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Math Help - Binomial/Poisson/Normal

  1. #1
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    Binomial/Poisson/Normal

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    (a) For each of the Binomial, Poisson and Normal distributions, write down a formula

    This is what I did for binomial:



    Then for the normal distribution I did the binomial/normal approximation:


    However using my calculator the value I get from the binomial is P(3<=X<=7) = 0.2041

    But from the normal approximation I get P(3<=Y<=7) = 0.1488

    The value from the normal approximation is a bit low... so I thought I might have done something wrong, is it because the approximation can only work if p is not too close to 0 or 1? And since p = 0.1 it is fairly close to 0...

    Any help would be appreciated, thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    That's exactly why you need the correction factor.

    P(3\le Bin\le 7)\approx P(2.5<Normal<7.5)=P\left({2.5-10\over 3}<Z<{7.5-10\over 3}\right)
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  3. #3
    MHF Contributor matheagle's Avatar
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    What about the Poisson approximation?

    e^{-10}\sum_{k=3}^7 {10^k\over k!}
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  4. #4
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    Yeah the poisson was alright the approximation was ~ 0.2175

    However I haven't learnt this "correction factor", what is it exactly?

    Thanks for your help!!
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  5. #5
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by usagi_killer View Post
    Yeah the poisson was alright the approximation was ~ 0.2175

    However I haven't learnt this "correction factor", what is it exactly?

    Thanks for your help!!
    I don't remember learning this but was intrigued and found it explained on Wikipedia. Here is the link for Normal approximation and very close to the beginning of that is a link to Continuity correction.
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  6. #6
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    Ah thanks I get what it is now

    But what I'm now interested in is that, why do we need this continuity correction? What is the reason behind it?

    Thanks again guys, learnt heaps today ^_^
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