Question

a survey found that womens heights are normally distributed with mean 63.5 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 99.38%.
(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. no, because the percentage of women who meet the height requirement is fairly small.
c. yes, because a large percentage of women are not allowed to join this branch of the military because of their height.
d. yes, because the percentage of women who meet the height requirement is fairly large.
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: Find the z - score for the shortest 1%

The area to the left of the z - score for the shortest 1% ($0.01$) is $0.01$. Looking up in the standard normal table, the z - score $z_1$ corresponding to an area of $0.01$ is approximately $z_1=- 2.33$.
Using the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 63.5$ and $\sigma = 2.2$.
We solve for $x$: $-2.33=\frac{x - 63.5}{2.2}$
Multiply both sides by $2.2$: $x-63.5=-2.33\times2.2$
$x-63.5=-5.126$
Add $63.5$ to both sides: $x = 63.5-5.126=58.4$

Step2: Find the z - score 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 in the standard normal table, the z - score $z_2$ corresponding to an area of $0.98$ is approximately $z_2 = 2.05$.
Using the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 63.5$ and $\sigma = 2.2$.
We solve for $x$: $2.05=\frac{x - 63.5}{2.2}$
Multiply both sides by $2.2$: $x-63.5=2.05\times2.2$
$x-63.5 = 4.51$
Add $63.5$ to both sides: $x=63.5 + 4.51=68.0$

Answer:

The new height requirements are at least $58.4$ in and at most $68.0$ in.