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Some mean/variance problems

  1. Mar 21, 2006 #1
    Ok I am stuck up deriving the 'variance for Binomial Distribution' and mean for the 'Hypergeometric distribution '

    For variance part , I first derived that variance can be written as =(second moment about origin) - (square of mean)

    But I am having trouble calaculating the second moment about the origin .Please can sum1 tell me sum site which can help me?
     
  2. jcsd
  3. Mar 21, 2006 #2

    HallsofIvy

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    to get the "second moment about the mean" you need to sum i2P(i) for i= 0 to n. For the binomial distribution, with probabilities p, 1-p, P(i)= nCipi(1-p)n-i. That is, you are summing
    [tex]\Sum_{i=0}^n _nC_i i^2 p^i (1-p)^{n-i}[/tex]
    Can you relate that to the binomial theorem?
     
  4. Mar 21, 2006 #3
    thats where I am stuck , I dont know how to solve this binomial further , its been a long time since I did Binomial, maybe lack of practice..
     
  5. Mar 22, 2006 #4
    please sum1 help.
     
  6. Mar 23, 2006 #5

    HallsofIvy

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  7. Mar 23, 2006 #6
    ok I read it , thats a good way to prove the mean of Binomial Distribution , but I want the proof for variance of Binomial , I want to solve it the same way as I told above.
     
  8. Mar 23, 2006 #7

    HallsofIvy

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    Then keep reading! The first half of the page gives a very simple way of deriving the mean (Since one trial the value is either 0 or 1, the mean is 0*(1-p)+ 1(p)= p. Since trials are independent, the mean of n trials is the sum of the means of each: np) the second half of the page derives the variance in the same way.
     
  9. Mar 23, 2006 #8
    ok thanx , I got hold of that idea.

    One more thing , can u pls tell me how ot solve this thing:

    [tex]\Sum_{i=0}^n _nC_i i^2 p^i (1-p)^{n-i}[/tex]

    I am interested to know this.!
     
  10. May 7, 2006 #9
    More detail please?

    Hi, I think that web page is too trivial. Do you know a more detail page? Thanks!
     
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