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

Men's heights are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches.

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Just as in another, similar question you asked, let's start by filling a gap in the problem. The standard door in America has the following dimensions: 36" x 80" (width x height). In other words, in order for a woman to get through a door without trouble, she would need to be maximally 80" tall.

Now, we can recognize that in order for a man to not bend over as he goes through a door, he must be 80" tall or less. To figure out the probability that a man is shorter than 80" tall, we will use the population mean and standard deviation:

`mu = 69.0`

`sigma = 2.8`

Here, `mu` is the mean for men's height, and `sigma` ` ` ` ` is the standard deviation. Now, we can calculate the "z-score" to determine what percentage of men fall below the 80" mark for height.

We use the following equation:

`z = (h-mu)/sigma`

Here, `h` is the height that we are testing, which is 80". We can now substitute numbers in for each variable in this equation, allowing us to calculate `z` :

`z = (80 - 69)/2.8 = 11/2.8 = 3.93`

Thus, we are 3.93 standard deviations above the mean. You can use a computer or z-score table to determine that the probability of having a height with a z-score less than 3.93 is 0.999958.

In other words, **the probability that a man can fit through a door without bending over is 0.999958.**

**Sources:**

You have given that the men's heights are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. But you have not given the height of the door which has been assumed to be 80 inches. Your question goes : "What percentage of adult men can fit through *the* door without bending over?" implying that the height of the door is given. I am sure in any statistics test, the examiner will not expexct you to know the height of a standard American door. Check on the question you have posted.

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