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Discrete random variable

  • #1
438
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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?
 

Answers and Replies

  • #2
351
1
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.
 
  • #3
ehild
Homework Helper
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1,876
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
 
  • #4
438
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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?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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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 "[itex]X\ge 80[/itex]".
 
  • #6
ehild
Homework Helper
15,494
1,876
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
 
  • #7
438
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Thanks all for helping me out.
 
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