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We require 60% of the population distribution of men to fit under the door without having to stoop. Hence we need to match the 60% point of the standard Normal (Gaussian) to the equivalent point on the distribution of men's heights in the population.
We are told that men's heights in the population are distributed approximately Normal(mean = 69.5in, sd = 2.4in).
The 60% point on this distribution of men's heights should match the 60% point on the standard Normal, or Normal(0,1) distribution.
First we convert a general point X on the men's height distribution to a standard Normal variate Z by using the formula
Z = (X - mean(X))/sd(X)
where we can take mean(X) to be 69.5in as given and sd(X) to be 2.4in as given.
Therefore we have
Z = (X - 69.5)/2.4
Now, the 60% point of the Normal(0,1) distribution is Z(0.6) = 0.2533471 (from statistical tables of the standard Normal distribution or, more accurately, software such as the free package called 'R').
Hence the particular X point we want (call it x) corresponds to z = 0.2533471 so that
(x - 69.5)/2.4 = 0.2533471
x = 70.10803 = 70.11 to 2dp
So we require the door height to be 70.11 inches (to 2dp) for 60% of the men in the population to walk through it without having to stoop.
Around 73 inches
You want to find the area which is 60% and less on a density curve. The simplest way to do this would be to use your calculator for this. If you have a calculator that lets you find the number.
For me, it is invnorm(0.6,69.5,2.4). The first number is the percent in a decimal. The second number is the mean and the third number is the standard deviation.
This yields 70.1 inches for the doorway height would be needed.
Another way would be look on your z-score chart for the value closest to 0.600. This value on the chart is 0.59843, which corresponds to the z value of 0.24.
Set up an equation now:
Solve for x,
As you can see, both methods yield nearly the same answer.
Therefore, a doorway height of 70.1 inches would be needed to allow 60% of men to fit without bending.
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