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 120 lb and 181 lb. weights of women are now normally distributed with a mean of 165 lb and a standard deviation of 41 lb. complete parts (a) through (c) below.
a. if 1 woman is randomly selected, find the probability that her weight is between 120 lb and 181 lb.
the probability is approximately 0.5154. (round to four decimal places as needed.)
b. if 35 different women are randomly selected, find the probability that their mean weight is between 120 lb and 181 lb.
the probability is approximately \square. (round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard deviation of the sample mean

The formula for the standard deviation of the sample mean ($\sigma_{\bar{x}}$) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 41$ (population standard deviation) and $n = 35$ (sample size).
$\sigma_{\bar{x}}=\frac{41}{\sqrt{35}}\approx\frac{41}{5.916}\approx6.93$

Step2: Calculate the z - scores

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\mu = 165$ (population mean).
For $\bar{x}=120$:
$z_1=\frac{120 - 165}{6.93}=\frac{- 45}{6.93}\approx - 6.49$
For $\bar{x}=181$:
$z_2=\frac{181 - 165}{6.93}=\frac{16}{6.93}\approx2.31$

Step3: Find the probability using the standard normal distribution

We want to find $P(-6.49Looking up in the standard normal table, $P(Z < 2.31)=0.9896$

So $P(-6.49

Answer:

$0.9896$