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Thread: PDF of discrete RV sum?

  1. #1
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    Question PDF of discrete RV sum?

    Let's say I choose a number between 1-4 and the probability of choosing each number is as such:

    p(1)=0.35
    p(2)=0.25
    p(3)=0.15
    p(4)=0.25

    I do this 100 times. Let X be the sum of all the numbers I picked. What is the approximate PDF of X?

    Also, what is the probability that X > 200?

    I think I have to use the normal distribution for the last question, but I'm not entirely sure how to go about setting this up and solving this. Any help would be greatly appreciated! Thanks.
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  2. #2
    MHF Contributor matheagle's Avatar
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    The central limit theorem says that the sum of iid copies a random variable will be approximately normal.

    I quickly calculated (check my work) E(X)=2.3 and V(X)=1.41.

    So X=\sum_{i=1}^{100}X_i\approx N(100(2.3), 1.41(100))=N(230,141)

    NOW, we really should use a correction factor here, since the rv's in question
    are discrete and the normal is continuous. Same as we do in the normal approximation to
    the binomial.

    P(X>200)\approx P\Biggl(Z>{200-230\over \sqrt{141}}\Biggr)=P(Z>-2.526455763)

    But since this is discrete I would argue that you should go with...

    P(X>200)\approx P\Biggl(Z>{200.5-230\over \sqrt{141}}\Biggr)=P(Z>-2.484348167)
    Last edited by matheagle; Oct 25th 2009 at 10:48 PM.
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  3. #3
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    Smile

    Thanks so much!
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  4. #4
    MHF Contributor matheagle's Avatar
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    There should be some kind of correction factor for overlaying a continuous rv (normal) on top of your discrete rv.
    The plus or minus one-half is correct when we are dealing with sums of Bernoullis. But here the value of each rv is 1,2,3 or 4 not 0,1.
    So this is off a bit.
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