How to Calculate Expectation and Variance for a Discrete Random Variable?

[thereddevils]

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

Homework Equations

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?

 

[hgfalling]

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]

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

 

[thereddevils]

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]

Yes, that is correct- if there are n possible outcomes and they are all equally likely, then the probability of anyone 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]

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

 

[thereddevils]

Thanks all for helping me out.