Non-Standard Normal distribution

Question

A survey found that​ women's heights are normally distributed with mean 62.2 in. and standard deviation 2.2 in. The survey also found that​ men's heights are normally distributed with mean 69.1in. and standard deviation 3.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 62 in. Complete parts​ (a) and​ (b) below.

a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement​ park?

The percentage of men who meet the height requirement is blank​%.

​(Round to two decimal places as​ needed.)

Answer 1

We need to work out a pair of Z scores (Z₁ and Z₂) for men and women.

Men: Z₁=(57-69.1)/3.1=-3.903, p=0.00475%,

Z₂=(62-69.1)/3.1=-2.29=1.1%

Women: Z₁=(57-62.2)/2.2=-2.36=1.18%,

Z₂=(62-62.2)/2.2=-0.091=46.38%.

So the percentage of men meeting the requirements: 1.1%-0.00475=1.1% approx.

Percentage of women meeting the requirements: 46.38-1.18=45.2%.

Clearly, the women have the advantage over the men. Only 1.10% of men meet the requirements.