Question

an airliner carries 200 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 formula for the z - score 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).

$$z=\frac{75 - 69.0}{2.8}=\frac{6}{2.8}\approx2.14$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(X\leq75)\), which is equivalent to \(P(Z\leq2.14)\) using the standard normal distribution table or a calculator with a normal - distribution function.
Using a standard normal table or a calculator (e.g., in Excel: NORM.S.DIST(2.14, TRUE)), we find that \(P(Z\leq2.14)\approx0.9838\)

Answer:

\(0.9838\)