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 28 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.)

To determine the probability that a randomly selected male passenger can fit through the doorway (height of 70 inches) without bending, we need to find the proportion of the male height distribution that is less than or equal to 70 inches. Given that the heights of men are normally distributed with a mean (μ) of 69.0 inches and a standard deviation (σ) of 28 inches, we can standardize the height using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where: - \( X \) is the height of the doorway (70 inches) - \( \mu \) is the mean height (69 inches) - \( \sigma \) is the standard deviation (28 inches) Calculating the Z-score: \[ Z = \frac{70 - 69}{28} = \frac{1}{28} \approx 0.0357 \] Next, we can use the standard normal distribution table or a calculator to find the probability associated with this Z-score. Using a standard normal distribution table or a calculator: The probability (P) that a randomly selected male passenger can pass through the doorway without bending is approximately: P(Z ≤ 0.0357) ≈ 0.5145 Thus, the probability that a randomly selected male passenger can fit through the doorway without bending is approximately **0.5145** (rounded to four decimal places).