Prob. for average value or less in binomial distribution?

[Gerenuk]

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

P=\sum_{k=0}^{np} \begin{pmatrix} n\\ k
\end{pmatrix} p^k(1-p)^{n-k}


I just figured that for the simplified case p=\frac{1}{n} this probability converges to 63% for large n. What about more general cases?

[mathman]

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.

[mXSCNT]

For the simplified case of p=1/n, your sum goes to 2/e (about 73%) as n->infty.

[Gerenuk]

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% (1-e^{-1}) sure that at least 1 of a N people is "positive".

[Enuma_Elish]

For Poisson I'm not sure where to put in my 2 variables n and pMean(Poisson) = np.