Question

an airliner carries 300 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.9838. (round to four decimal places as needed.)
b. if half of the 300 passengers are men, find the probability that the mean height of the 150 men is less than 75 in. the probability is ■. (round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard deviation for 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 $\sigma = 2.8$ (population standard deviation) and $n = 150$ (sample size).
$$\sigma_{\bar{x}}=\frac{2.8}{\sqrt{150}}\approx\frac{2.8}{12.25}\approx0.2286$$

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}=75$, $\mu = 69.0$.
$$z=\frac{75 - 69}{0.2286}=\frac{6}{0.2286}\approx26.25$$

Step3: Find the probability

Using the standard normal distribution table (or a calculator with a normal - distribution function), for $z = 26.25$, the probability $P(Z<26.25)\approx1.0000$

Answer:

$1.0000$