Question

when women were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. the ejection seats were designed for men weighing between 140 lb and 181 lb. weights of women are now normally distributed with a mean of 171 lb and a standard deviation of 47 lb. complete parts (a) through (c) below.
a. if 1 woman is randomly selected, find the probability that her weight is between 140 lb and 181 lb.
the probability is approximately .3286. (round to four decimal places as needed.)
b. if 28 different women are randomly selected, find the probability that their mean weight is between 140 lb and 181 lb.
the probability is approximately . (round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - scores for part (a)

For a single - value \(x\) from a normal distribution \(N(\mu,\sigma)\), the z - score is \(z=\frac{x - \mu}{\sigma}\). Here, \(\mu = 171\), \(\sigma=47\), \(x_1 = 140\) and \(x_2 = 181\).
For \(x = 140\): \(z_1=\frac{140 - 171}{47}=\frac{- 31}{47}\approx - 0.66\)
For \(x = 181\): \(z_2=\frac{181 - 171}{47}=\frac{10}{47}\approx0.21\)
Using the standard normal distribution table \(P(Z\lt z_1)\) and \(P(Z\lt z_2)\), \(P(-0.66\lt Z\lt0.21)=P(Z\lt0.21)-P(Z\lt - 0.66)\)
From the standard normal table, \(P(Z\lt0.21) = 0.5832\), \(P(Z\lt - 0.66)=0.2546\)
\(P(-0.66\lt Z\lt0.21)=0.5832 - 0.2546=0.3286\)

Step2: Calculate the z - scores for part (b)

For the sample mean \(\bar{x}\) of a sample of size \(n\) from a normal distribution \(N(\mu,\sigma)\), the sampling distribution of \(\bar{x}\) is \(N(\mu,\frac{\sigma}{\sqrt{n}})\). Here, \(n = 28\), \(\mu = 171\), \(\sigma = 47\), so \(\sigma_{\bar{x}}=\frac{47}{\sqrt{28}}\approx\frac{47}{5.2915}\approx8.88\)
For \(\bar{x}_1 = 140\): \(z_1=\frac{140 - 171}{8.88}=\frac{-31}{8.88}\approx - 3.5\)
For \(\bar{x}_2 = 181\): \(z_2=\frac{181 - 171}{8.88}=\frac{10}{8.88}\approx1.13\)
Using the standard normal distribution table \(P(-3.5\lt Z\lt1.13)=P(Z\lt1.13)-P(Z\lt - 3.5)\)
From the standard normal table, \(P(Z\lt1.13)=0.8708\), \(P(Z\lt - 3.5)=0.0002\)
\(P(-3.5\lt Z\lt1.13)=0.8708 - 0.0002=0.8706\)

Answer:

a. \(0.3286\)
b. \(0.8706\)