normal distribution question? I was wondering if anyone could explain how to use the mean and standard deviation of a normal distribution to determine the top 5% of population?
-It is for homework, but I wont post the actual questions since I want to do them myself. Just needing an explanation that isn't a mathbook that I can understand, thanks
 A: If a certain random variable $X$ has normal distribution, with mean $\mu$ and standard deviation $\sigma$, then with probability $0.05$ we have $X\gt \mu+ 1.645\sigma$ (the $1.645$ is approximate).
A: The result will, of course, depend on the mean and standard deviation of your population.
Almost everything you need to know about the Normal Distribution is obtained by knowing how far away you are from the mean; this is the $Z-$ value; a $Z-$ value of , e.g., 1,
tells you your value is $1$ unit (meaning one standard deviation) away from the mean ( if you get $+1$, you are right of the mean, and if you get a $-1$, you are
 one unit ( i.e., one standard deviation) left of the mean.
Strictly-speaking, you could obtain this value from the Normal  distribution function ,
as in , e.g.:http://mathworld.wolfram.com/NormalDistributionFunction.html 
But this is not easy to do. Instead of that, you can find tables like :
http://www.mathsisfun.com/data/standard-normal-distribution-table.html
Where the $Z$-value given represents the number of standard deviations from the mean
you have to "travel" to get a certain percentile. In your case, if you move the ticker
so that your % equals $45%$ , you will get a $Z-$ value of around $1.65$. This means
that, if you have a Normal Distribution with mean $m$ and standard deviation $\sigma$,
all values above $m+1.65\sigma$ have probability less than $0.95%$ of occurring.
