# 4.5: The Poisson Probability Distribution

**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 \begin{align*}n\end{align*}

## 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

\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*}

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

**Solution:**

1. The mean value number is \begin{align*}3\end{align*}

\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 \begin{align*}0, 3,\end{align*}

\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 \begin{align*}7\;\mathrm{hours}\end{align*}

\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 \begin{align*}\lambda = 10\end{align*}

3. The probability that you will catch \begin{align*}4\end{align*}

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

Using the complement rule,

\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 \begin{align*}99\%\end{align*}

**Example:**

A zoologist is studying the number of times a rare kind of bird has been sighted. The random variable \begin{align*}x\end{align*}

a. Find the mean and standard deviation of \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 \begin{align*}1-\end{align*}

**Solution:**

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

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

Then the standard deviation is,

\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,

\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,

\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,

\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 \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 \begin{align*}\mu\end{align*} and \begin{align*}x\end{align*} separated by commas and press**[ENTER]**. - Use
**poissoncdf**(for probability of at most \begin{align*}x\end{align*} successes.

**Note:** it is not necessary to close the parentheses.

Using EXCEL

- In a cell, enter the function =poisson( \begin{align*}\mu\end{align*} ,\begin{align*}x\end{align*}, false), where \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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