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Part 1

An airliner carries 

5050 

passengers and has doors with a height of 

7878 

in. Heights of men are normally distributed with a mean of 

69.069.0 

in and a standard deviation of 

2.82.8 

in. Complete parts​ (a) through​ (d).

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Part 1

**a.** If a male passenger is randomly​ selected, find the probability that he can fit through the doorway without bending.

The probability is 

enter your response here.

​(Round to four decimal places as​ needed.) 




We are tasked with finding the probability that a randomly selected male passenger can fit through the doorway of an airliner without bending. This can be modeled using a normal distribution.

### Given dataWe are tasked with finding the probability that a randomly selected male passenger can fit through the doorway of an airliner without bending. This can be modeled using a normal distribution.

### Given 

- Mean height of men: \(\mu = 69.0 \text{ inches}\)
- Standard deviation of heights: \(\sigma = 2.8 \text{ inches}\)
- Height of the door: \(78 \text{ inches}\)

The problem is asking for the probability that the height of a randomly selected male passenger is less than or equal to \(78\) inches, which corresponds to finding the cumulative probability in a normal distribution.

### Step 1: Standardize the height
We can standardize the value of 78 inches using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \(X = 78 \text{ inches}\) (the height of the door),
- \(\mu = 69.0 \text{ inches}\) (mean height of men),
- \(\sigma = 2.8 \text{ inches}\) (standard deviation).

Substituting the values:
\[
Z = \frac{78 - 69.0}{2.8} = \frac{9}{2.8} \approx 3.21
\]

### Step 2: Find the cumulative probability
Using the Z-score table (or a normal distribution calculator), a Z-score of \(3.21\) corresponds to a cumulative probability of approximately 0.9993.

Thus, the probability that a randomly selected male passenger can fit through the doorway without bending is:

\[
\boxed{0.9993}
\]

This means there is a 99.93% chance that a male passenger can fit through the door without bending.

---

Would you like to explore more details or have any other questions?  
Here are 5 related questions to deepen your understanding:

1. How do we compute Z-scores for values below the mean?
2. What is the significance of the Z-score in the context of normal distribution?
3. How would the result change if the doorway height were lower, say 75 inches?
4. Can we apply the same method for female passengers if given their mean height and standard deviation?
5. How does increasing the standard deviation affect the probability in a normal distribution?

**Tip:** Always use the Z-score formula to standardize values before looking them up in a normal distribution table.

An airliner carries 5050 passengers and has doors with a height of 7878 in. Heights of men are normally distributed with a mean of 69.0 in and a standard deviation of 2.8 in. Find the probability that a randomly selected male passenger can fit through the doorway without bending.

Calculate the probability that a randomly selected male passenger can fit through an airliner doorway of 78 inches without bending. Given that men's heights are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches, the Z-score formula is used to solve this problem. Suitable for Grades 10-12.

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