# Normal Distribution using Z - Score Rules

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


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

• Change " 68-95-99.7 rule" to "50- 68-95-99.7 rule", you will get correct answer. – user158565 Oct 16 '18 at 22:53
• Sorry, little mistake, Should be "0-68-95-99.7 rule". in "68-96-99.7 rule" it means if you get 1, you have 68%, 2 for 95%, 3 for 99.7%. So you need to find x for 0% to construct "0-68-95-99.7 rule". – user158565 Oct 16 '18 at 23:06
• Whatever you call it, the Empirical Rule says that for normal (or nearly normal) data, about half of the observations are above the mean $\mu = 69.$ So the first answer is 50%. // 74 is two standard deviations ($2 \times 2.5 = 5$) inches above the mean. The Empirical rule says about 95% of the men are in $\mu \pm 2\sigma = 69 \pm 5$ or $[64,74].$ That means 5% are outside that interval; by symmetry, that must be 2.5% above and 2.5% below. So about 2.5% should be taller than 74 inches. – BruceET Oct 16 '18 at 23:54

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)

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%.