An airliner carries 300 passengers and has doors with a height of 70 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 4 of 1 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 decimal places as needed)

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Below is the step-by-step solution:

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### Step 1. Define the Random Variable

Let $$ X $$ represent the height (in inches) of a male passenger, which is normally distributed with mean
$$
\mu = 69.0 \text{ in}
$$
and standard deviation
$$
\sigma = 2.8 \text{ in}.
$$
Thus, we write:
$$
X \sim N(69.0,\, 2.8^2).
$$

---

### Step 2. Formulate the Probability Statement

A male passenger can fit through the doorway without bending if his height is at most $$ 70 $$ in. The probability we want to compute is:
$$
P(X \le 70).
$$

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### Step 3. Standardize the Random Variable

We standardize $$ X $$ to the standard normal variable $$ Z $$ using the formula:
$$
Z = \frac{X - \mu}{\sigma}.
$$

For $$ X = 70 $$, we have:
$$
z = \frac{70 - 69.0}{2.8} = \frac{1.0}{2.8} \approx 0.3571.
$$

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### Step 4. Find the Probability Using the Standard Normal Distribution

The probability that a randomly selected male can fit through the doorway is:
$$
P(X \le 70) = P(Z \le 0.3571).
$$

Using a standard normal distribution table or a calculator, we find:
$$
P(Z \le 0.3571) \approx 0.6394.
$$

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### Final Answer

The probability that a randomly selected male passenger can fit through the doorway without bending is approximately:
$$
\boxed{0.6394}.
$$
The probability that a randomly selected male passenger can fit through the doorway without bending is approximately 0.6394.

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