Question

a survey found that womens heights are normally distributed with mean 63.8 in and standard deviation 2.3 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 58 in and 80 in

For \(x = 58\), the z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 63.8\) and \(\sigma=2.3\). So \(z_1=\frac{58 - 63.8}{2.3}=\frac{- 5.8}{2.3}\approx - 2.52\). For \(x = 80\), \(z_2=\frac{80 - 63.8}{2.3}=\frac{16.2}{2.3}\approx7.04\).

Step2: Use the standard normal distribution table

The area to the left of \(z_1=-2.52\) from the standard - normal table is \(P(Z < - 2.52)=0.0059\). The area to the left of \(z_2 = 7.04\) is approximately \(1\) (since for \(z\) values greater than \(3.49\), the area to the left is very close to \(1\)). The percentage of women meeting the height requirement is \(P(-2.52<Z<7.04)=P(Z < 7.04)-P(Z < - 2.52)=1 - 0.0059 = 0.9941\) or \(99.41\%\).

Answer:

\(99.41\)