normal distribution question?

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

See, for example, this answer for details on how to do calculations involving normal random variables. — Dilip Sarwate, Oct 31 '13 at 00:28

score 1 · Answer 1 · answered Oct 31 '13 at 00:30

1

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).

answered Oct 31 '13 at 00:30

André Nicolas

507,029

did you just randomly put in 1.645? – shari Oct 31 '13 at 00:32
I know the number, it comes up often. But to find it, look at a table of the standard normal $Z$, and look in the table for the point $z$ such that $\Pr(Z\le z)=0.95$. – André Nicolas Oct 31 '13 at 00:35

DBFdalwayse · Answer 2 · 2013-10-31T01:51:13.563

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.

normal distribution question?

2 Answers2