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Prob. for average value or less in binomial distribution?

  1. May 30, 2009 #1
    Hello!
    Is there a closed form expression or a good estimate for the probability that a binomial distribution yield the average np or less. Basically I'm asking for a good way to evaluate
    [tex]
    P=\sum_{k=0}^{np} \begin{pmatrix} n\\ k
    \end{pmatrix} p^k(1-p)^{n-k}
    [/tex]

    I just figured that for the simplified case [itex]p=\frac{1}{n}[/itex] this probability converges to 63% for large n. What about more general cases?
     
    Last edited: May 30, 2009
  2. jcsd
  3. May 30, 2009 #2

    mathman

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    Science Advisor
    Gold Member

    There is no closed form expression for Prob. (other than the exact formula). However for large n, the binomial can be approximated by the normal distribution when p is fixed. In the case you are describing (np fixed), the Poisson distribution is a good approximation.
     
  4. May 30, 2009 #3
    For the simplified case of p=1/n, your sum goes to 2/e (about 73%) as n->infty.
     
  5. May 30, 2009 #4
    Unfortunately using the normal distribution yields 50%, which is not true when the discrete character isn't lost. For Poisson I'm not sure where to put in my 2 variables n and p :(

    Btw, this problem I thought of when trying to think of how likely a "statistical statement" would be. With the odds 1:N for example you can be 63% ([itex]1-e^{-1}[/itex]) sure that at least 1 of a N people is "positive".
     
    Last edited: May 30, 2009
  6. May 31, 2009 #5
    Mean(Poisson) = np.
     
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