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 □.
(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 table

We want to find \(P(X < 75)\), which is equivalent to \(P(Z<2.14)\) in the standard normal distribution (\(X\) is the height of men, \(Z\) is the standard normal variable).
Looking up the value of \(z = 2.14\) in the standard - normal distribution table (the cumulative probability for \(z\) values), we find that \(P(Z < 2.14)=0.9838\)

Answer:

\(0.9838\)