a survey found that womens heights are normally distributed with mean 63.9 in and standard deviation 2.2 in. a branch of the military requires womens 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?
click to view page 1 of the table. click to view page 2 of the table.
a. the percentage of women who meet the height requirement is 99.63 %.(round to two decimal places as needed.)
are many women being denied the opportunity to join this branch of the military because they are too short or too tall?
a. no, because only a small percentage of women are not allowed to join this branch of the military because of their height.
b. yes, because the percentage of women who meet the height requirement is fairly large.
c. yes, because a large percentage of women are not allowed to join this branch of the military because of their height.
d. no, because the percentage of women who meet the height requirement is fairly small.
b. for the new height requirements, this branch of the military requires womens heights to be at least
and at most
(round to one decimal place as needed.)

Question

Accepted Answer

# Explanation:
## Step1: Standardize the lower - bound height
We use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 63.9$ (mean), $\sigma=2.2$ (standard deviation), and $x = 58$.
$z_1=\frac{58 - 63.9}{2.2}=\frac{- 5.9}{2.2}\approx - 2.68$

## Step2: Standardize the upper - bound height
Using the same formula with $x = 80$, we have $z_2=\frac{80 - 63.9}{2.2}=\frac{16.1}{2.2}\approx7.32$

## Step3: Find the probability
We want $P(-2.68<Z<7.32)$. Since the standard normal distribution table gives $P(Z < - 2.68)$ and $P(Z<7.32\approx1)$. From the standard - normal table, $P(Z < - 2.68)=0.0037$. So $P(-2.68<Z<7.32)=1 - 0.0037 = 0.9963$ or $99.63\%$. And since only $1 - 0.9963 = 0.0037$ (a small percentage) of women are not allowed due to height, the answer to part a second - question is A.

## Step4: Find the z - scores for the new requirements
For the shortest $1\%$, the z - score $z_{lower}$ such that $P(Z<z_{lower})=0.01$. Looking up in the standard - normal table, $z_{lower}\approx - 2.33$. For the tallest $2\%$, the z - score $z_{upper}$ such that $P(Z > z_{upper})=0.02$, so $P(Z<z_{upper})=1 - 0.02 = 0.98$. Looking up in the table, $z_{upper}\approx2.05$

## Step5: Find the new height requirements
Using the z - score formula $x=\mu+z\sigma$. For the lower height, $x_{lower}=63.9+(-2.33)	imes2.2=63.9 - 5.126\approx58.8$ inches. For the upper height, $x_{upper}=63.9 + 2.05	imes2.2=63.9+4.51\approx68.4$ inches

# Answer:
a. A. No, because only a small percentage of women are not allowed to join this branch of the military because of their height.
b. At least 58.8 in and at most 68.4 in

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QUESTION IMAGE

Question

Explanation:

Step1: Standardize the lower - bound height

Step2: Standardize the upper - bound height

Step3: Find the probability

Step4: Find the z - scores for the new requirements

Step5: Find the new height requirements

Answer: