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 0.8577
(round to four decimal places as needed.)
b. if half of the 400 passengers are men, find the probability that the mean height of the 200 men is less than 72 in.
the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 2.8$ in and $n = 200$, then $\sigma_{\bar{x}}=\frac{2.8}{\sqrt{200}}\approx\frac{2.8}{14.1421}\approx0.1979$.

Step2: Calculate the z - score

The z - score formula for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$.
Here, $\bar{x} = 72$ in, $\mu=69.0$ in, and $\sigma_{\bar{x}}\approx0.1979$. So $z=\frac{72 - 69}{0.1979}=\frac{3}{0.1979}\approx15.16$.

Step3: Find the probability

Using the standard normal distribution table (or a calculator with a normal - distribution function, such as the normalcdf function on a TI - 84: normalcdf(-1000,15.16)), the probability $P(Z < 15.16)\approx1.0000$.

Answer:

$1.0000$