# Stats: Normal distribution, std dev, mean z score -- find x

## Homework Statement

So the question is, given a set of random numbers, find the mean and the value that will be >= 99% of the occurances. So for a set of random numbers between say, 1-100, if the mean is 50, how do I find out what number will be >= 99% of all observations of the time.

Havent taken stats in a long time. No book provided.

## Homework Equations

99% has a z score of 2.576
Somehow I have to turn that into a number using standard deviation.

## The Attempt at a Solution

z=(x-m)/stdev
x=(stdev*z)+mean
is x the number I am looking for???

Thanks all!

HallsofIvy
Homework Helper
What do you mean by "random numbers between say, 1-100"? That is, what probability distribution are you assuming for choosing the random numbers? Most likely, you are assuming all numbers are "equally likely" so that you are using the "uniform distribution" on numbers 1-100. According to Quora (https://www.quora.com/What-is-the-mean-and-variance-of-uniform-distribution) the uniform distribution on 1-100 has mean (100+1)/2= 50.5 and variance (100- 1)^2/12= 816.75, so standard deviation sqrt(816.75).

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