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As read on Wikipedia, the binomial distribution $B(n, p)$ is approximately normal with mean $np$ and variance $np(1−p)$ for large $n$ and for $p$ not too close to zero or one. Why ? Why this condition on $p$ ?

I know that the sum of Bernoulli distributions gives a binomial distribution, but this is not enough to apply the central limit theorem. Indeed the central limit theorem involves the mean of some variables not only the sum of these variables. Can somebody show me at which point one obtains a mean of variables ?

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  • $\begingroup$ Probably, if $p$ is close to zero or one, then $n$ has to be really large, as opposed to just large. $\endgroup$ – Rasmus Mar 1 '13 at 15:15
  • $\begingroup$ I am also asking for a sketch of the proof. $\endgroup$ – user63008 Mar 1 '13 at 16:06
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If $p$ is close to $0$ or $1$ you get perturbations because the binomial can't go beyond $0$ or $n$, while the normal distribution goes on forever. When $p$ is "reasonable" the missing tails are ignorably small, but if you have a significant chance of hitting the end, it won't be so close. Of course, if $p$ is exactly $0$ or $1$ you are fine, as the variance becomes zero.

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  • $\begingroup$ Thank you Ross. And what about the theoretical reason why a binomial distribution can be approximated by a normal distribution ? $\endgroup$ – user63008 Mar 1 '13 at 15:41
  • $\begingroup$ I am asking just for the sketch of the proof. $\endgroup$ – user63008 Mar 1 '13 at 15:50
  • $\begingroup$ @user63008: the easiest way I know is to use Stirling's approximation to express the factorials in the binomial distribution. You can then see that it approaches the normal distribution. I haven't run through it in some years, but I remember it going smoothly. $\endgroup$ – Ross Millikan Mar 1 '13 at 16:41
  • $\begingroup$ Thank you very much Ross. And does it have something to do with the fact that a binomial is a sum of Bernoullis ? $\endgroup$ – user63008 Mar 1 '13 at 16:47
  • $\begingroup$ @user63008: I don't think so, unless you think the binomial expression for ${n \choose k}=\frac {n!}{k!(n-k)!}$ comes from that. $\endgroup$ – Ross Millikan Mar 1 '13 at 16:49
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Yes it is because for binomial distribution the shape remains mesokurtic when n is large and p is not near 0 an 1. If n is very large like say 1000, p can be as extreme as 0.9 as well and normal approximation will be a good fit. Normal distribution is mesokurtic.

The formula for checking Kurtosis in case of binomial is available here. http://mathworld.wolfram.com/Kurtosis.html

Now put n and p in that formula and check if result is almost 0 like 0.005 or -0.005, just an e.g. When result is zero the curve is mesokurtic just like normal distribution is.

Large n and non extreme values of p when put in that Kurtosis formula will give you small very small value of Kurtosis. I.e. Almost mesokurtic distribution.

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