# 4.5 The Poisson Distribution

## LEARNING OBJECTIVES

• Recognize the Poisson probability distribution and apply it appropriately.

There are two main characteristics of a Poisson experiment:

1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event.  For example, a book editor might be interested in the number of words spelled incorrectly in a particular book.  It might be that, on the average, there are five words spelled incorrectly in 100 pages.  The interval is the 100 pages.
2. The Poisson distribution may be used to approximate the binomial distribution if the probability of success is “small” (such as 0.01) and the number of trials is “large” (such as 1,000).

The random variable $X$ associated with a Poisson experiment is the number of occurrences in the interval of interest.  In a Poisson distribution, $\lambda$ is the average number of occurrences in an interval.  The mean of a Poisson probability distribution is $\displaystyle{\mu=\lambda}$ and the standard deviation is $\displaystyle{\sigma=\sqrt{\lambda}}$.

## EXAMPLE

The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12.  Of interest is the number of loaves of bread put on the shelf in five minutes.  The time interval of interest is five minutes.  What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?

Solution:

Let $X$ be the number of loaves of bread put on the shelf in five minutes.  If the average number of loaves put on the shelf in 30 minutes (half an hour) is 12, then the average number of loaves put on the shelf in five minutes is $\displaystyle{\frac{5}{30} \times 12=2}$ loaves of bread.

The probability question asks you to find $\displaystyle{P(x=3)}$.

## CALCULATING POISSON PROBABILITIES IN EXCEL

To calculate probabilities associated with a Poisson experiment in Excel, use the Poisson.dist(x, λ, logic operator) function.

• For x, enter the number of successes over the interval.
• For λ, enter the average number of successes over the interval.
• For the logic operator, enter false to find the probability of exactly x successes and enter true the find the probability of at most (less than or equal to) x successes.

The output from the Poisson.dist function is:

• the probability of getting exactly x successes over the interval when the logic operator is false.
• the probability of at most x successes over the interval when the logic operator is true.

### NOTE

Because we can only enter false or true into the logic operator, the Poisson.dist function can only directly calculate the probability of getting exactly x successes or getting at most x success over the interval.  In order to calculate other Poisson probabilities, such as fewer than x successes, more than x successes or at least x successes, we need to manipulate how we use the Poisson.dist function by changing what we enter into the Poisson.dist function, using the complement rule, or both.

## EXAMPLE

1. What is the probability that Leah receives exactly 4 calls in the next two hours?
2. What is the probability that Leah receives at most 9 calls in the next two hours?
3. What is the probability that Leach receives at most 2 calls in the next hour?

Solution:

1. The average number of calls in any two hour period is 6, so $\lambda=6$.
 Function Poisson.dist Answer Field 1 4 0.1339 Field 2 6 Field 3 false

The probability that Leah receives 4 calls in the next two hours is 13.39%.

2. The average number of calls in any two hour period is 6, so $\lambda=6$.
 Function Poisson.dist Answer Field 1 9 0.9161 Field 2 6 Field 3 true

The probability that Leah receives at most 6 calls in the next two hours is 91.61%.

3. The average number of calls in any two hour period is 6.  So the average number of calls in one hour is $\displaystyle{\frac{6}{2}=3}$.
 Function Poisson.dist Answer Field 1 2 0.4232 Field 2 3 Field 3 true

The probability that Leah receives at most 6 calls in the next two hours is 42.32%.

## TRY IT

The customer service department of a technology company receives an average of 10 phone calls every hour.

1. What is the probability that the customer service department receives exactly 7 phone calls in an hour?
2. What is the probability that the customer service department receives exactly 2 phone calls in a 15 minute period?
3. What is the probability that the customer service department receives at most 4 phone calls in a 30 minute period?
4. What is the probability that the customer service department receives at most 20 phone calls in a three hour period?
Click to see Solution
1.  Function Poisson.dist Answer Field 1 7 0.0901 Field 2 10 Field 3 false
2.  Function Poisson.dist Answer Field 1 2 0.2565 Field 2 2.5 Field 3 false
3.  Function Poisson.dist Answer Field 1 4 0.4405 Field 2 5 Field 3 true
4.  Function Poisson.dist Answer Field 1 20 0.0353 Field 2 30 Field 3 true

## EXAMPLE

According to Baydin, an email management company, an email user gets, on average, 147 emails over a six hour period.

1. What is the probability that an email user receives fewer than 160 emails over an six hour period?
2. What is the probability that an email user receives more than 40 emails over a two hour period?
3. What is the probability that an email user receives at least 600 emails over a 24 hour period?
4. What is the probability that an email user receives between 150 and 200 emails over a six hour period?

Solution:

1. The average over a six hour period is 147.  We want to find $\displaystyle{P(x \lt 160)}$.  We cannot find this probability directly in Excel because the Poisson.dist function can only calculate $=$ or $\leq$ probabilities.  Because $x$ must be an integer (it is the number of emails), $x \lt 160$ is the same as $x \leq 159$.  So $\displaystyle{P(x \lt 160)=P(x \leq 159)}$ and $\displaystyle{P(x \leq 159)}$ is a probability we can calculate with the Poisson.dist functon.
 Function Poisson.dist Answer Field 1 159 0.8486 Field 2 147 Field 3 true

The probability a user receives fewer than 160 emails over a six hour period is 84.86%.

2. The average over a two hour period is $\displaystyle{\frac{147}{3}=49}$.  We want to find $\displaystyle{P(x \gt 40)}$.  We cannot find this probability directly in Excel because the Poisson.dist function can only calculate $=$ or $\leq$ probabilities.  The complement of $\gt$ is $\leq$, so $\displaystyle{P(x \gt 40)=1-P(x \leq 40)}$ and $\displaystyle{P(x \leq 40)}$ is a probability we can calculate with the Poisson.dist function.
 Function 1-Poisson.dist Answer Field 1 40 0.8902 Field 2 49 Field 3 true

The probability a user receives more than 40 emails over a two hour period is 89.02%.

3. The average over a 24 hour period is $147 \times 4=588$.  We want to find $\displaystyle{P(x \geq 600)}$.  We cannot find this probability directly in Excel because the Poisson.dist function can only calculate $=$ or $\leq$ probabilities.  The complement of $\geq$ is $\lt$, so $\displaystyle{P(x \geq 600)=1-P(x\lt600)}$.  Because $x$ must be an integer (it is the number of emails), $x\lt600$ is the same as $x \leq 599$.  So $\displaystyle{P(x \geq 600)=1-P(x \lt 600)=1-P(x \leq 599)}$ and $\displaystyle{P(x \leq 599)}$ is a probability we can calculate with the Poisson.dist function.
 Function 1-Poisson.dist Answer Field 1 599 0.3158 Field 2 588 Field 3 true

The probability a user receives at least 600 emails over a 24-hour period is 31.58%.

4. We want to find $\displaystyle{P(150 \leq x \leq 200)}$.  We cannot find this probability directly in Excel because the Poisson.dist function can only calculate $=$ or $\leq$ probabilities.  But, $\displaystyle{P(150 \leq x \leq 200)=P(x \leq 200)-P(x \leq 149)}$.  So we can calculate $\displaystyle{P(150 \leq x \leq 200)}$ as the difference of two Poisson.dist functions.
 Function Poisson.dist -Poisson.dist Answer Field 1 200 149 0.4132 Field 2 147 147 Field 3 true true

The probability a user receives between 150 and 200 emails over a six hour period is 41.32%.

## TRY IT

A car parts manufacturer can produce an average of 25 parts from 100 meters of sheet metal.

1. What is the probability that more than 30 parts can be made from 100 meters of sheet metal?
2. What is the probability that between 10 and 20 parts can be made from 50 meters of sheet metal?
3. What is the probability that fewer than 5 parts can be made from 25 meters of sheet metal?
4. What is the probability that at least 80 parts can be made from 400 meters of sheet metal?
Click to see Solution
1.  Function 1-Poisson.dist Answer Field 1 30 0.1367 Field 2 25 Field 3 true
2.  Function Poisson.dist -Poisson.dist Answer Field 1 20 9 0.7813 Field 2 12.5 12.5 Field 3 true true
3.  Function Poisson.dist Answer Field 1 4 0.2530 Field 2 6.25 Field 3 true
4.  Function 1-Poisson.dist Answer Field 1 79 0.9825 Field 2 100 Field 3 true

Watch this video: The Poisson Distribution by Dr. Nic’s Math and Stats [7:48]

## Concept Review

A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event.  The Poisson distribution may be used to approximate the binomial, if the probability of success is “small” (less than or equal to 0.05) and the number of trials is “large” (greater than or equal to 20).