I am having trouble with this question, please help.

Question

The height of an adult male is known to be normally distributed with a mean of 69 inches and a standard deviation of 2.5 inches. How high should a doorway be so that 99.5% of the adult males can pass through it without having to bend? (show work with z scores)

Answer 1

Ok, so we have to look at the height of the males, which is 69 inches, with a standard deviation of 2.5 inches. So with that being known, we would have to do some work with the Z score. So first:

We have to find the 99.5 percentile of the Z scores. So looking at the table, we find that the z-score is 2.58.

So we know that Z=x-u/sd, so we can just plug in the numbers. So now we have 2.58=x-69/2.5

6.45=x-69

75.45=x

So the door would have to be 75.45 inches tall so that 99.5% of men can fit through the door without bending.

Answer 2

Hello, thank you for taking the time to post your question!

The underlying relationship that you want to use here in the context of a normal distribution calculation is that

Z = (x – mean) / (standard deviation)

We’re given the mean and standard deviation values directly, so that makes it

Z = (x – 69) / (2.5)

The variable “x” is what we’re trying to solve for

“Z” we can actually locate based on the percentage … so it’s saying that 99.5% of adult males can pass through it … that percentage corresponds to a Z score of 2.58. so the full equation here is

2.58 = (x – 69) / 2.5

2.58(2.5) = x – 69

x = 2.58(2.5) + 69

x = 75.45

so the doorway should be 75.45 inches high so that 99.5% of adult males can pass through without bending

I hope that helps get you moving in the right direction! Feel free to reach out if you have any additional questions beyond that :)