Question

an airliner carries 250 passengers and has doors with a height of 76 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 \(x = 76\) (door height), \(\mu=69.0\) (mean height of men), and \(\sigma = 2.8\) (standard deviation of men's heights).

$$z=\frac{76 - 69.0}{2.8}=\frac{7}{2.8}=2.5$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(X<76)\), which is equivalent to \(P(Z < 2.5)\) in the standard normal distribution (\(X\) is the height of men, \(Z\) is the standard normal variable).
Looking up the value of \(P(Z < 2.5)\) in the standard - normal table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: normalcdf(-100,2.5)), we get \(P(Z < 2.5)=0.9938\)

Answer:

\(0.9938\)