Question

an airliner carries 300 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 .9838. (round to four decimal places as needed.)
b. if half of the 300 passengers are men, find the probability that the mean height of the 150 men is less than 75 in. the probability is 1. (round to four decimal places as needed.)
c. when considering the comfort and safety of passengers, which result is more relevant: the probability from part (a) or the probability from part (b)? why?
o a. the probability from part (a) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.
o b. the probability from part (a) is more relevant because it shows the proportion of male passengers that will not need to bend.
o c. the probability from part (b) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.
o d. the probability from part (b) is more relevant because it shows the proportion of male passengers that will not need to bend.

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 75\), \(\mu=69.0\), and \(\sigma = 2.8\).

$$z=\frac{75 - 69.0}{2.8}=\frac{6}{2.8}\approx2.1429$$

Step2: Find the probability for part (a)

Using the standard normal distribution table, \(P(Z<2.1429)\). Looking up the value in the table, \(P(Z < 2.14)=0.9838\) (this is given as the probability for part (a)).

Step3: Calculate the mean and standard deviation for the sample mean in part (b)

For a sample of size \(n = 150\), the mean of the sample means \(\mu_{\bar{x}}=\mu = 69.0\), and the standard deviation of the sample means \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{2.8}{\sqrt{150}}\approx\frac{2.8}{12.2474}\approx0.2286\)

Step4: Calculate the z - score for the sample mean in part (b)

Using the formula \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\), with \(\bar{x}=75\), \(\mu_{\bar{x}} = 69.0\), and \(\sigma_{\bar{x}}\approx0.2286\)

$$z=\frac{75 - 69.0}{0.2286}=\frac{6}{0.2286}\approx26.24$$

Step5: Find the probability for part (b)

Using the standard normal distribution table, \(P(Z < 26.24)\approx1\) (since for large z - values, the probability approaches 1)

Step6: Analyze part (c)

When considering the comfort and safety of passengers, the probability from part (b) is more relevant. Because part (b) gives the probability about the mean height of a group of passengers. If the mean height of a group of passengers (in this case, half of the 300 passengers, so \(n = 150\)) is less than the door height, it is related to the overall situation of multiple passengers' comfort and safety in a flight. While part (a) is about a single passenger.

Answer:

a. \(0.9838\)
b. \(1\)
c. B. The probability from part (b) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.