5.4: Poisson Distribution

Learning Objectives

• Define the Poisson distribution to model the number of times an event occurs in a fixed interval of time or space under conditions of independence and a constant rate.
• Apply the Poisson distribution to rare events such as phone calls per hour or accidents per day.
• Calculate probabilities using the average number of occurrences within the specified interval.

Introduction

The Poisson distribution is popular for modeling the number of times an event occurs in an interval of time or space. It is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event.

two main characteristics of a Poisson experiment

1. The Poisson probability distribution gives the probability of several 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 average, there are five words spelled incorrectly in 100 pages. The interval is 100 pages.
2. The Poisson distribution may be used to approximate the binomial if the probability of success is "small" (such as 0.01) and the number of trials is "large" (such as 1,000). You will verify the relationship in the homework exercises. \(n\) is the number of trials, and \(p\) is the probability of a "success."

The Formula for Poisson Distribution

The formula to compute the Poisson Distribution is presented below.

\(P\left(X\right)=\dfrac{e^{-\lambda}\lambda^X}{X!}\)

• X = 0, 1, 2, 3, ...
• \(\lambda\) = mean number of occurrences in the interval.
• \(e\) = Euler's constant \(\approx 2.718280\)

Examples

Example \(\PageIndex{1}\)

Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah will receive more than one call in the next 15 minutes?

Solution

Let \(X\) = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or \(\frac{1}{4}\) hour.)

\[x = 0, 1, 2, 3, \dotsc\]

If Leah receives, on average, six telephone calls in two hours, and there are eight 15-minute intervals in two hours, then Leah receives

\(\left(\frac{1}{8}\right)(6) = 0.75\) calls in 15 minutes, on average. So, \(\lambda = 0.75\) for this problem.

\(X \sim P(0.75)\)

Find \(P(x > 1)\). \(P(x > 1) = 0.1734\) (calculator or computer)

• Press 1 – and then press 2^nd DISTR.
• Arrow down to poissoncdf. Press ENTER.
• Enter (.75,1).
• The result is \(P(x > 1) = 0.1734\).

\(P\left(X\right)=\dfrac{e^{-\lambda}\lambda^X}{X!}\)

Recall that the sum of all the probabilities in a discrete distribution is 1. To use the formula, we must find P(0) and P(1) and subtract the value from 1. The result is P(x > 1).

\(p\left(0\right)=\dfrac{e^{-0.75}\cdot0.75^0}{0!}\) = 0.4724

\(p\left(1\right)=\dfrac{e^{-0.75}\cdot0.75^1}{1!}\) = 0.3542

P(x > 1) = 1 - P(0) - P(1) = 1 - 0.4724 - 0.3542

P(x > 1) = 0.1734

The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:

\(P(x > 1) = 1 − \text{poissoncdf}(0.75, 1)\).

The graph of \(X \sim P(0.75)\) is:

Figure \(\PageIndex{1}\):Graph of Poisson Distribution

The y-axis contains the probability of \(x\) where \(X =\) the number of calls in 15 minutes.

Example \(\PageIndex{2}\)

A customer service center receives about ten emails every half-hour. What is the probability that the customer service center will receive more than four emails in the next six minutes?

Solution

\(P(x > 4) = 0.0526\)

Recall that the sum of all the probabilities in a discrete distribution is 1. To use the formula, we must find P(5), P(6), P(7), and other values until the probabilities become close to zero and then subtract the values from 1. The result is P(x > 4).

\(\lambda\) = \(\frac{10}{30}\cdot6=2\)

\(P\left(5\right)=\dfrac{e^{-2}\cdot2^5}{5!}=0.0361\)

\(P\left(6\right)=\dfrac{e^{-2}\cdot2^6}{6!}=0.0120\)

\(P\left(7\right)=\dfrac{e^{-2}\cdot2^7}{7!}=0.0034\)

\(P\left(8\right)=\dfrac{e^{-2}\cdot2^8}{8!}=0.0009\)

\(P\left(9\right)=\dfrac{e^{-2}\cdot2^9}{9!}=0.0002\)

All probabilities beyond P(9) are small values that can be treated as 0.

P(x > 4) = P(5) + P(6) + P(7) + P(8) + P(9) = 0.0361 + 0.0120 + 0.0034 + 0.0009 + 0.0002 = 0.0526

P(x > 4) =0.0526.

Example \(\PageIndex{3}\)

According to Baydin, an email management company, an email user gets, on average, 144 emails per day. Let \(X =\) be the number of emails an email user receives per hour. The discrete random variable \(X\) takes on the values \(x = 0, 1, 2 \dotsc\). The random variable \(X\) has a Poisson distribution: \(X \sim P(147)\). Assume the interval of time is for 1 hour. The mean is \(\dfrac{144}{24}\) = 6 emails per hour.

1. What is the probability that an email user receives exactly 2 emails per hour?
2. What is the probability that an email user receives at most 2 emails per hour?

Solutions

Using a TI-84+ use the following functions.

1. \(P(x = 160) = \text{poissonpdf}(6, 2) \approx 0.0446\)
2. \(P(x \leq 160) = \text{poissoncdf}(6, 2) \approx 0.0620\)

Another approach is to use the formula to calculate the answers.

a. For this part \(\lambda\) = 6 and x = 2.

\(P\left(2\right)=\dfrac{e^{-6}\cdot6^2}{2!}=0.0446\)

b. For this part \(\lambda\) = 6 and the probabilities P(0), P(1), and P(2) must be computed and added together.

\(P\left(0\right)=\dfrac{e^{-6}\cdot6^0}{0!}=0.0025\)

\(P\left(1\right)=\dfrac{e^{-6}\cdot6^1}{1!}=0.0149\)

\(P\left(2\right)=\dfrac{e^{-6}\cdot6^2}{2!}=0.0446\)

\(P(x \le 2)\)= P(0) + P(1) + P(2) = 0.0025 + 0.0149 + 0.0446 = 0.0620

\(P(x \le 2)\) =0.0620

Example \(\PageIndex{4}\)

An office building contains 100 office spaces of 350 square feet. The carpets in the office space have around 2 stains per 350 square feet. For a sample of 3 office spaces of 350 square feet, compute the following probabilities.

1. What is the probability that the 3 office spaces have 8 stains?
2. What is the probability that the office spaces have at most 3 stains?

Solution

Using a TI-84+ use the following functions.

The average number of stains in the 3 office spaces is \(\lambda\) = \(\frac{2}{350}\cdot1050\) = 6

1. \(P(x = 8) = \text{poissonpdf}(6, 8) \approx 0.1033\)
2. \(P(x \leq 3) = \text{poissoncdf}(6, 3) \approx 0.1512\)

Another approach is to use the formula to calculate the answers.

a. Using the formula, compute p(8).

\(P\left(8\right)=\dfrac{e^{-6}\cdot6^8}{8!}=0.1033\)

b. Using the formula compute P(0), P(1), P(2), and P(3), and then add them together.

\(P\left(0\right)=\dfrac{e^{-6}\cdot6^0}{0!}=0.0025\)

\(P\left(1\right)=\dfrac{e^{-6}\cdot6^1}{1!}=0.0149\)

\(P\left(2\right)=\dfrac{e^{-6}\cdot6^2}{2!}=0.0446\)

\(P\left(3\right)=\dfrac{e^{-6}\cdot6^3}{3!}=0.0892\)

\(P(x \le 3)\)= P(0) + P(1) + P(2) + P(3) = 0.0025 + 0.0149 + 0.0446 + 0.0892 = 0.1512

\(P(x \le 3)\) =0.1512.

Example \(\PageIndex{5}\)

A publisher notices that there are 2 small typos in 500 pages of a typical book.

1. How many small typos does a 1,000-page book have on average?
2. What is the probability that a 1000-page book will have 5 small typos?
3. What is the probability that a 1000-page book will have more than 5 small typos?

Solutions

Using a TI-84+ use the following functions.

1. Let \(X =\) be the number of texts that a user sends or receives in one hour. The average number of small typos per 1000 pages of a book is \(\frac{2}{500}\cdot1000=4\).
2. \(X \sim P(4)\), so \(P(x = 5) = \text{poissonpdf}(4, 5) \approx 0.1563\)
3. \(P(x > 5) = 1 – P(x \leq 5) = 1 – \text{poissoncdf}(4, 4) \approx 1 – 0.6288 = 0.3712\)

Another approach is to use the formula to calculate the answers.

a. See part a above.

b. For the formula, compute p(5).

\(P\left(5\right)=\dfrac{e^{-4}\cdot4^5}{5!}=0.1563\)

c. Compute P(0), P(1), P(2), P(3), and P(4). Add them together and subtract the sum from 1.

\(P\left(0\right)=\dfrac{e^{-4}\cdot4^0}{0!}=0.0183\)

\(P\left(1\right)=\dfrac{e^{-4}\cdot4^1}{1!}=0.0733\)

\(P\left(2\right)=\dfrac{e^{-4}\cdot4^2}{4!}=0.1465\)

\(P\left(3\right)=\dfrac{e^{-4}\cdot4^3}{3!}=0.1954\)

\(P\left(4\right)=\dfrac{e^{-4}\cdot4^4}{4!}=0.1954\)

\(P(x > 5)\) = 1 - P(0) - P(1) - P(2) - P(3) - P(4) = 1 - 0.0183 - 0.0733 - 0.1465 - 0.1954 - 0.1954 = 1 - 0.6288 = 0.3712

\(P(x > 5)\) =0.3712

Example \(\PageIndex{6}\)

Atlanta’s Hartsfield-Jackson International Airport is the busiest in the world. On average there are 2,500 arrivals and departures each day.

1. How many airplanes arrive and depart the airport per hour?
2. What is the probability that there are exactly 100 arrivals and departures in one hour?
3. What is the probability that there are at most 100 arrivals and departures in one hour?

Solution

1. Let \(X =\) the number of airplanes arriving and departing from Hartsfield-Jackson in one hour. The average number of arrivals and departures per hour is \(\frac{2,500}{24} \approx 104.1667\).
2. \(X \sim P(104.1667)\), so \(P(x = 100) = \text{poissonpdf}(104.1667, 100) \approx 0.0366\).
3. \(P(x \leq 100) = \text{poissoncdf}(104.1667, 100) \approx 0.3651\).

The Poisson distribution can be used to approximate probabilities for a binomial distribution. This next example demonstrates the relationship between the Poisson and the binomial distributions. Let \(n\) represent the number of binomial trials and let \(p\) represent the probability of success for each trial. If \(n\) is large enough and \(p\) is small enough then the Poisson approximates the binomial very well. In general, \(n\) is considered “large enough” if it is greater than or equal to 20. The probability \(p\) from the binomial distribution should be less than or equal to 0.05. When the Poisson is used to approximate the binomial, we use the binomial mean \(\mu = np\). The variance of \(X\) is \(\sigma^{2} = \sqrt{\mu}\) and the standard deviation is \(\sigma = \sqrt{\mu}\). The Poisson approximation to a binomial distribution was commonly used in the days before technology made both values very easy to calculate.

Example \(\PageIndex{7}\)

On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?

Solution

Let \(X\) = the number of days with low seismic activity.

Using the binomial distribution:

• \(P(x = 10) = \text{binompdf}(200, .0102, 10) \approx\ 0.000039\)

Using the Poisson distribution:

• Calculate \(\mu = np = 200(0.0102) \approx 2.04\)
• \(P(x = 10) = \text{poissonpdf}(2.04, 10) \approx 0.000045\)

We expect the approximation to be good because \(n\) is large (greater than 20) and \(p\) is small (less than 0.05). The results are close—both probabilities reported are almost 0.

Example \(\PageIndex{8}\)

Add example text here.

Solution

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Example \(\PageIndex{9}\)

On May 13, 2013, starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?

Solution

Let \(X =\) the number of days with moderate seismic activity.

Using the binomial distribution: \(P(x = 5) = \text{binompdf}(100, 0.0143, 5) \approx 0.0115\)

Using the Poisson distribution:

• Calculate \(\mu = np = 100(0.0143) = 1.43\)
• \(P(x = 5) = \text{poissonpdf}(1.43, 5) = 0.0119\)

We expect the approximation to be good because \(n\) is large (greater than 20) and \(p\) is small (less than 0.05). The results are close—the difference between the values is 0.0004.

Authors

"5.4: Poisson Distribution" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY 4.0

Attributions

"4.7: Poisson Distribution", OpenStax is licensed under CC BY 4.0 / A derivative from the original work