The Poisson Random Variable


Suppose an event has a small probability of occurring and a large number of independent trials take place. Suppose further that you know the average number of occurrences µ over a period of time. Then the Poisson random variable, denoted X ~ Poi(µ), counts the total number of occurrences during a given time period.

The probability of having exactly k occurrences during this time, for k >= 0, is given by

P(X = k) = e ^(- µ) * µ^k / k! .

The average (or mean) number of occurrences is given by E[X] = µ.

The variance of the number of occurrences is given by Var(X) = µ.

The most likely number of occurrences, or mode, will be the largest integer k such that k <= µ. The random variable is bimodal when µ is an integer. In that case, this value of k = µ and the previous integer k - 1 will be the modes.

There are no closed-form formulas for the cumulative probability P(X <= k) or for computing probabilities such as P(j <= X <= k).

Using the POISSON Program

The POISSON program can be used to compute probabilites such as P(j <= X <= k), P(X = k), and P(X <= k). To execute the program, we enter the average value and the lower and upper bounds of j and k. (Enter the same value k for both the lower and upper bound to compute a pdf value P(X = k).) The program also asks if you want a complete distribution to be entered into the STAT Edit screen. If so, then enter 1. If not, then enter 0. The program then displays P(j <= X <= k) along with the average value, standard deviation, and mode.

If you entered 1, then most of the distribution is entered into the STAT Edit screen. Under L1, the possible number of occurrences 0, 1, . . . , n are listed. Adjacent under L2, the pdf values of P(X = k), for 0 <= k <= n, are listed. Under L3, the cdf values P(X <=k) are listed for 0 <= k <= n. The upper bound n is chosen so that P(X <= n) exceeds 0.975.

Click here for info on the TI-86 and TI-89 Stat Edit displays.

If you pressed 1 for a complete distribution, then press GRAPH to see a scatterplot of the distribution.

Example. Traffic engineers often use a Poisson distribution to model the flow of cars in light traffic. Suppose then that during the hour from 9:00 am to 10:00 am each weekday, an average of 4.7 cars pass through a particular stop sign without making a complete stop. During a given such hour, what are the probabilities that

(a) from 3 to 7 cars pass through without stopping?
(b) exactly 5 cars pass through?
(c) at least 4 cars pass through?

Solution. After calling up the POISSON program, enter 4.7 for AVERAGE, enter 3 for LOWER BOUND, and enter 7 for UPPER BOUND. Also enter 1 for a complete distribution.

(a) We see that P(3 <= X <= 7) = 0.74373.

(b) P(X = 5) = 0.17383 (from list L2, yStat on the TI-86, or c2 in APPS, 6, 1 on the TI-89).

(c) P(X >= 4) = 1 - P(X <= 3) = 1 - 0.30968 = 0.69032.


On May 8, 1994 The New York Times published an article under the headline "Lies, Damn Lies, and Baseball Statistics." It was stated that during April of the 1994 Major League Baseball season, there was an average of 2.22 home runs per game. Assuming that this average continued to hold throughout the first few days of May, find the probability of a game having had at least 5 home runs.


Here we can assume that X ~ Poi(2.22). We wish to find P(X >= 5) = 1 - P(X <= 4). So we first find P(X <= 4). In the POISSON program, enter 2.22 for AVERAGE, then enter 0 for the LOWER BOUND and 4 for UPPER BOUND. We find that P(0 <= X <= 4) = 0.9253; thus, P(X >= 5) = 1 - 0.9253 = 0 .0747.

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