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.)
Answer
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 \(\boxed{0.9993}\).
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 \(\boxed{0.9993}\).
