# Central limit theorem to approximate a Poisson distribution.

Considerable controversy has arisen over the possible aftereffects of a nuclear weapons test conducted in Nevada in $1957$. Included as part of the test were some $3000$ military and civilian "observers." Now, more than $50$ years later, eight cases of leukemia have been diagnosed among those $3000$. The expected number of cases, based on the demographic characteristics of the observers, was three. Assess the statistical significance of those findings. Calculate both an exact answer using the Poisson distribution as well as an approximation based on the central limit theorem.

I was able to calculate the answer using the Poisson distribution such that I found that $\frac{3}{3000}=p$ so $\lambda = (3000)(0.001)$ and then $P(X=8)=\frac{e^{-3}(3^8)}{8!}=0.008101$. (If I am incorrect please correct me.) I am now having trouble using the central limit theorem to approximate.

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You're probably expected to replace the Poisson distribution with a normal distribution of the same mean and standard deviation. However, note that computing $P(X=8)$ is not the same as "assessing the statistical significance of the findings". There's more for you to do in the exact case, which should bring you closer to a situation where you can see something analogous to do with the normal distribution. –  Henning Makholm Feb 3 '13 at 15:57

You've almost got it. Normally one would reject the null hypothesis that $\lambda\le3$ if $X$ is bigger than some cut-off point, and how big that is depends on how big a probability of Type I error one is willing to allow. So what you need is $$\Pr(X\ge 8) = 1 - \sum_{x=0}^7 \Pr(X=x).$$ The latter sum may (possibly?) be the quickest way to compute the number. That probability is the $p$-value.