28.2 - Normal Approximation to Poisson

Just as the Central Limit Theorem can be applied to the sum of independent Bernoulli random variables, it can be applied to the sum of independent Poisson random variables. Suppose \(Y\) denotes the number of events occurring in an interval with mean \(\lambda\) and variance \(\lambda\). Now, if \(X_1, X_2,\ldots, X_{\lambda}\) are independent Poisson random variables with mean 1, then:

\(Y=\sum\limits_{i=1}^\lambda X_i\)

is a Poisson random variable with mean \(\lambda\). (If you're not convinced of that claim, you might want to go back and review the homework for the lesson on The Moment Generating Function Technique, in which we showed that the sum of independent Poisson random variables is a Poisson random variable.) So, now that we've written \(Y\) as a sum of independent, identically distributed random variables, we can apply the Central Limit Theorem. Specifically, when \(\lambda\) is sufficiently large:

\(Z=\dfrac{Y-\lambda}{\sqrt{\lambda}}\stackrel {d}{\longrightarrow} N(0,1)\)

We'll use this result to approximate Poisson probabilities using the normal distribution.

Example 28-2 Section

building collapsed from an earthquake

The annual number of earthquakes registering at least 2.5 on the Richter Scale and having an epicenter within 40 miles of downtown Memphis follows a Poisson distribution with mean 6.5. What is the probability that at least 9 such earthquakes will strike next year? (Adapted from An Introduction to Mathematical Statistics, by Richard J. Larsen and Morris L. Marx.)


We can, of course use the Poisson distribution to calculate the exact probability. Using the Poisson table with \(\lambda=6.5\), we get:

\(P(Y\geq 9)=1-P(Y\leq 8)=1-0.792=0.208\)

Now, let's use the normal approximation to the Poisson to calculate an approximate probability. First, we have to make a continuity correction. Doing so, we get:

\(P(Y\geq 9)=P(Y>8.5)\)

Once we've made the continuity correction, the calculation again reduces to a normal probability calculation:

\begin{align} P(Y\geq 9)=P(Y>8.5)&= P(Z>\dfrac{8.5-6.5}{\sqrt{6.5}})\\ &= P(Z>0.78)=0.218\\ \end{align}

So, in summary, we used the Poisson distribution to determine the probability that \(Y\) is at least 9 is exactly 0.208, and we used the normal distribution to determine the probability that \(Y\) is at least 9 is approximately 0.218. Not too shabby of an approximation!