Questions: An airliner carries 350 passengers and has doors with a height of 74 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 0.9629 . (Round to four decimal places as needed.) b. If half of the 350 passengers are men, find the probability that the mean height of the 175 men is less than 74 in. The probability is (Round to four decimal places as needed.)

Solution Steps

To determine the probability that a randomly selected male passenger can fit through the doorway without bending, we first calculate the Z-score for the door height of 74 inches using the formula:

\[ z = \frac{X - \mu}{\sigma} = \frac{74 - 69.0}{2.8} = 1.7857 \]

Next, we find the probability that a randomly selected male passenger has a height less than 74 inches. This is given by the cumulative distribution function (CDF) of the standard normal distribution:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.7857) - \Phi(-\infty) = 0.9629 \]

Now, we calculate the probability that the mean height of 175 men is less than 74 inches. Since we are dealing with a sample mean, we use the standard error of the mean, which is calculated as:

\[ \sigma_{mean} = \frac{\sigma}{\sqrt{n}} = \frac{2.8}{\sqrt{175}} \approx 0.2111 \]

However, since the Z-score for the mean height is significantly high, we find:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(23.6228) - \Phi(-\infty) \approx 1.0 \]

Final Answer