Question

A survey found that​ women's heights are normally distributed with mean 63.5 in and standard deviation 2.5 in. A branch of the military requires​ women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%, what are the new height​ requirements?

Answer 1

Answer:

a) 98.51% women meet the height requirement. This is a low percentage, so there are not many women being denied the opportunity to join this branch of the military because they are too short or too​ tall.

b) The new height requirements would be between 57.7in and 68.6 in.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 63.5, \sigma = 2.5$

a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall?

This is the pvalue of Z when X = 80 subtracted by the pvalue of Z when X = 58.

X = 80

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

$Z = \frac{80 - 63.5}{2.5}$

$Z = 6.6$

$Z = 6.6$ has a pvalue of 1.

X = 58

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

$Z = \frac{58 - 63.5}{2.5}$

$Z = -2.2$

$Z = -2.2$ has a pvalue of 0.0139.

1 - 0.0139 = 0.9861

98.51% women meet the height requirement. This is a low percentage, so there are not many women being denied the opportunity to join this branch of the military because they are too short or too​ tall.

b. If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%, what are the new height​ requirements?

Between the 1st and the 100-2 = 98th percentile

1st percentile

value of X when Z has a pvalue of 0.01. So X when Z = -2.325.

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

$-2.325 = \frac{X - 63.5}{2.5}$

$X - 63.5 = -2.325*2.5$

$X = 57.7$

98th percentile

value of X when Z has a pvalue of 0.98. So X when Z = 2.055.

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

$2.055 = \frac{X - 63.5}{2.5}$

$X - 63.5 = 2.055*2.5$

$X = 68.6$

The new height requirements would be between 57.7in and 68.6 in.