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
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 75$ (door height), $\mu=69.0$ (mean height of men), and $\sigma = 2.8$ (standard deviation of men's heights). So $z=\frac{75 - 69.0}{2.8}=\frac{6}{2.8}\approx2.1429$.

Step2: Find the probability

We want to find $P(X<75)$, which is equivalent to $P(Z < 2.1429)$ using the standard normal distribution. Looking up the value in the standard - normal table or using a calculator with a normal distribution function, we find that $P(Z < 2.1429)\approx0.9839$.

Answer:

$0.9839$