Question

A survey found that women's heights are normally distributed with mean 63.8in. and standard deviation 3.5in. The survey also found that men's heights are normally distributed with mean 67.5in. and standard deviation 3.5in. Most of the live characters employed at an amusement park have height requirements of a minimum of 56in. and a maximum of 64in. 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?  (Round to two decimal places as needed.)

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. (Round to one decimal place as needed.)

Answer 1

Given, women's heights(in.) ~ Normal(63.8,3.5) & men's heights(in.) ~ Normal(67.5,3.5)

Now, P(women's heights will be minimum of 56in. and a maximum of 64in) = P() = P(-2.22857 Z 0.05714) = 0.5099, {where Z is the standard Normal variable, and the probability values can obtain from any standard normal table}

& P(men's heights will be minimum of 56in. and a maximum of 64in) = P() = P(-3.28571 Z -1) = 0.1581

a) 15.81% of men meeting the height requirement. The result suggests that, maximum are women of the people who are employed as characters at the amusement park.

b)For The new height requirements,

P(men's heights will be higher than maximum height) = 0.5

=> P() = 0.5 => => max. height = 67.5 in.

and, P(men's heights will be lower than minimum height) = 0.05

=> P() = 0.05 => => min. height = 61.8 in.

The new height requirements are a minimum of __61.8__ in. and a maximum of __67.5__ in. (Rounded to one decimal place as needed.)