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# 4.6: The Poisson Probability Distribution

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

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

## Introduction

In this lesson, you will be introduced to Poisson distributions. Not only will you learn how to describe a Poisson distribution, but you will also learn how to apply the formula used with this type of distribution. Many real-world problems will be shown.

### Poisson Distributions

A Poisson probability 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 are 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 a binomial distribution, if the number of trials, n\begin{align*}n\end{align*}, gets larger and larger as the probability of success, p\begin{align*}p\end{align*}, gets smaller and smaller, we obtain a Poisson distribution. The section below lists some of the basic characteristics of a Poisson distribution.

### Characteristics of a 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

The probability distribution, mean, and variance of a Poisson random variable are given as follows:

p(x)μσ2=λxeλx!x=0,1,2,3,=λ=λ\begin{align*}p(x) &= \frac{\lambda^x e^{-\lambda}}{x!} \quad 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 in the time, distance, volume, or area

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

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

a. What is the probability distribution for X\begin{align*}X\end{align*}, the number of fish that you will catch in 7 hours?

b. What is the probability that you will catch 0 fish in seven hours of fishing? What is the probability of catching 3 fish? How about 10 fish?

c. What is the probability that you will catch 4 or more fish in 7 hours?

a. The mean number of fish is 3 fish in 2 hours, or 1.5 fish/hour. This means that over seven hours, the mean number of fish will be λ=1.5 fish/hour7 hours=10.5 fish\begin{align*}\lambda=1.5 \ \text{fish/hour} \bullet 7 \ \text{hours}=10.5 \ \text{fish}\end{align*}. Thus, the equation becomes:

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*}

b. To calculate the probabilities that you will catch 0, 3, or 10 fish, perform the following calculations:

p(0)p(3)p(10)=(10.5)0e10.50!0.0000270%=(10.5)3e10.53!0.00530.5%=(10.5)10e10.510!0.123612%\begin{align*}p(0) &= \frac{(10.5)^0 e^{-10.5}}{0!} \approx 0.000027 \approx 0 \%\\ p(3) &= \frac{(10.5)^3 e^{-10.5}}{3!} \approx 0.0053 \approx 0.5 \%\\ p(10) &= \frac{(10.5)^{10} e^{-10.5}}{10!} \approx 0.1236 \approx 12 \%\end{align*}

This means that it is almost guaranteed that you will catch some fish in 7 hours.

c. The probability that you will catch 4 or more fish in 7 hours is equal to the sum of the probabilities that you will catch 4 fish, 5 fish, 6 fish, and so on, as is shown below:

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

The Complement Rule can be used to find this probability as follows:

p(x4)=1[p(0)+p(1)+p(2)+p(3)]10.0000270.0002890.001520.00530.9929\begin{align*}p(x \ge 4) &= 1-[p(0)+p(1)+p(2)+p(3)]\\ & \approx 1-0.000027 - 0.000289 - 0.00152-0.0053\\ & \approx 0.9929\end{align*}

Therefore, there is about a 99% chance that you will catch 4 or more fish within a 7 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.

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-month period.

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

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

This means that 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. For this, we use the Poisson distribution formula:

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

c. The probability of two or more sightings is an infinite sum and is impossible to compute directly. However, we can use the Complement Rule as follows:

p(x2)=1p(x1)=1[p(0)+p(1)]=1(2.5)0e2.50!(2.5)1e2.51!0.713\begin{align*}p(x \ge 2) & = 1-p(x \le 1)\\ & = 1-[p(0)+p(1)]\\ & = 1-\frac{(2.5)^0 e^{-2.5}}{0!}-\frac{(2.5)^1 e^{-2.5}}{1!}\\ & \approx 0.713\end{align*}

Therefore, according to the Poisson model, the probability that two or more sightings are made in a month is 0.713.

Technology Note: Calculating Poisson Probabilities on the TI-83/84 Calculator

Press [2ND][DIST] and scroll down 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 the probability of at most x\begin{align*}x\end{align*} successes.

Note: It is not necessary to close the parentheses.

Technology Note: Using Excel

In a cell, enter the function =Poisson(μ,x,\begin{align*}\mu,x,\end{align*}false), where μ\begin{align*}\mu\end{align*} and x\begin{align*}x\end{align*} are numbers. Press [ENTER], and the probability of x\begin{align*}x\end{align*} successes will appear in the cell.

For the probability of at least x\begin{align*}x\end{align*} successes, replace 'false' with 'true'.

## Lesson Summary

Characteristics of a 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.

Poisson Random Variable:

The probability distribution, mean, and variance of a Poisson random variable are given as follows:

p(x)μσ2=λxeλx!x=0,1,2,3,=λ=λ\begin{align*}p(x) &= \frac{\lambda^x e^{-\lambda}}{x!} \quad 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 in the time, distance, volume or area

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

For a discussion on the Poisson distribution and how to calculate probabilities (4.0)(7.0), see ExamSolutions, Statistics: Poisson Distribution - Introduction (12:34).

For an example of finding probability in a Poisson situation (7.0), see EducatorVids, Statistics: Poisson Probability Distribution (1:54).

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