An airliner carries 50 passengers and has doors with a height of 78 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 decimal places as needed.)

Step 1 ：The problem is asking for the probability that a randomly selected male passenger can fit through the doorway without bending. This is a problem of normal distribution. The height of the door is 78 inches, and we know that the mean height of men is 69.0 inches with a standard deviation of 2.8 inches. We need to find the probability that a randomly selected man's height is less than or equal to 78 inches.

Step 2 ：To solve this, we can use the z-score formula to standardize the height of the door. The z-score is calculated as \((X - μ) / σ\), where X is the value we're interested in (the height of the door), μ is the mean, and σ is the standard deviation.

Step 3 ：Substituting the given values into the z-score formula, we get \(z = (78 - 69.0) / 2.8 = 3.2142857142857144\).

Step 4 ：Once we have the z-score, we can use a z-table or a statistical function to find the probability that a randomly selected value from a normal distribution is less than or equal to that z-score. The probability corresponding to a z-score of 3.2142857142857144 is approximately 0.9993461525959454.

Step 5 ：This means that almost all male passengers can fit through the doorway without bending. Therefore, the probability that a randomly selected male passenger can fit through the doorway without bending is approximately \(\boxed{0.9993}\).