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Suppose 8% of the tires manufactured at a particular plant are defective. To illustrate the use of the Poisson approximation for the binomial, the probability of obtaining exactly one defective tire from a sample of 20 is calculated as follows:

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in this case, the n is only 20!

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up vote 1 down vote accepted

It is a approximation at the end of the day - the natural question is for what values of $n$ and $p$ is the approximation reasonable. You might initially think that $n$ should be huge for this to work, but as the examples in show, it can work even for $n=20$ and $p=0.08$.

If you look at how the poisson distribution is obtained from the binomial, as derived in, you see there are two main approximations.


\begin{equation} \frac{n(n-1)\ldots(n-k+1)}{n^k} \approx 1 \end{equation}

The above is not a bad approximation for $n=20$ and $k=1$. As you increased k you would see begin to see a problem soon. It is also dependent on how much accuracy you need.

Secondly, you also approximate $(1-\lambda/n)^{n-k} \approx (1-\lambda/n)^{n} \approx e^{-\lambda}$ which is a pretty good approximation for moderately large $n$ values and small $k$.

Since these are the two approximations involved, we can be confident about using the poisson distribution whenever they are satisfactory, which in this case is true even though $n$ is only 20.

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are you sure $\begin{equation} \frac{n(n-1)\ldots(n-k+1)}{n^n} \approx 1 \end{equation}$? because (by looking at how it's derived) the denominator is $k!$ – user133466 Oct 7 '12 at 0:56
The denominator should be $n^k$ and I've edited it accordingly. The $k!$ is part of the final poisson distribution expression. I brought the $n^k$ in from the $(\lambda/n)^k$ term. – svenkatr Oct 7 '12 at 1:15

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