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.8706. (round to four decimal places as needed.)
c. when redesigning the ejection seat, which probability is more relevant?
a. the part (a) probability is more relevant because the seat performance for a sample of pilots is more important.
b. the part (a) probability is more relevant because the seat performance for a single pilot is more important.
c. the part (b) probability is more relevant because the seat performance for a sample of pilots is more important.
d. the part (b) probability is more relevant because the seat performance for a single pilot is more important.

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 table \(P(-0.66<Z<0.21)=\Phi(0.21)-\Phi(-0.66)\)
\(\Phi(0.21) = 0.5832\), \(\Phi(-0.66)=1 - \Phi(0.66)=1 - 0.7454 = 0.2546\)
\(P(-0.66<Z<0.21)=0.5832-0.2546 = 0.3286\)

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 z - score is \(z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Here, \(n = 28\), \(\mu = 171\), \(\sigma = 47\), \(\bar{X}_1 = 140\) and \(\bar{X}_2 = 181\)
For \(\bar{X}=140\): \(z_1=\frac{140 - 171}{\frac{47}{\sqrt{28}}}=\frac{-31}{\frac{47}{5.2915}}\approx\frac{-31}{8.882}\approx - 3.5\)
For \(\bar{X}=181\): \(z_2=\frac{181 - 171}{\frac{47}{\sqrt{28}}}=\frac{10}{\frac{47}{5.2915}}\approx\frac{10}{8.882}\approx1.13\)
Using the standard normal table \(P(-3.5<Z<1.13)=\Phi(1.13)-\Phi(-3.5)\)
\(\Phi(1.13)=0.8708\), \(\Phi(-3.5)=1-\Phi(3.5)=1 - 0.9998 = 0.0002\)
\(P(-3.5<Z<1.13)=0.8708 - 0.0002=0.8706\)

Step3: Determine the more relevant probability

When redesigning the ejection seat, the performance for a single pilot is more important. Because the ejection seat is used for one pilot at a time.

Answer:

a. \(0.3286\)
b. \(0.8706\)
c. B. The part (a) probability is more relevant because the seat performance for a single pilot is more important.