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### Given Information:
- The airliner has doors with a height of 72 inches.
- Heights of men are normally distributed with:
  - Mean (\(\mu\)) = 69.0 inches
  - Standard deviation (\(\sigma\)) = 2.8 inches
- There are 400 passengers, and half of them (200) are men.

### Part (a): Probability that a randomly selected male can fit through the doorway without bending.
We need to find \( P(X < 72) \), where \( X \) is a normally distributed variable with mean 69.0 and standard deviation 2.8.

1. **Calculate the Z-score:**
   \[
   Z = \frac{X - \mu}{\sigma} = \frac{72 - 69.0}{2.8} = \frac{3.0}{2.8} \approx 1.07
   \]

2. **Find the probability using the Z-table:**
   - The Z-score of 1.07 corresponds to a cumulative probability of approximately 0.8577.
   - Therefore, the probability that a randomly selected male can fit through the doorway without bending is about **0.8577**.

### Part (b): Probability that the mean height of 200 men is less than 72 inches.
For the mean of a sample, the standard deviation (standard error) is calculated as:
   \[
   \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{2.8}{\sqrt{200}} \approx 0.198
   \]

1. **Calculate the Z-score for the sample mean:**
   \[
   Z = \frac{72 - 69.0}{0.198} \approx 15.15
   \]

2. **Find the probability:**
   - A Z-score of 15.15 is extremely high, indicating a probability very close to 1.
   - Thus, the probability that the mean height of the 200 men is less than 72 inches is essentially **1.0000**.

### Part (c): More relevant result for passenger comfort and safety.
- **Option A** is more relevant because it considers the proportion of flights where the mean height of male passengers will be less than the door height, which relates directly to overall comfort during boarding.

### Part (d): Why women are ignored in this case.
- **Option C** is appropriate because men tend to be taller than women on average, so ensuring a design suitable for most men will also likely accommodate a larger proportion of women.

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Would you like more detailed explanations or have any questions?

### Related Questions:
1. What is the probability that a randomly selected female passenger would fit through the door, assuming a different mean and standard deviation?
2. How does changing the door height to 70 inches affect the probabilities calculated in parts (a) and (b)?
3. How would these probabilities change if the standard deviation of men's heights increased?
4. What is the probability that at least one of the 200 men is taller than the doorway?
5. How does using a different sample size affect the standard error of the mean?

### Tip:
When working with normal distributions, converting values to Z-scores helps standardize the data, making it easier to find probabilities using statistical tables.

An airliner carries 400 passengers and has doors with a height of 72 inches. Heights of men are normally distributed with a mean of 69 inches and a standard deviation of 2.8 inches. (a) Find the probability that a randomly selected male can fit through the doorway without bending. (b) If half of the passengers are men, find the probability that the mean height of the 200 men is less than 72 inches. (c) Which result is more relevant for passenger comfort and safety? (d) Why are women ignored in this case?

Learn how to calculate probabilities for an airliner carrying passengers with normally distributed heights. This problem focuses on the Z-score formula, standard error, and understanding the relevance of probability in ensuring passenger safety, particularly for men fitting through 72-inch doors. Suitable for college students and advanced high school learners.

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