COMPLETELY CONFUSED

Question

A survey found that women's heights are normally distributed with mean 63.6 in. and standard deviation 2.9 in. The survey also found that men's heights are normally distributed with a mean 67.4 in. and standard deviation 2.8.
a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 4 in. Find the percentage of women meeting the height requirement.

 

aThe percentage of women who meet the height requirement is ? %.

b. Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is ? %.

c. If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women, what are the new height requirements? The new height requirements are at least ? in. and at most ? in.

Answer 1

Wendy,

 

This involves finding z values which tell you percentage from 0 to x.

z=(x-μ)/σ

 

In this case our minimum sample point is 4 ft 9 inches or 4*12+9=57 inches. The maximum sample point is 6 ft 4 inches or 6*12+4=76.

 

For the minimum point z=(57-63.6)/2.9=-2.28 which equates to .01. The maximum z=(76-63.6)/2.9=4.27 which is pretty much 100%. So there is a 99% chance of meeting the height requirement for women.

 

For men the minimum z=(57-67.4)/2.8=-3.71 which is essentially 0 (0.0001). The maximum z=(76-67.4)/2.8=3.07 which is 99.9% So there is a 99.9% chance of meeting the height requirement for men.

 

To exclude the tallest 5% of men, we need to find the z for .95 and then solve for x. Here z=1.645=(x-67.4)/2.8

4.606=x-67.4

x=72 inches for men max or 6 ft. Current minimum is fine.

 

To exclude the shortest 5% of women, we need to find the z for .05 which I just happen to know is -1.645. (symmetric around the z=0 axis). So -1.645=(x-63.6)/2.9

-4.7705=x-63.6

x=58.83 inches minimum or 4 ft 10.83 inches. Current maximum is fine.

 

Hope that helps!