Question

A survey found that​ women's heights are normally distributed with mean 62.3 in. and standard deviation 2.8 in. The survey also found that​ men's heights are normally distributed with mean 67.9 in. and standard deviation 3.9 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in.

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

- The percentage of men who meet the height requirement is 10.12%

b. If the height requirements are changed to exclude only the tallest​ 50% of men and the shortest​ 5% of​ men, what are the new height​ requirements?

The new height requirements are a minimum of __ in. and a maximum of __ in.

Answer 1

x : women's heights

µx = mean 62.3 in , standard deviation = σx = 2.8 in

y : men's heights

µy = mean 67.9 in , standard deviation = σy = 3.9 in

Park height requirements of a minimum of 57 in. and a maximum of 63 in.

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

z = (x-µx)/σx

P( 57 < x < 63 ) = P ( -2.79 < z < -1.26 )

= P ( z < -1.26 ) - P ( x < -2.79 )

= 0.1038 - 0.0026

= 0.1012

- The percentage of men who meet the height requirement is 10.12%

b. If the height requirements are changed to exclude only the tallest​ 50% of men and the shortest​ 5% of​ men, what are the new height​ requirements?

To exclude tallest 50% , z = 0.00 , x = zσx + µx = 0.00*2.8+62.3 = 62.3

To exclude shortest 5% , z = -1.645 , x = zσx + µx = -1.645*2.8+62.3 = 57.69

The new height requirements are a minimum of 57.7 in. and a maximum of 62.3 in.