A survey found that women's heights are normally distributed with mean $63.5 \mathrm{in}$. and standard deviation $2.6 \mathrm{in}$. The survey also found that men's heights are normally distributed with mean 69.8 in. and standard deviation 3.1 in. Consider an executive jet that seats six with a doorway height of 55.8 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.)

Step 1 ：Given that the mean height of men is 69.8 inches and the standard deviation is 3.1 inches, we need to find the percentage of men who can fit through a door that is 55.8 inches tall without bending.

Step 2 ：To do this, we first calculate the z-score, which is a measure of how many standard deviations an element is from the mean. In this case, we want to find out how many standard deviations below the mean the height of the doorway is for men.

Step 3 ：The formula for the z-score is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are comparing to the mean (the height of the door), \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 4 ：Substituting the given values into the formula, we get \(Z = \frac{55.8 - 69.8}{3.1} = -4.516129032258064\).

Step 5 ：This means that the height of the door is approximately 4.52 standard deviations below the mean height of men.

Step 6 ：We then use a z-table or a statistical function to find the percentage of men who are shorter than the doorway. The z-score of -4.52 corresponds to a percentage of approximately 0.0003149014862667157, or 0.0003% when rounded to four decimal places.

Step 7 ：Thus, the percentage of adult men who can fit through the door without bending is extremely low, less than 0.001%. This is because the height of the door is more than 4 standard deviations below the mean height of men, which is quite rare in a normal distribution.

Step 8 ：Final Answer: The percentage of adult men who can fit through the door without bending is approximately \(\boxed{0.0003\%}\).