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an airliner carries 300 passengers and has doors with a height of 75 in…

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?
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.
b. the probability from part (a) is more relevant because it shows the proportion of male passengers that will not need to bend.
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.
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 (b)

The formula for the z - score of the sample mean \(\bar{x}\) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\mu = 69.0\), \(\sigma = 2.8\), \(n = 150\), and \(\bar{x}=75\).

$$ z=\frac{75 - 69.0}{\frac{2.8}{\sqrt{150}}}=\frac{6}{\frac{2.8}{12.2474}}\approx\frac{6}{0.2286}\approx26.25 $$

Step2: Find the probability using the z - score

Using the standard normal distribution table or a calculator, \(P(Z\lt26.25)\approx1\) (since for a standard normal distribution \(Z\), when \(z\) is very large, the cumulative probability approaches 1).

Answer:

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.