Prob. for average value or less in binomial distribution?

  • Thread starter Gerenuk
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  • #1
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Main Question or Discussion Point

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:

Answers and Replies

  • #2
mathman
Science Advisor
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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.
 
  • #3
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For the simplified case of p=1/n, your sum goes to 2/e (about 73%) as n->infty.
 
  • #4
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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.
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:
  • #5
For Poisson I'm not sure where to put in my 2 variables n and p
Mean(Poisson) = np.
 

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