Question

A survey found that women's heights are normally distributed with mean 62.7 in, and standard deviation 2.8 in. The survey also found that men's heights are normally distributed with mean
69.3 in. and standard deviation 3.9 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 %.
(Round to two decimal places as needed)

Answer 1

Using the normal distribution, we have that:

The percentage of men who meet the height requirement is 3.06%. This suggests that the majority of employees at the park are females.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation of men's heights are given as follows:

$\mu = 69.3, \sigma = 3.9$.

The proportion of men who meet the height requirement is is the p-value of Z when X = 62 subtracted by the p-value of Z when X = 55, hence:

X = 62:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{62 - 69.3}{3.9}$

Z = -1.87

Z = -1.87 has a p-value of 0.0307.

X = 55:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55 - 69.3}{3.9}$

Z = -3.67

Z = -3.67 has a p-value of 0.0001.

0.0307 - 0.0001 = 0.0306 = 3.06%.

The percentage of men who meet the height requirement is 3.06%. This suggests that the majority of employees at the park are females.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1