Question

an airliner carries 400 passengers and has doors with a height of 72 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 \(x = 72\) (door height), \(\mu=69.0\) (mean height of men), and \(\sigma = 2.8\) (standard deviation of men's heights).

$$z=\frac{72 - 69.0}{2.8}=\frac{3}{2.8}\approx1.07$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(X<72)\), which is equivalent to \(P(Z < 1.07)\) in the standard normal distribution.
Looking up the value of \(z = 1.07\) in the standard - normal table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: normalcdf(-100,1.07)), we find that \(P(Z < 1.07)=0.8577\)

Answer:

\(0.8577\)