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# 4.5: The Poisson Probability Distribution

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Know the definition of the Poisson distribution.
• Identify the characteristics of the Poisson distribution.
• Identify the type of statistical situation to which the Poisson distribution can be applied.
• Use the Poisson distribution to solve statistical problems.

The Poisson distribution is useful for describing the number of events that will occur during a specific interval of time or in a specific distance, area, or volume. Examples of such random variables are:

• The number of traffic accidents at a particular intersection.
• The number of house fire claims per month that is received by an insurance company.
• The number of people who are infected with the AIDS virus in a certain neighborhood.
• The number of people who walk into a barber shop without an appointment.

In relation to the binomial distribution, if the number of trials n\begin{align*}n\end{align*} gets larger and larger as the probability of successes p\begin{align*}p\end{align*} gets smaller and smaller, we obtain the Poisson distribution. The box below shows some of the basic characteristics of the Poisson distribution.

## Characteristics of the Poisson Distribution

• The experiment consists of counting the number of events that will occur during a specific interval of time or in a specific distance, area, or volume.
• The probability that an event occurs in a given time, distance, area, or volume is the same.
• Each event is independent of all other events. For example, the number of people who arrive in the first hour is independent of the number who arrive in any other hour.

## Poisson Random Variable

### Mean and Variance

p(x)μσ2=λxeλx!   x=0,1,2,3,=λ=λ\begin{align*}p(x) & = \frac{\lambda^x e^{-\lambda}} {x!} \ \ \ x = 0, 1, 2, 3,\ldots \\ \mu & = \lambda \\ \sigma^2 & = \lambda \end{align*}

where

λ=\begin{align*}\lambda =\end{align*} The mean number of events during the time, distance, volume or area.

e=\begin{align*}e =\end{align*} the base of the natural logarithm =2.718281828\begin{align*}= 2.718281828 \ldots\end{align*}

Example:

A lake, popular among boat fishermen, has an average catch of three fish every two hours during the month of October.

• What is the probability distribution for x\begin{align*}x\end{align*}, the number of fish that you will catch, in 7hours\begin{align*}7\;\mathrm{hours}\end{align*} ?
• What is the probability that you will catch 0,3,\begin{align*}0, 3,\end{align*} or 10\begin{align*}10\end{align*} fish in seven hours of fishing?
• What is the probability that you will catch 4\begin{align*}4\end{align*} or more fish in 7\begin{align*}7\end{align*} hours?

Solution:

1. The mean value number is 3\begin{align*}3\end{align*} fish in 2hours\begin{align*}2\;\mathrm{hours}\end{align*} or 1.5fish/1hour\begin{align*}1.5\;\mathrm{fish}/1\;\mathrm{hour}\end{align*} . This means, over seven hours, the mean number events will be λ=1.5fish/hour\begin{align*}\lambda = 1.5\;\mathrm{fish}/\mathrm{hour}\end{align*} * 7hours\begin{align*}7\;\mathrm{hours}\end{align*} = 10.5\begin{align*}10.5\end{align*} fish. Thus,

p(x)=λxeλx!=(10.5)xe10.5x!\begin{align*}p(x) = \frac{\lambda^x e^{-\lambda}} {x!} = \frac{(10.5)^x e^{-10.5}} {x!}\end{align*}

2. To calculate the probabilities that you will catch 0,3,\begin{align*}0, 3,\end{align*} or 10\begin{align*}10\end{align*}:

p(0)=(10.5)0e10.50!0.0000270%\begin{align*}p(0) = \frac{(10.5)^0 e^{-10.5}} {0!} \approx 0.000027 \approx 0\%\end{align*}

This says that it is almost guaranteed that you will catch fish in 7hours\begin{align*}7\;\mathrm{hours}\end{align*} .

p(3)p(10)=(10.5)3e10.53!0.00520.5%=(10.5)10e10.510!0.121212%\begin{align*}p(3) & = \frac{(10.5)^3 e^{-10.5}} {3!} \approx 0.0052 \approx 0.5\% \\ p(10) & = \frac{(10.5)^{10} e^{-10.5}} {10!} \approx 0.1212 \approx 12\%\end{align*}

The curve in the figure with λ=10\begin{align*}\lambda = 10\end{align*} is very similar graph of the function for this problem.

3. The probability that you will catch 4\begin{align*}4\end{align*} or more fish in 7\begin{align*}7\end{align*} hours is,

p(x4)=p(4)+p(5)+p(6)+\begin{align*}p(x \ge 4) = p(4) + p(5) + p(6) + \ldots\end{align*}

Using the complement rule,

P(x4)=1[p(0)+p(1)+p(2)+p(3)]10.0000270.0002890.001520.00520.9903\begin{align*}P(x \ge 4) & = 1 - [p(0) + p(1) + p(2) + p(3)]\\ & \approx 1 - 0.000027 - 0.000289 - 0.00152 - 0.0052 \\ & \approx 0.9903\end{align*}

Therefore there is about 99%\begin{align*}99\%\end{align*} chance that you will catch 4\begin{align*}4\end{align*} or more fish within a 7\begin{align*}7-\end{align*}hour period during the month of October.

Example:

A zoologist is studying the number of times a rare kind of bird has been sighted. The random variable x\begin{align*}x\end{align*} is the number of times the bird is sighted every month. We assume that x\begin{align*}x\end{align*} has a Poisson distribution with a mean value of 2.5\begin{align*}2.5\end{align*}.

a. Find the mean and standard deviation of x\begin{align*}x\end{align*}.

b. Find the probability that exactly five birds are sighted in one month.

c. Find the probability that two or more birds are sighted in a 1\begin{align*}1-\end{align*}month period.

Solution:

a. The mean and the variance are both equal to λ\begin{align*}\lambda\end{align*}. Thus,

μσ2=λ=2.5=λ=2.5\begin{align*}\mu & = \lambda = 2.5 \\ \sigma^2 & = \lambda = 2.5\end{align*}

Then the standard deviation is,

σ=1.58\begin{align*}\sigma = 1.58\end{align*}

b. Now we want to calculate the probability that exactly five birds are sighted in one month. We use the Poisson distribution formula,

p(x)p(5)p(x)=λxeλx!=(2.5)5e2.55!=0.067=0.067\begin{align*}p(x) & = \frac{\lambda^x e^{-\lambda}} {x!} \\ p(5) & = \frac{(2.5)^5 e^{-2.5}} {5!} = 0.067 \\ p(x) & = 0.067\end{align*}

c. To find the probability of two or more sightings,

p(x2)=p(2)+p(3)+=x2p(x)\begin{align*}p(x \ge 2) = p(2) + p(3) +\ldots = \sum_{x - 2}^\infty p(x)\end{align*}

This is of course an infinite sum and it is impossible to compute. However, we can use the complementation rule,

Undefined control sequence \intertext\begin{align*}p(x \ge 2) & = 1 - p(x \le 1)\\ & = 1 - [p(0) + p(1)]\\ \intertext{---Calculating,---}\\ & = 1 - \frac{(2.5)^0 e^{-2.5}} {0!} - \frac{(2.5)^1 e^{-2.5}} {1!}\\ & \approx 0.713\end{align*}

So, according to the Poisson model, the probability that two or more sightings are made in a month is 71.3%\begin{align*}71.3\%\end{align*}

## Technology Note

The TI-83/84 calculators and the EXCEL spreadsheet have commands for the Poisson distribution.

Using the TI-83/84 Calculators

• Press [DIST] and scroll down (or up) to poissonpdf ( Press [ENTER] to place poissonpdf on your home screen.) Type values of μ\begin{align*}\mu\end{align*} and x\begin{align*}x\end{align*} separated by commas and press [ENTER].
• Use poissoncdf (for probability of at most x\begin{align*}x\end{align*} successes.

Note: it is not necessary to close the parentheses.

Using EXCEL

• In a cell, enter the function =poisson( μ\begin{align*}\mu\end{align*} ,x\begin{align*}x\end{align*}, false), where μ\begin{align*}\mu\end{align*} and \begin{align*}x\end{align*} are numbers. Press [Enter] and the probability of \begin{align*}x\end{align*} successes will appear in the cell.
• For probability of at least \begin{align*}x\end{align*} successes, replace “false” with “true”

## Lesson Summary

1. Characteristics of the Poisson Distribution:

• The experiment consists of counting the number of events that will occur during a specific interval of time or in a specific distance, area, or volume.
• The probability that an event occurs in a given time, distance, area, or volume is the same.
• Each event is independent of all other events.

2. Poisson Random Variable

Mean and Variance

\begin{align*}p(x) & = \frac{\lambda^x e^{-\lambda}} {x!}\ \ \ x = 0, 1, 2, 3,\ldots \\ \mu & = \lambda \\ \sigma^2 & = \lambda\end{align*}

where

\begin{align*}\lambda =\end{align*} The mean number of events during the time, distance, volume or area.

\begin{align*}e=\end{align*} the base of the natural logarithm \begin{align*}= 2.718281828\ldots\end{align*}

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