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Discrete random variable

  1. Jul 27, 2010 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    3. 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


    How can i continue from here?
  2. jcsd
  3. Jul 27, 2010 #2
    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.
  4. Jul 27, 2010 #3


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    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?

  5. Jul 28, 2010 #4
    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?
  6. Jul 28, 2010 #5


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    Science Advisor

    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]".
  7. Jul 28, 2010 #6


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    Homework Helper

    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?

  8. Jul 28, 2010 #7
    Thanks all for helping me out.
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