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 130 lb and 201 lb. weights of women are now normally distributed with a mean of 169 lb and a standard deviation of 35 lb. complete parts (a) through (c) below.
a. if 1 woman is randomly selected, find the probability that her weight is between 130 lb and 201 lb.
the probability is approximately 0.6851 (round to four decimal places as needed.)
b. if 33 different women are randomly selected, find the probability that their mean weight is between 130 lb and 201 lb.
the probability is approximately 1.0000 (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 single pilot is more important.
b. the part (b) 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 (a) probability is more relevant because the seat performance for a sample of pilots is more important.

Explanation:

Step1: Recall the z - score formula for a single value

The z - score formula for a single value $x$ from a normal distribution with mean $\mu$ and standard deviation $\sigma$ is $z=\frac{x - \mu}{\sigma}$. For $\mu = 169$ and $\sigma=35$, when $x = 130$, $z_1=\frac{130 - 169}{35}=\frac{- 39}{35}\approx - 1.11$. When $x = 201$, $z_2=\frac{201 - 169}{35}=\frac{32}{35}\approx0.91$.

Step2: Find the probability for a single - selected woman

We want $P(130

Step3: Recall the z - score formula for the sample mean

The z - score formula for the sample mean $\bar{x}$ of a sample of size $n$ from a normal distribution with mean $\mu$ and standard deviation $\sigma$ is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$. Here, $n = 33$, $\mu = 169$, $\sigma = 35$. When $\bar{x}=130$, $z_1=\frac{130 - 169}{\frac{35}{\sqrt{33}}}=\frac{-39}{\frac{35}{\sqrt{33}}}\approx\frac{-39}{6.11}\approx - 6.38$. When $\bar{x}=201$, $z_2=\frac{201 - 169}{\frac{35}{\sqrt{33}}}=\frac{32}{\frac{35}{\sqrt{33}}}\approx\frac{32}{6.11}\approx5.24$.

Step4: Find the probability for the sample mean

We want $P(130<\bar{X}<201)$. Using the standard normal distribution table, $P(130<\bar{X}<201)=P(-6.38

Step5: Determine the more relevant probability

When redesigning the ejection seat, the performance for a single pilot is more important. So the probability in part (a) is more relevant because the seat is designed for each individual pilot.

Answer:

a. $0.6851$
b. $1.0000$
c. A. The part (a) probability is more relevant because the seat performance for a single pilot is more important.