Question

A survey found that​ women's heights are normally distributed with mean

62.2

in. and standard deviation

3.2

in. The survey also found that​ men's heights are normally distributed with mean

67.2

in. and standard deviation

3.5

in. Most of the live characters employed at an amusement park have height requirements of a minimum of

55

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 what %?

​(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 height and maximum height?

Answer 1

Solution:

women's height

mean 62.2 in.

standard deviation 3.2 in

​ men's heights

mean 67.2 in.

standard deviation 3.5 inches

a) probability that a man meets the height requirements

b)

Z score in the normal probability table in the appendis that corresponds with a probability of 0.05 and 0.50

z= -1.645
z= 0

x = µ + z σ = 67.2 + (-1.645)*3.2 = 61.936

x = µ + z σ = 67.2 + 0*3.2 = 67.2

The smallest 5 % have height below 61.936

largest 50% have height 68.6 inches