Questions: An airliner carries 50 passengers and has doors with a height of 76 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 (Round to four

An airliner carries 50 passengers and has doors with a height of 76 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 (Round to four
Transcript text: An airliner carries 50 passengers and has doors with a height of 76 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 $\square$ (Round to four
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Solution

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Solution Steps

Step 1: Define the Problem

We need to find the probability that a randomly selected male passenger can fit through a doorway with a height of \(76\) inches. The heights of men are normally distributed with a mean (\(\mu\)) of \(69.0\) inches and a standard deviation (\(\sigma\)) of \(2.8\) inches.

Step 2: Set Up the Normal Distribution

The height of a randomly selected male passenger follows the normal distribution \(X \sim N(\mu, \sigma^2)\), where:

  • \(\mu = 69.0\)
  • \(\sigma = 2.8\)

We want to find the probability \(P(X \leq 76)\).

Step 3: Calculate the Z-Score

To find this probability, we first calculate the Z-score for the upper limit of the doorway height:

\[ Z = \frac{X - \mu}{\sigma} = \frac{76 - 69.0}{2.8} = \frac{7.0}{2.8} \approx 2.5 \]

Step 4: Find the Probability

Using the Z-score, we can find the probability:

\[ P(X \leq 76) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(2.5) - \Phi(-\infty) \]

Since \(\Phi(-\infty) = 0\), we have:

\[ P(X \leq 76) = \Phi(2.5) \approx 0.9938 \]

Final Answer

The probability that a randomly selected male passenger can fit through the doorway without bending is approximately \(0.9938\).

Thus, the final answer is:

\[ \boxed{0.9938} \]

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