# What percentage of adult men can fit through the doors without bending over?Men's heights are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. The mark vi...

What percentage of adult men can fit through the doors without bending over?

Men's heights are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. The mark vi monorail used at disney world have doors with a height of 72''.

The height of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. The mark vi monorail used at Disney world has doors with a height of 72 inches.

Men would not have to bend over if their height is less than 72 inches.

Given the mean height of men as 69 inches and the standard deviation of 2.8 inches, the z-score for x = 72 is `(x - mu)/(sigma) = (72 - 69)/2.8 = 1.07`

From a normal distribution table the area corresponding to this value is 0.85769.

**This gives the percentage of men that can enter without bending over as 85.769%**

This question has been answered here: https://www.enotes.com/math/q-and-a/what-percentage-adult-women-can-fit-through-341141

As the mean and standard deviation of the height of only men was provided in both the questions, it was not possible to estimate the percentage of women that can fit the door without bending over and the other question was edited. The mean and standard deviation of the height of women would be required to determine the percentage of women that can fit the door without bending over.