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Probability Mass Function

  1. Aug 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Let X be a random variable with PMF

    x 0 2 4 5
    p(x) 0.15 0.30 0.30 0.25

    (The x values above are supposed to match up with the p(x) values)

    Find P[2 < X < 4], P[X [tex]\leq[/tex] 3]

    3. The attempt at a solution

    Okay, I kinda have an idea on how to do this, but I just want to make sure. To find P[2 < X < 4], I think it's basically just .3 + .3, since the PMF must equal 1 then you'd basically just have to add both p(x) values together. For P[X [tex]\leq[/tex] 3], I think it's .15 + .2 (since both are less than 3). Are these answers correct, or am I completely wrong? (I have a bad feeling I am...)
     
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  3. Aug 4, 2009 #2

    Dick

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    For the first one neither 2 nor 4 satisfy 2<x<4, do they? For the other one do you mean .15+.3? Or are you more confused than I think?
     
  4. Aug 4, 2009 #3
    What you are trying to find is the cumulative distribution function (cdf) which is defined by:

    [tex]P(X \leq x) = \sum_{y \leq x} p(y) [/tex]

    To use different numbers... say you want to find [tex] P(X \leq 2) [/tex], it would be

    [tex] P(X \leq 2) = p(0) + p(2) =... [/tex]

    but [tex] P(X < 2) = p(0) \neq p(0) + p(2) = P(X \leq 2) [/tex].

    You're confusing the two symbols (<) less than or greater than and (<=) less than or equal to.

    Edit: Also,

    This is really irrelevant. By definition of a pmf, the sum of all possible p(x_i) = 1. Here you are confusing the symbols.
     
    Last edited: Aug 4, 2009
  5. Aug 4, 2009 #4
    Yup, I was meant to write .15+.3 for the second one.

    Sorry, my class actually hasn't covered cdfs just yet (we've only done pmfs and are learning about binomial and poisson random variables at the moment). Do you know another way to do it with pmfs, or is it irrelevant in this case?
    Also, from what I understand, in order to find P[2 < X < 4], you have to find P(3)? (Which is zero in this case?)

    Thanks for all the help.
     
  6. Aug 4, 2009 #5

    Dick

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    Right. It's zero. But you don't have to 'find P(3)'. There just aren't any points in your distribution that are strictly between 2 and 4.
     
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