A survey found that women's heights are normally distributed with mean 63.9 in. and standard deviation 3.1 in. The survey also found that men's heights are normally distributed with mean 67.7 in. and standard deviation 3.8 in. Consider an executive jet that seats six with a doorway height of 56 in. Complete parts (a) through (c) below. a. What percentage of adult men can fit through the door without bending? The percentage of men who can fit without bending is (Round to two decimal places as needed.)

To find the percentage of adult men who can fit through the 56-inch doorway without bending, we can use the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Where: - \( X \) is the height of the doorway (56 in.) - \( \mu \) is the mean height of men (67.7 in.) - \( \sigma \) is the standard deviation for men's height (3.8 in.) Calculating the Z-score: \[ Z = \frac{56 - 67.7}{3.8} = \frac{-11.7}{3.8} \approx -3.08 \] Next, we can use the Z-table to find the percentage of men below this Z-score. A Z-score of -3.08 is extremely low, and typically, we find that this corresponds to a percent very close to 0% of men fitting through the door. Thus, rounding to two decimal places, the percentage of adult men who can fit through the doorway without bending is approximately **0.00%.**