A survey found that women’s heights are normally distributed with mean 63.8 in and a standard deviation 2.3 in. A branch of the military requires women’s heights to be between 58 in and 80 in.

AFind 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?
BIf 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?

The percentage of women who meet the height requirement is %.

For the new height requirements, this branch of the military requires women’s heights to be at least in and at most __ in.

Base on the questions, which is asking to many questions, ans the following would be the answers,
A. the women that fall between 58 and 80 inches tall would have a probability of 99.416%,
B. the new heights would 58,45 inches and 68,52 inches.

I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications

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joaobezerra joaobezerra · Answer 2 · 2019-12-22T16:53:11-05:00

Answer:

a) 99.41% of women meeting the height requirement. 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 requirement is between 58.441 in and 68.515 in.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

\(\mu = 63.8, \sigma = 2.3\)

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?

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

If the probability that a woman is accepted on the military is lower than 5%, we say that a woman being accepted into the military is unlikely.

So

X = 80

\(Z = \frac{X - \mu}{\sigma}\)

\(Z = \frac{80 - 63.8}{2.3}\)

\(Z = 7.04\)

\(Z = 7.04\) has a pvalue of 1.

X = 58

\(Z = \frac{X - \mu}{\sigma}\)

\(Z = \frac{58 - 63.8}{2.3}\)

\(Z = -2.52\)

\(Z = -2.52\) has a pvalue of 0.0059

So there is a 1-0.0059 = 0.9941 = 99.41% of women meeting the height requirement. 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?

Tallest 2%: X when Z has a pvalue of 0.98. So X when \(Z = 2.05\)

\(Z = \frac{X - \mu}{\sigma}\)

\(2.05 = \frac{X - 63.8}{2.3}\)

\(X - 63.8 = 2.3*2.05\)

\(X = 68.515\)

Shortest 1%: X when Z has a pvalue of 0.01. So X when \(Z = -2.33\)

\(Z = \frac{X - \mu}{\sigma}\)

\(-2.33 = \frac{X - 63.8}{2.3}\)

\(X - 63.8 = -2.33*2.3\)

\(X = 58.441\)

The new requirement is between 58.441 in and 68.515 in.