### Poisson Probability Distributions

**Describing** 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, \begin{align*}n\end{align*}, gets larger and larger as the probability of success, \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. INSERT
Press

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

Note: It is not necessary to close the parentheses.

**Poisson Random Variable**

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

\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:

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

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

#### Calculating the Probability Distribution

1. 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 \begin{align*}X\end{align*}, the number of fish that you will catch in 7 hours?

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 \begin{align*}\lambda=1.5 \ \text{fish/hour} \bullet 7 \ \text{hours}=10.5 \ \text{fish}\end{align*}. Thus, the equation becomes:

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

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?

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

\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*}

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

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

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:

\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:

\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.

#### Calculating the Mean, Standard Deviation, and Probabilities

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

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

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

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

This means that the standard deviation is \begin{align*}\sigma=1.58\end{align*}.

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

Now we want to calculate the probability that exactly five birds are sighted in one month. For this, 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!}\\ p(5)& =0.067\end{align*}

c. Find the probability that two or more birds are sighted in a 1-month period.

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:

\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.

#### Calculating the Distribution

Suppose that customers enter a store according to a Poisson distribution with an average of 40 customers for a whole day. Suppose that 3 out of 5 customers result in a sale. What is the distribution for the number of sales in half a day?

On average, 40 customers enter a store during a whole day. 3 out of 5, or 60% of these result in a sale. That is on average during a whole day there are \begin{align*}40(.6) = 24\end{align*} sales. The distribution of the number of sales during a half day is Poisson with mean 12.

### Technology Note: Using Excel

In a cell, enter the function =Poisson(\begin{align*}\mu,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 the probability of at least \begin{align*}x\end{align*} successes, replace 'false' with 'true'.

### Examples

Suppose at a speed checkpoint the number of cars caught speeding follows a Poisson distribution with an average of 2.1 cars caught speeding per hour.

#### Example 1

What is the average number of cars caught in \begin{align*}t\end{align*} hours?

The average number of cars caught in \begin{align*}t\end{align*} hours is \begin{align*}2.1t\end{align*}.

#### Example 2

What is P(no cars caught speeding in 15 minutes)?

On average the number of cars caught speeding in 15 minutes would be \begin{align*}2.1/4 = .525\end{align*}.

#### Example 3

P (at least 3 caught in 1.5 hours)?

On average \begin{align*}2.1(1.5)=3.15\end{align*} is the number of cars caught in 1.5 hours.

### Review

- A Poisson distribution has a standard deviation of 3.76. What is its mean?
- For the distribution in problem (1) find:
- \begin{align*}P(X=4)\end{align*}
- \begin{align*}P(X\le 4)\end{align*}
- \begin{align*}P(X\ge 7)\end{align*}
- \begin{align*}P(X\ge 3| X\ge 1)\end{align*}

- Rent-A-Car rents cars to tourists. They have five cars to rent out on a daily basis. The number of requests each day is distributed according to a Poisson distribution with a mean of 4. Determine each of the following probabilities:
- None of its cars are rented
- At least 4 of its cars are rented
- Some requests will have to be refused

- A random variable \begin{align*}X\end{align*} is distributed as a Poisson with mean equal to \begin{align*}m\end{align*}. Find \begin{align*}m\end{align*} given that \begin{align*}P(X=1)+P(X=2)=P(X=3)\end{align*}
- A random variable is distributed as a Poisson with mean equal to 2.7. Find
- \begin{align*}P(X\ge 2)\end{align*}
- \begin{align*}P(X\le 4|X\ge 2)\end{align*}

- A typist makes on average 4 mistakes per page. What is the probability of a particular page having 5 mistakes?
- A computer crashes once every 3 days on average. What is the probability of their being 2 crashes in one week?
- Components are packed in boxes of 25. The probability of a component being defective is 0.2. What is the probability of a box containing 2 defective components?
- The mean number of construction defects in a new house is 9. What is the probability of buying a new house with exactly 2 construction defects?
- On an average Saturday evening a waitress gets no tip from 6 customers. Find the probability that she will get no tip from 8 customers this Saturday evening.
- During a typical football game, a coach can expect 4.1 injuries. Find the probability that the team should have at most 2 injuries in this game.
- A life insurance company claims that on average it gets 8 death claims per day. Find the probability the insurance company will get at least 6 death claims on a randomly selected day.
- Let \begin{align*}Y\end{align*} be a Poisson distribution. If \begin{align*}P(Y=2)=P(Y=3)\end{align*} find the value of \begin{align*}P(Y=2)\end{align*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 4.6.