage of men meeting the height requirement.

C. If the Disney World height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women, what are the new height requirements?

Answer 1

Solution :

Given that ,

A) mean = = 65.7 inches. ( women's)

standard deviation = = 2.6 inches.

P(58 < x < 77 ) = P[( 58 - 65.7 )/ 2.6 ) < (x - ) /  < ( 77 - 65.7 ) / 2.6) ]

= P( -2.96 < z < 4.35 )

= P(z < 4.35 ) - P(z < -2.96 )

= 1 - 0.0015

= 0.9985.

The percentage is = 99.85%

B) mean = = 71.3 inches. ( men's)

standard deviation = = 2.8 inches.

P( 58 < x < 77 ) = P[( 58 - 71.3 )/ 2.8 ) < (x - ) /  < ( 77 - 71.3 ) / 2.8 ) ]

= P( -4.75 < z < 2.04)

= P(z < 2.04 ) - P(z < - 4.75 )

= 0.9793 - 0

= 0.9793

The percentage is = 97.93%

C) Using standard normal table

P(Z > z ) = 5%

1 - P(Z < z ) = 0.05

P ( Z < z ) = 1 - 0.05 = 0.95

P( Z < 1.645) = 0.95

z = 1.645

Using z-score formula,

x = z * +

x = 1.645 * 2.8 + 71.3

x = 75.91 inches.( men)

Using standard normal table

P(Z < z) = 5%

P( Z < z) = 0.05

P(Z< -1.645 ) = 0.05

z = -1.645

Using z-score formula,

x = z * +

x = -1.645 * 2.6 + 65.7

x = 61.42 inches.( women)