Question

(30 points) The MARK VI monorail at Disney World has doors with a height of 72 inches. Heights of men are normally distributed with a mean of 68.6 in. and a standard deviation of 2.8 in. 4. a. What percentage of adult men can fit through the door without bending? Explain. b. If a car is loaded with 60 randomly selected men, what is the probability that their mean height is less than 72 in.? Explain.

Answer 1

Q.4) The MARK VI monorail at Disney world has doors with a height of 72 inches.

For men height : mean $(\mu)$ = 68.6 in.

standard deviation $(\sigma)$ = 2.8 in.

a) We want to find, the percentage of adults men can fit through the door without bending. That means we want to find, the percentage of adults men having height 72 inches or less.

$P(X \leq 72) = P(\frac {X -\mu}{\sigma} \leq \frac {72 -68.6}{2.8}) = P(Z \leq 1.21) = 0.8869$

In percentage : 0.8869 * 100% = 88.69%

Therefore, required percentage is 88.69%

b) sample size ( n ) = 60

We want to find, $P(\bar X \leq 72)$ $P(\bar X \leq 72) = P(\frac {\bar X -\mu}{\frac {\sigma}{\sqrt n }} \leq \frac {72 -68.6}{\frac {2.8}{\sqrt {60}}} ) = P(Z \leq 9.41 ) = 1.0000$

Therefore, required probability is 1