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 □. (round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 165\), \(\sigma=41\).
For \(x = 120\):
\(z_1=\frac{120 - 165}{41}=\frac{- 45}{41}\approx - 1.0976\)
For \(x = 181\):
\(z_2=\frac{181-165}{41}=\frac{16}{41}\approx0.3902\)

Step2: Find the probabilities using the standard normal distribution table

We know that \(P(120\lt X\lt181)=P(-1.0976\lt Z\lt0.3902)\)
\(P(-1.0976\lt Z\lt0.3902)=P(Z\lt0.3902)-P(Z\lt - 1.0976)\)
From the standard normal distribution table, \(P(Z\lt0.3902)\approx0.6517\) and \(P(Z\lt - 1.0976)\approx0.1363\)

Step3: Calculate the final probability

\(P(-1.0976\lt Z\lt0.3902)=0.6517 - 0.1363=0.5154\)

Answer:

\(0.5154\)