z = (score -mean)/sd
a. z = (74-69)/2.8
z = 5/2.8 = 1.79 is .0367
The percentage of men that are taller than in is 3.67%
b. z = (score -mean)/sd
z = ((70-63.6)/2.5
z = 6.4/2.5 = 2.56 is .0052
The percentage of women that are taller than in is .52%.
Tall clubs international is a social organization for tall people. It has a height requirement that men must be at least 74 in. and women must be at least 70 in. tall. If men's heights have a normal mean of 69.0 in and a deviation of 2.8 and women's normal heights have a mean of 63.6 in. and a deviation of 2.5 in.
a) what percentage of men meet that requirement?
b) what percentage of women meet that requirement?
C)Are the requirements for men and women fair? why and why not?
1 answer
4 answers