Question

need help with both (a) and (b) please!

A survey found that women's heights are normally distributed with mean 63.8 in and standard deviation 2.4 in. 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? 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? 0 A. 0 B. O C. O D. No, because only a small percentage of women are not allowed to join this branch of the military because of their height. Yes, because a large percentage of women are not allowed to join this branch of the military because of their height. Yes, because the percentage of women who meet the height requirement is fairly large. No, because the percentage of women who meet the height requirement is fairly small. in and at mostin. b. For the new height requirements, this branch of the military requires women's heights to be at least Round to one decimal place as needed.)

Answer 1

Let the RV height be represented by .

a) The required probability is

$P\left ( 58<X<80 \right )=P\left ( \frac{58-63.8}{2.4}<Z<\frac{80-63.8}{2.4} \right )\\ P\left ( 58<X<80 \right )=P\left ( -2.417<Z<6.75 \right )\\ P\left ( 58<X<80 \right )=0.9841$

The percentage of women who meet the height requirement is ${\color{Blue} 98.41\%}$ .

b)

We need the lowest height c

$P\left ( X<c \right )=0.01\\ P\left ( \frac{X-63.8}{2.4}<\frac{c-63.8}{2.4} \right )=0.01\\ P\left ( Z<\frac{c-63.8}{2.4} \right )=0.01\\ \Phi \left ( \frac{c-63.8}{2.4} \right )=0.01\\ \frac{c-63.8}{2.4} =\Phi ^{-1}\left (0.01 \right )\\ \frac{c-63.8}{2.4}=-2.327\\ c=58.2152$

We need the largest height c

$P\left ( X>c \right )=0.02\\ P\left ( \frac{X-63.8}{2.4}>\frac{c-63.8}{2.4} \right )=0.02\\ P\left ( Z>\frac{c-63.8}{2.4} \right )=0.02\\ \Phi \left ( \frac{c-63.8}{2.4} \right )=1-0.02\\ \ \frac{c-63.8}{2.4} =\Phi ^{-1}\left (0.98 \right )\\ \frac{c-63.8}{2.4}=2.054\\ c=68.7296$

The women height should be at least ${\color{Blue} 58.21 \textup{ in}}$ and at most ${\color{Blue} 68.73 \textup{ in}}$ .