# Discrete random variable

## Homework Statement

A random variable X takes values 1,2,...,n with equal probabilities. Determine the expectation, R for X and show that the variance, Q^2 is given by 12Q^2=n^2-1. Hence, find
P(|X-R|>Q) in the case n=100

## The Attempt at a Solution

I can show that 12Q^2=n^2-1 but not the latter.

Substituting n=100, R=50.5 and Q=28.87

P(|X-50.5|>28.87)=P(X-50.5>28.87)+P(x-50.5<-28.87)

How can i continue from here?

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What is P(X - 50.5 > 28.87)? You might start by figuring out in words just what this expression means. It's pretty simple to get the answer once you understand the expression.

ehild
Homework Helper
All values for x are of equal probability. There are 100 possible value. What is the probability that x takes a special value, like x=10?

ehild

All values for x are of equal probability. There are 100 possible value. What is the probability that x takes a special value, like x=10?

ehild
If this is a discrete random variable, how can it take values which is not discrete ie with decimals?

To answer your question, all the probabilities add up to 1 so any value would have a probability of 1/100.

Am i correct?

HallsofIvy
Homework Helper
Yes, that is correct- if there are n possible outcomes and they are all equally likely, then the probability of any one outcome is 1/n. Since there are only integer outcomes, "X- 50.5> 28.87", which is the same as "X> 79.37" is really "$X\ge 80$".

ehild
Homework Helper
If this is a discrete random variable, how can it take values which is not discrete ie with decimals?
X takes integer values. What is the probability that x takes the value 1 or 2 or 3... up to 21 or 80 or 81 ... up to 100?

ehild

Thanks all for helping me out.