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

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Homework Help Overview

The problem involves calculating the expectation and variance of a discrete random variable X that takes integer values from 1 to n with equal probabilities. The original poster attempts to demonstrate the relationship between variance and n, and to find the probability of a specific condition involving the expectation and variance when n=100.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the meaning of the probability expression P(X - 50.5 > 28.87) and its implications. There is also a question about the nature of discrete random variables and their possible values, particularly regarding the use of decimal points in the context of integer outcomes.

Discussion Status

Some participants provide clarifications regarding the probabilities associated with discrete outcomes and confirm the correctness of the original poster's understanding of equal probabilities. There is ongoing exploration of the implications of the conditions set in the problem.

Contextual Notes

The discussion includes considerations about the properties of discrete random variables and the constraints of the problem, such as the requirement for integer outcomes and the implications of the calculations involving expectation and variance.

thereddevils
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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?
 
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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.
 
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
 
ehild said:
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?
 
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 "[itex]X\ge 80[/itex]".
 
thereddevils said:
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.
 

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