Solved – Normal Distribution using Z – Score Rules

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The distribution of heights of adult men in the U.S. is approximately normal with mean 69 inches and standard deviation 2.5 inches. Use what you know about a normal distribution and the 68-95-99.7 rule to answer the following.

About what percent of men are taller than 69 inches? 

 69 - 69 = 0
 0 / 2.5 = 0

What could you really do with 0 as a Z-Score? Why am I getting the answer 0 incorrect? I cannot subtract 100 from anything, I'm really confused on this part.

If percent of men are taller than 74 then it would be

74 - 69 = 5
5/2.5 = 2 so 2 is 95%

100 - 95 = 5 
5/2 = 2.5 (Final Answer)

I follow the same exact steps for problem 1 as I did with problem 2 but it's wrong.

Best Answer

Comment continued: Here is a histogram of heights of 10,000 hypothetical men sampled from the distribution $\mathsf{Norm}(\mu = 69,\, \sigma=2.5).$ Let's see what happens.

set.seed(1016);  x = rnorm(10^4, 69, 2.5)
cutp = seq(69-10, 69+10, by=2.5)  # to make histogram bins 2.5" wide
hist(x, br=cutp, ylim=c(0,4000), label=T, col="skyblue2")

enter image description here

So out of the 10,000 simulated men we got $203+13 = 216$ or 2.16% above 74, which is reasonably close to 2.5%.

Also, we got $3492 + 1384 + 203 + 13 = 5092$ or 50.92% above 69, which is close to 50%.

The Empirical Rule is exact for a normal population and it works pretty well for moderately large samples from a normal population. [Examples given in elementary statistics textbooks for small samples are sometimes contrived to give results that are a little better than one can expect in practice.]

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