Questions: An airliner carries 50 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 9839. (Round to four decimal places as needed.) b. If half of the 50 passengers are men, find the probability that the mean height of the 25 men is less than 75 in. The probability is . (Round to four decimal places as needed.)

Solution Steps

To determine if a randomly selected male passenger can fit through the doorway without bending, we calculate the Z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \(X = 75\) in (height of the doorway),
• \(\mu = 69.0\) in (mean height of men),
• \(\sigma = 2.8\) in (standard deviation of men's height).

Substituting the values, we have:

\[ z = \frac{75 - 69.0}{2.8} = 2.1429 \]

Next, we find the probability that a single male passenger can fit through the doorway. This is given by:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(2.1429) - \Phi(-\infty) \]

Using the cumulative distribution function \( \Phi \), we find:

\[ P = 0.9839 \]

Now, we calculate the probability that the mean height of 25 men is less than 75 inches. The Z-score for the sample mean is calculated as follows:

\[ z = \frac{X - \mu}{\sigma / \sqrt{n}} = \frac{75 - 69.0}{2.8 / \sqrt{25}} = \frac{75 - 69.0}{0.56} = 10.7143 \]

Then, we find the probability:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(10.7143) - \Phi(-\infty) \]

Since \( \Phi(10.7143) \) approaches 1, we have:

\[ P = 1.0 \]

Final Answer