Discrete Random Variables

24 Poisson Distribution

Another useful probability distribution is the Poisson distribution, or waiting time distribution. This distribution is used to determine how many checkout clerks are needed to keep the waiting time in line to specified levels, how may telephone lines are needed to keep the system from overloading, and many other practical applications. A modification of the Poisson, the Pascal, invented nearly four centuries ago, is used today by telecommunications companies worldwide for load factors, satellite hookup levels and Internet capacity problems. The distribution gets its name from Simeon Poisson who presented it in 1837 as an extension of the binomial distribution which we will see can be estimated with the Poisson.

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.
  2. The events are 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 and it is assumed that there is no relationship between when misspellings occur.
  3. The random variable X = the number of occurrences in the interval of interest.

A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let X = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question.

P(x < 5)

You notice that a news reporter says “uh,” on average, two times per broadcast. What is the probability that the news reporter says “uh” more than two times per broadcast.

This is a Poisson problem because you are interested in knowing the number of times the news reporter says “uh” during a broadcast.

a. What is the interval of interest?

a. one broadcast measured in minutes

b. What is the average number of times the news reporter says “uh” during one broadcast?

b. 2

c. Let X = ____________. What values does X take on?

c. Let X = the number of times the news reporter says “uh” during one broadcast.
x = 0, 1, 2, 3, …

d. The probability question is P(______).

d. P(x > 2)

Notation for the Poisson: P = Poisson Probability Distribution Function

X ~ P(μ)

Read this as “X is a random variable with a Poisson distribution.” The parameter is μ (or λ); μ (or λ) = the mean for the interval of interest. The mean is the number of occurrences that occur on average during the interval period.

The formula for computing probabilities that are from a Poisson process is:

P\left(x\right)=\frac{{\mu }^{x}{e}^{-\mu }}{x!}

where P(X) is the probability of X successes, μ is the expected number of successes based upon historical data, e is the natural logarithm approximately equal to 2.718, and X is the number of successes per unit, usually per unit of time.

In order to use the Poisson distribution, certain assumptions must hold. These are: the probability of a success, μ, is unchanged within the interval, there cannot be simultaneous successes within the interval, and finally, that the probability of a success among intervals is independent, the same assumption of the binomial distribution.

In a way, the Poisson distribution can be thought of as a clever way to convert a continuous random variable, usually time, into a discrete random variable by breaking up time into discrete independent intervals. This way of thinking about the Poisson helps us understand why it can be used to estimate the probability for the discrete random variable from the binomial distribution. The Poisson is asking for the probability of a number of successes during a period of time while the binomial is asking for the probability of a certain number of successes for a given number of trials.

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

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, …

If Leah receives, on the 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, μ = 0.75 for this problem.

X ~ P(0.75)

Find P(x > 1). P(x > 1) = 0.1734

Probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734.

The graph of X ~ P(0.75) is:

This graphs shows a poisson probability distribution. It has 5 bars that decrease in height from left to right. The x-axis shows values in increments of 1 starting with 0, representing the number of calls Leah receives within 15 minutes. The y-axis ranges from 0 to 0.5 in increments of 0.1.

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

According to a survey a university professor gets, on average, 7 emails per day. Let X = the number of emails a professor receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P(7). The mean is 7 emails.

  1. What is the probability that an email user receives exactly 2 emails per day?
  2. What is the probability that an email user receives at most 2 emails per day?
  3. What is the standard deviation?
  1. P\left(x=2\right)=\frac{{\mu }^{x}{e}^{\mathrm{-\mu }}}{\mathrm{x!}}=\frac{{7}^{2}{e}^{-7}}{2!}=0.022
  2. P\left(x\le 2\right)=\frac{{7}^{0}{e}^{-7}}{0!}+\frac{{7}^{1}{e}^{-7}}{1!}+\frac{{7}^{2}{e}^{-7}}{2!}=0.029
  3. Standard Deviation = \sigma =\sqrt{\mu }=\sqrt{7}\approx 2.65

Text message users receive or send an average of 41.5 text messages per day.

  1. How many text messages does a text message user receive or send per hour?
  2. What is the probability that a text message user receives or sends two messages per hour?
  3. What is the probability that a text message user receives or sends more than two messages per hour?
  1. Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is \frac{41.5}{24} ≈ 1.7292.
  2. P\left(x=2\right)=\frac{{\mu }^{x}{e}^{\mathrm{-\mu }}}{\mathrm{x!}}=\frac{{1.729}^{2}{e}^{-1.729}}{2!}=0.265
  3. P\left(x>2\right)=1-P\left(x\le 2\right)=1-\left[\frac{{7}^{0}{e}^{-7}}{0!}+\frac{{7}^{1}{e}^{-7}}{1!}+\frac{{7}^{2}{e}^{-7}}{2!}\right]=0.250

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?

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

Using the binomial distribution:

  • P\left(x=10\right)=\frac{200!}{10!\left(200-10\right)!}×{.0102}^{10}=0.000039

Using the Poisson distribution:

  • Calculate μ = np = 200(0.0102) ≈ 2.04
  • P\left(x=10\right)=\frac{{\mu }^{x}{e}^{\mathrm{-\mu }}}{\mathrm{x!}}=\frac{{2.04}^{10}{e}^{-2.04}}{10!}=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.

Estimating the Binomial Distribution with the Poisson Distribution

We found before that the binomial distribution provided an approximation for the hypergeometric distribution. Now we find that the Poisson distribution can provide an approximation for the binomial. We say that the binomial distribution approaches the Poisson. The binomial distribution approaches the Poisson distribution is as n gets larger and p is small such that np becomes a constant value. There are several rules of thumb for when one can say they will use a Poisson to estimate a binomial. One suggests that np, the mean of the binomial, should be less than 25. Another author suggests that it should be less than 7. And another, noting that the mean and variance of the Poisson are both the same, suggests that np and npq, the mean and variance of the binomial, should be greater than 5. There is no one broadly accepted rule of thumb for when one can use the Poisson to estimate the binomial.

As we move through these probability distributions we are getting to more sophisticated distributions that, in a sense, contain the less sophisticated distributions within them. This proposition has been proven by mathematicians. This gets us to the highest level of sophistication in the next probability distribution which can be used as an approximation to all of those that we have discussed so far. This is the normal distribution.

A survey of 500 seniors in the Price Business School yields the following information. 75% go straight to work after graduation. 15% go on to work on their MBA. 9% stay to get a minor in another program. 1% go on to get a Master’s in Finance.

What is the probability that more than 2 seniors go to graduate school for their Master’s in finance?

This is clearly a binomial probability distribution problem. The choices are binary when we define the results as “Graduate School in Finance” versus “all other options.” The random variable is discrete, and the events are, we could assume, independent. Solving as a binomial problem, we have:

Binomial Solution

n·p=500·0.01=5=µ
P\left(0\right)=\frac{500!}{0!\left(500-0\right)!}{0.01}^{0}\left(1-0.01{\right)}^{{500}^{{-}^{0}}}=0.00657
P\left(1\right)=\frac{500!}{1!\left(500-1\right)!}{0.01}^{1}\left(1-0.01{\right)}^{{500}^{{-}^{1}}}=0.03318
P\left(2\right)=\frac{500!}{2!\left(500-2\right)!}{0.01}^{2}\left(1-0.01{\right)}^{{500}^{{-}^{2}}}=0.08363

Adding all 3 together = 0.12339

1-0.12339=0.87661

Poisson approximation

n·p=500·0.01=5=\mu
n·p·\left(1-p\right)=500·0.01·\left(0.99\right)\approx 5={\sigma }^{2}=\mu
P\left(X\right)=\frac{{e}^{{\mathrm{-np}}^{}}\left(np{\right)}^{x}}{x!}=\left\{P\left(0\right)=\frac{{e}^{-5}·{5}^{0}}{0!}\right\}+\left\{P\left(1\right)=\frac{{e}^{-5}·{5}^{1}}{1!}\right\}+\left\{P\left(2\right)=\frac{{e}^{-5}·{5}^{2}}{2!}\right\}
0.0067+0.0337+0.0842=0.1247
1-0.1247=0.8753

An approximation that is off by 1 one thousandth is certainly an acceptable approximation.

References

“ATL Fact Sheet,” Department of Aviation at the Hartsfield-Jackson Atlanta International Airport, 2013. Available online at http://www.atl.com/about-atl/atl-factsheet/ (accessed February 6, 2019).

Center for Disease Control and Prevention. “Teen Drivers: Fact Sheet,” Injury Prevention & Control: Motor Vehicle Safety, October 2, 2012. Available online at http://www.cdc.gov/Motorvehiclesafety/Teen_Drivers/teendrivers_factsheet.html (accessed May 15, 2013).

“Children and Childrearing,” Ministry of Health, Labour, and Welfare. Available online at http://www.mhlw.go.jp/english/policy/children/children-childrearing/index.html (accessed May 15, 2013).

“Eating Disorder Statistics,” South Carolina Department of Mental Health, 2006. Available online at http://www.state.sc.us/dmh/anorexia/statistics.htm (accessed May 15, 2013).

“Giving Birth in Manila: The maternity ward at the Dr Jose Fabella Memorial Hospital in Manila, the busiest in the Philippines, where there is an average of 60 births a day,” theguardian, 2013. Available online at http://www.theguardian.com/world/gallery/2011/jun/08/philippines-health#/?picture=375471900&index=2 (accessed May 15, 2013).

“How Americans Use Text Messaging,” Pew Internet, 2013. Available online at http://pewinternet.org/Reports/2011/Cell-Phone-Texting-2011/Main-Report.aspx (accessed May 15, 2013).

Lenhart, Amanda. “Teens, Smartphones & Testing: Texting volum is up while the frequency of voice calling is down. About one in four teens say they own smartphones,” Pew Internet, 2012. Available online at http://www.pewinternet.org/~/media/Files/Reports/2012/PIP_Teens_Smartphones_and_Texting.pdf (accessed May 15, 2013).

“One born every minute: the maternity unit where mothers are THREE to a bed,” MailOnline. Available online at http://www.dailymail.co.uk/news/article-2001422/Busiest-maternity-ward-planet-averages-60-babies-day-mothers-bed.html (accessed May 15, 2013).

Vanderkam, Laura. “Stop Checking Your Email, Now.” CNNMoney, 2013. Available online at http://management.fortune.cnn.com/2012/10/08/stop-checking-your-email-now/ (accessed May 15, 2013).

“World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. http://www.world-earthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013).

Chapter 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 0.01) and the number of trials is “large” (greater than or equal to 25). Other rules of thumb are also suggested by different authors, but all recognize that the Poisson distribution is the limiting distribution of the binomial as n increases and p approaches zero.

The formula for computing probabilities that are from a Poisson process is:

P\left(x\right)=\frac{{\mu }^{x}{e}^{-\mu }}{x!}

where P(X) is the probability of successes, μ (pronounced mu) is the expected number of successes, e is the natural logarithm approximately equal to 2.718, and X is the number of successes per unit, usually per unit of time.

Formula Review

X ~ P(μ) means that X has a Poisson probability distribution where X = the number of occurrences in the interval of interest.

X takes on the values x = 0, 1, 2, 3, …

The mean μ or λ is typically given.

The variance is σ2 = μ, and the standard deviation is
\sigma \text{ = }\sqrt{\mu }.

When P(μ) is used to approximate a binomial distribution, μ = np where n represents the number of independent trials and p represents the probability of success in a single trial.

P\left(x\right)=\frac{{\mu }^{x}{e}^{-\mu }}{x!}

Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day.

Assume the event occurs independently in any given day. Define the random variable X.

<!– <solution id=”fs-idm218580928″> X ~ P(120) –>

What values does X take on?

0, 1, 2, 3, 4, …

What is the probability of getting 150 customers in one day?

<!– <solution id=”fs-idm98751248″> 0.0010 –>

What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day.

0.0485

What is the probability that the store will have more than 12 customers in the first hour?

<!– <solution id=”fs-idm122562880″> 0.2084 –>

What is the probability that the store will have fewer than 12 customers in the first two hours?

0.0214

Which type of distribution can the Poisson model be used to approximate? When would you do this?

<!– <solution id=”fs-idm58434144″> The Poisson distribution can approximate a binomial distribution, which you would do if the probability of success is small and the number of trials is large. –>


Use the following information to answer the next six exercises: On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age.

Assume the event occurs independently in any given day. In words, define the random variable X.

X = the number of U.S. teens who die from motor vehicle injuries per day.

X ~ _____(_____,_____)

<!– <solution id=”fs-idp14476864″> P(8) –>

What values does X take on?

0, 1, 2, 3, 4, …

For the given values of the random variable X, fill in the corresponding probabilities.

<!– <solution id=”fs-idp29073952″>

–>

Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S? Justify your answer numerically.

No

Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.

<!– <solution id=”id10039877″> No –>

HOMEWORK

The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let X = the number of calls received at noon.

  1. Find the mean and standard deviation of X.
  2. What is the probability that the office receives at most six calls at noon on Monday?
  3. Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon?
  4. What is the probability that the office receives more than eight calls at noon?
  1. X ~ P(5.5); μ = 5.5; \sigma \text{ = }\sqrt{5.5} ≈ 2.3452
  2. P(x ≤ 6) ≈ 0.6860
  3. There is a 15.7% probability that the law staff will receive more calls than they can handle.
  4. P(x > 8) = 1 – P(x ≤ 8) ≈ 1 – 0.8944 = 0.1056

The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let X = the number of births in an hour.

  1. Find the mean and standard deviation of X.
  2. Sketch a graph of the probability distribution of X.
  3. What is the probability that the maternity ward will deliver three babies in one hour?
  4. What is the probability that the maternity ward will deliver at most three babies in one hour?
  5. What is the probability that the maternity ward will deliver more than five babies in one hour?

<!– <solution id=”fs-idm62476208″> X ~ P(2.5); μ = 2.5; σ = 2.5 ≈ 1.5811

P(x = 3) = poissonpdf(2.5, 3) ≈ 0.2138 P(x ≤ 3) = poissoncdf(2.5, 3) ≈ 0.7576 P(x > 5) = 1 – P(x ≤ 5) = 1 – poissoncdf(2.5, 5) ≈ 0.0420 –>

A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions.

Let X = the number of defective bulbs in a string.

Using the Poisson distribution:

  • μ = np = 100(0.03) = 3
  • X ~ P(3)
  • P(x ≤ 4) ≈ 0.8153

Using the binomial distribution:

  • X ~ B(100, 0.03)
  • P(x ≤ 4) = 0.8179

The Poisson approximation is very good—the difference between the probabilities is only 0.0026.

The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen.

  1. In words, define the random variable X.
  2. List the values that X may take on.
  3. Find the probability that she has no children.
  4. Find the probability that she has fewer children than the Japanese average.
  5. Find the probability that she has more children than the Japanese average.

The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen.

  1. In words, define the Random Variable X.
  2. List the values that X may take on.
  3. Find the probability that she has no children.
  4. Find the probability that she has fewer children than the Spanish average.
  5. Find the probability that she has more children than the Spanish average .
  1. X = the number of children for a Spanish woman
  2. 0, 1, 2, 3,…
  3. 0.2299
  4. 0.5679
  5. 0.4321

Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces:

  1. In words, define the random variable X.
  2. List the values that X may take on.
  3. Give the distribution of X. X ~ _______
  4. Find the probability that she has no litters in one year.
  5. Find the probability that she has at least two litters in one year.
  6. Find the probability that she has exactly three litters in one year.

<!– <solution id=”fs-idp96709152″> X = the number of litters a fertile, female cat produces per year 0, 1, 2, 3, … X ~ P(3) 0.0498 0.8009 0.2240 –>

The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem.

  1. In words, define the random variable X.
  2. List the values that X may take on.
  3. How many cookies do we expect to have an extra fortune?
  4. Find the probability that none of the cookies have an extra fortune.
  5. Find the probability that more than three have an extra fortune.
  6. As n increases, what happens involving the probabilities using the two distributions? Explain in complete sentences.
  1. X = the number of fortune cookies that have an extra fortune
  2. 0, 1, 2, 3,… 144
  3. 4.32
  4. 0.0124 or 0.0133
  5. 0.6300 or 0.6264
  6. As n gets larger, the probabilities get closer together.

According to the South Carolina Department of Mental Health web site, for every 200 U.S. women, the average number who suffer from anorexia is one. Out of a randomly chosen group of 600 U.S. women determine the following.

  1. In words, define the random variable X.
  2. List the values that X may take on.
  3. Give the distribution ofX. X ~ _____(_____,_____)
  4. How many are expected to suffer from anorexia?
  5. Find the probability that no one suffers from anorexia.
  6. Find the probability that more than four suffer from anorexia.

<!– <solution id=”id13728739″> X = the number of women that suffer from anorexia 0, 1, 2, 3, …, 600 X ~ P(3) 3 0.0498 0.1847 –>

The chance of an IRS audit for a tax return with over ?25,000 in income is about 2% per year. Suppose that 100 people with tax returns over ?25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions.

  1. In words, define the random variable X.
  2. List the values that X may take on.
  3. How many are expected to be audited?
  4. Find the probability that no one was audited.
  5. Find the probability that at least three were audited.
  1. X = the number of people audited in one year
  2. 0, 1, 2, …, 100
  3. 2
  4. 0.1353
  5. 0.3233

Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school.

  1. In words, define the random variable X.
  2. List the values that X may take on.
  3. How many seniors are expected to have participated in after-school sports all four years of high school?
  4. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
  5. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.

<!– <solution id=”id18997392″> X = the number of seniors who participated in after-school sports all four years of high school 0, 1, 2, 3, … 60 4.8 Yes 4 –>

On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes.

  1. In words, define the random variable X.
  2. List the values that X may take on.
  3. On average, how many pieces of egg shell do you expect to be in the cake?
  4. What is the probability that there will not be any pieces of egg shell in the cake?
  5. Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any egg shell in any of the cakes?
  6. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why?
  1. X = the number of shell pieces in one cake
  2. 0, 1, 2, 3,…
  3. 1.5
  4. 0.2231
  5. 0.0001
  6. Yes

Use the following information to answer the next two exercises: The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week.

In words, the random variable X = _________________

  1. the number of times Mrs. Plum’s cats wake her up each week.
  2. the number of times Mrs. Plum’s cats wake her up each hour.
  3. the number of times Mrs. Plum’s cats wake her up each night.
  4. the number of times Mrs. Plum’s cats wake her up.

<!– <solution id=”id13453446″> a –>

Find the probability that her cats will wake her up no more than five times next week.

  1. 0.5000
  2. 0.9329
  3. 0.0378
  4. 0.0671

d

Glossary

Poisson Probability Distribution
a discrete random variable (RV) that counts the number of times a certain event will occur in a specific interval; characteristics of the variable:

  • The probability that the event occurs in a given interval is the same for all intervals.
  • The events occur with a known mean and independently of the time since the last event.

The distribution is defined by the mean μ of the event in the interval. The mean is μ = np. The standard deviation is \sigma \text{ = }\sqrt{\mu }. The probability of having exactly x successes in r trials is P\left(x\right)=\frac{{\mu }^{x}{e}^{-\mu }}{x!}. The Poisson distribution is often used to approximate the binomial distribution, when n is “large” and p is “small” (a general rule is that np should be greater than or equal to 25 and p should be less than or equal to 0.01).

License

Icon for the Creative Commons Attribution 4.0 International License

Introductory Business Statistics by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book