Question

an airliner carries 150 passengers and has doors with a height of 75 in. heights of men are normally distributed with a mean of 69.0 in and a standard deviation of 2.8 in. complete parts (a) through (d)
a. if a male passenger is randomly selected, find the probability that he can fit through the doorway without bending. the probability is 0.9830 (round to four decimal places as needed.)
b. if half of the 150 passengers are men, find the probability that the mean height of the 75 men is less than 75 in. the probability is (round to four decimal places as needed.)

Explanation:

Step1: Recall the Central Limit Theorem

For the sampling - distribution of the sample mean $\bar{X}$, if the population is normally distributed with mean $\mu$ and standard deviation $\sigma$, and the sample size is $n$, then $\bar{X}$ is normally distributed with mean $\mu_{\bar{X}}=\mu$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$. Here, $\mu = 69.0$ in, $\sigma = 2.8$ in, and $n = 75$.

Step2: Calculate the standard deviation of the sample - mean

$\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{2.8}{\sqrt{75}}\approx\frac{2.8}{8.66}\approx0.3233$

Step3: Calculate the z - score

The z - score is calculated using the formula $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$. We want to find $P(\bar{X}<75)$, so $z=\frac{75 - 69.0}{0.3233}=\frac{6}{0.3233}\approx18.56$

Step4: Find the probability

Since the standard normal distribution table (z - table) typically goes up to a z - score of around 3.49, and for a z - score of 18.56, the probability $P(Z < 18.56)\approx1.0000$

Answer:

$1.0000$