Question

a survey found that womens heights are normally distributed with mean 63.4 in and standard deviation 2.5 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 % (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 the percentage of women who meet the height requirement is fairly small. b. yes, because the percentage of women who meet the height requirement is fairly large. c. no, because only a small percentage of women are not allowed to join this branch of the military because of their height. d. yes, because a large percentage of women are not allowed to join this branch of the military because of their height. b. for the new height requirements, this branch of the military requires womens heights to be at least in and at most in. (round to one decimal place as needed.)

Explanation:

Step1: Calculate z - scores for the given heights

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 63.4$ (mean), $\sigma=2.5$ (standard deviation).
For $x = 58$, $z_1=\frac{58 - 63.4}{2.5}=\frac{- 5.4}{2.5}=-2.16$.
For $x = 80$, $z_2=\frac{80 - 63.4}{2.5}=\frac{16.6}{2.5}=6.64$.

Step2: Find the probabilities corresponding to the z - scores

Using the standard normal distribution table, $P(Z < - 2.16)=0.0154$ and $P(Z < 6.64)\approx1$.
The probability of a woman's height being between 58 in and 80 in is $P(-2.16So the percentage of women meeting the height requirement is $98.46\%$.
Since the percentage of women not meeting the requirement is $1 - 0.9846=0.0154 = 1.54\%$, not many women are being denied the opportunity. The answer to part a second - question is C.

Step3: Find the z - scores for the new height requirements

For the shortest 1%: Looking up the z - score in the standard normal distribution table for an area of 0.01, the z - score $z_{lower}\approx - 2.33$.
For the tallest 2%: The area to the left of the z - score for the tallest 2% is $1 - 0.02=0.98$. Looking up this area in the standard normal distribution table, the z - score $z_{upper}\approx2.05$.

Step4: Calculate the new height values

Using the z - score formula $x=\mu+z\sigma$.
For the lower height: $x_{lower}=63.4+( - 2.33)\times2.5=63.4 - 5.825 = 57.6$ in.
For the upper height: $x_{upper}=63.4+2.05\times2.5=63.4 + 5.125 = 68.5$ in.

Answer:

a. $98.46$
C. No, because only a small percentage of women are not allowed to join this branch of the military because of their height.
b. $57.6$
$68.5$