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 □ (round to four decimal places as needed.)

Explanation:

Step1: Recall the Central - Limit Theorem for sample means

The mean of the sampling distribution of the sample mean $\mu_{\bar{X}}=\mu$ and the standard deviation of the sampling distribution of the sample mean $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$, where $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Given $\mu = 169$, $\sigma=35$, and $n = 33$. So, $\mu_{\bar{X}}=169$ and $\sigma_{\bar{X}}=\frac{35}{\sqrt{33}}\approx6.077$.

Step2: Calculate the z - scores

The z - score formula is $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$. For $\bar{x}=130$, $z_1=\frac{130 - 169}{6.077}=\frac{- 39}{6.077}\approx - 6.42$. For $\bar{x}=201$, $z_2=\frac{201-169}{6.077}=\frac{32}{6.077}\approx5.27$.

Step3: Find the probability

We want to find $P(130<\bar{X}<201)$, which is equivalent to $P(-6.42

Answer:

$1.0000$