Re: binomial distribution, variance question.
- From: "David Jones" <dajxxx@xxxxxxxxx>
- Date: Wed, 30 Nov 2005 12:00:41 -0000
Surendra Singhi wrote:
> Hello,
> Assume that I have a coin with which p=0.75 probability, gives
head
> and q=0.25 probability gives tail.
>
> I toss this coin n=1000 times, the mean of the number of heads will
> be 'np', and the variance of the number of heads will be 'npq'.
>
> Now, if I rather plot a probability distribution function of the
mean
> of tosses, is it correct to assume that the plot will be a binomial
> distribution?
In the usual terminology, no. Values from a "binomial distribution"
have as possible values the integers 0, 1, to n. But the distribution
of the mean is a scaled version of this with possible values 0, 1/n,
2/n to 1.
>
> If I now approximate this plot by a normal distribution, the mean
> will be 'p', but the variance will be pq/n, is this correct?
>
This might be a good approximation depending on the size of n, and on
how close p or q is to zero. For your values it is probably OK. The
mean and variance you quote are exact values. Depending on what you
are trying to do, you might want to apply a "continuity correction"
when you apply approximation by the normal distribution. For example
to approximate the probability that the mean is greater than or equal
to j/n, you would look at the upper tail of the normal distribution
corresponding to a value (j-1/2)/n. This is a standard step to improve
the approximation, but it might have little effect with your sample
size. Of course, given the computing power typically assumed to be
available nowadays, some computer program should be available to
compute the probabilities directly from the underlying Binomial
distribution.
David Jones
.
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- From: Surendra Singhi
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