Question

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 % (round to two decimal places as needed.)

Explanation:

Step1: Calculate z - scores for height limits

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 63.9$ (mean), $\sigma=2.2$ (standard deviation). For $x = 58$, $z_1=\frac{58 - 63.9}{2.2}=\frac{- 5.9}{2.2}\approx - 2.68$. For $x = 80$, $z_2=\frac{80 - 63.9}{2.2}=\frac{16.1}{2.2}\approx7.32$.

Step2: Find the area between the z - scores

Using the standard normal distribution table, the area to the left of $z_1=-2.68$ is $A_1 = 0.0037$. The area to the left of $z_2 = 7.32$ is approximately $1$. The area between $z_1$ and $z_2$ is $A=1 - 0.0037=0.9963$. So the percentage of women meeting the height requirement is $99.63\%$. Since the percentage of women not meeting the requirement is $1 - 0.9963 = 0.0037$ or $0.37\%$, not many women are being denied.

Step3: Find z - scores for new height requirements

For the shortest $1\%$ or $0.01$ area to the left, the z - score $z_{lower}$ from the standard - normal table is approximately $z_{lower}=-2.33$. For the tallest $2\%$ or $0.02$ area to the right, the area to the left is $1 - 0.02=0.98$, and the z - score $z_{upper}$ from the standard - normal table is approximately $z_{upper}=2.05$.

Step4: Calculate new height limits

Using the formula $x=\mu+z\sigma$. For the lower limit, $x_{lower}=63.9+( - 2.33)\times2.2=63.9-5.126 = 58.774\approx58.77$ inches. For the upper limit, $x_{upper}=63.9 + 2.05\times2.2=63.9+4.51=68.41$ inches.

Answer:

a. $99.63\%$
b. The new height requirements are between $58.77$ inches and $68.41$ inches.