An important discrete probability distribution is the Poisson distribution.¹ We examine it because of its key role in complementing the exponential distribution in queuing theory in Chapter 12. The distribution describes situations in which customers arrive independently during a certain time interval, and the number of arrivals depends on the length of the time interval. Examples are patients arriving at a health clinic, customers arriving at a bank window, passengers arriving at an airport, and telephone calls going through a central exchange.

The formula for the Poisson distribution is

where

• $P\left(X\right)=\text{probability of exactly }X\text{ arrivals or occurrences}$

• $\lambda =$ average number of arrivals per unit of time (the mean arrival rate), pronounced “lambda”

• $e=2.718\text{, the base of the natural logarithm}$

• $X=\text{number of occurrences }\left(0\text{, }1\text{, }2\text{, }\dots \right)$

The mean and variance of the Poisson distribution are equal and are computed simply as

With the help of the table in Appendix C, the values of $e^{\mathrm{-}\lambda }$ are easy to find. We can use these in the formula to find probabilities. For example, if $\lambda =2,$ from Appendix C we find $e^{\mathrm{-}2}=\mathrm{0.1353.}$ The Poisson probabilities that X is 0, 1, and 2 when $\lambda =2$ are as follows:

• $P\left(X\right)=\frac{e^{\mathrm{-}\lambda }{\lambda }^x}{X!}$

• $P\left(0\right)=\frac{e^{\mathrm{-}2}{2}^0}{0!}=\frac{\left(0.1353\right)1}{1}=0.1353\approx 14\%$

• $P\left(1\right)=\frac{e^{\mathrm{-}2}{2}^1}{1!}=\frac{e^{\mathrm{-}2}2}{1}=\frac{0.1353\left(2\right)}{1}=0.2706\approx 27\%$

• $P\left(2\right)=\frac{e^{\mathrm{-}2}{2}^2}{2!}=\frac{e^{\mathrm{-}2}4}{2\left(1\right)}=\frac{0.1353\left(4\right)}{2}=0.2706\approx 27\%$

These probabilities, as well as others for $\lambda =2$ and $\lambda =4,$ are shown in Figure 2.18. Notice that the chances that 9 or more customers will arrive in a particular time period are virtually nil. Programs 2.6A and 2.6B illustrate how Excel can be used to find Poisson probabilities.

It should be noted that the exponential and Poisson distributions are related. If the number of occurrences per time period follows a Poisson distribution, then the time between occurrences follows an exponential distribution. For example, if the number of phone calls arriving at a customer service center followed a Poisson distribution with a mean of 10 calls per hour, the time between each phone call would be exponentially distributed with a mean time between calls of ${}^1{/}_{10}$ hour (6 minutes).