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Chapter 4: Discrete Random Variables

4.6 Poisson Distribution

Learning Objectives

By the end of this section, the student should be able to:

  • Identify the components of a Poisson experiment
  • Use the formulas for a Poisson random variable to compute the mean, variance, and standard deviation

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 number of events occurring has a constant average and the events are independent of the "time" since the last event. Continuing the book example, the average number of misspelled words is 5, and finding a word that is misspelled does not influence the page on which the next misspelled word lies. Additionally, two events shouldn't occur simultaneously.

The Poisson distribution is frequently used for experiments where there are a large number of possible but rare events. For instance, 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.

The random variable [latex]X =[/latex]the number of occurrences in the interval of interest. Because the average per interval is constant, we can find the average for any size interval by scaling to the size of the interval.

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, selected randomly. The time interval of interest is five minutes. What is the probability that the number of loaves put on the shelf in five minutes is three?

Let [latex]X =[/latex] 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-hour) is 12, then the average number of loaves put on the shelf in five minutes is [latex]\left(\frac{5}{30}\right)[/latex](12) = 2 loaves of bread.

The probability question asks you to find [latex]P(X = 3)[/latex].

Your Turn!

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.

  1. What is the interval of interest?
  2. What is the average number of times the news reporter says "uh" during one broadcast?
  3. Let [latex]X = \underline{\hspace{2cm}}[/latex]. What values does X take on?
  4. The probability question is [latex]P( \underline{\hspace{2cm}} )[/latex].
Solution
  1. one broadcast
  2. 2
  3. Let [latex]X =[/latex] the number of times the news reporter says "uh" during one broadcast. [latex]x = 0, 1, 2, 3, ...[/latex]
  4. [latex]P(x > 2)[/latex]

Your Turn!

An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reasons why this would be a Poisson distribution.

Solution

This problem wants to find the probability of events occurring in a fixed interval of time with a known average rate (5 per hour). The events are independent.

Your Turn!

The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. What is the average number of fish caught in 15 minutes?

Solution

[latex]\left( \frac{15}{60} \right) (8) = 2 \text{ fish}[/latex]

Notation and Mean

The notation for the Poisson Probability Distribution function is [latex]P[/latex], and we denote "[latex]X[/latex] is a random variable with a Poisson probability distribution as [latex]X \sim P(\lambda)[/latex]. The parameter is [latex]\lambda[/latex], which is also the mean [latex]\mu[/latex]. So, we may also write [latex]X \sim P(\mu)[/latex].

The mean is given as part of a Poisson experiment, [latex]\mu = \lambda[/latex].

The formula for the variance is [latex]\sigma^2 = \lambda = \mu[/latex], that is, the average is also the variance of a Poisson random variable.

The formula for the standard deviation is [latex]\sigma = \sqrt{\lambda}[/latex].

Example

Leah's school email receives an average of six emails between 8 a.m. and 10 a.m. each day. What is the probability that Leah receives more than one email in the next 15 minutes?

Let [latex]X =[/latex] the number of emails Leah receives in 15 minutes. (The interval of interest is 15 minutes or [latex]\frac{1}{4}[/latex] hour) and [latex]x = 0, 1, 2, 3, ...[/latex]

If Leah receives, on the average, six emails in two hours, and there are eight 15 minute intervals in two hours, then Leah receives [latex]\left(\frac{1}{8}\right)(6) = 0.75[/latex] emails in 15 minutes, on average. So, [latex]\lambda = \mu = 0.75[/latex] for this problem, and [latex]X \sim P(0.75)[/latex].

The Poisson distribution is discrete, and thus [latex]x > 1[/latex] includes all whole numbers through infinity.

The solution is to subtract the probability less than 1 thus [latex]1 - [P(x = 0) + P(x = 1)][/latex].

[latex]P(x)=\frac{\mu^{x}e^{-\mu}}{x!}[/latex]

[latex]P(x > 1) = 1 - P(x \lte 1) = 1 - (P(x = 0) + P(x = 1))[/latex]

[latex]= 1 - (\frac{0.75^{0}e^{-0.75}}{0!} + \frac{0.75^{1}e^{-0.75}}{1!})[/latex]

[latex]= 1 - (\frac{(1)(0.474)}{1} + \frac{(0.75)(0.4724)}{1})[/latex]

[latex]= 1 - (0.4724+0.3543)=0.1733[/latex]

A Poisson probability distribution histogram. The x-axis represents the number of calls Leah receives within 15 minutes.
Figure 1. Graph of [latex]X \sim P(0.75)[/latex]

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

Large values of [latex]X =[/latex] with large [latex]\lambda[/latex] make the formula of [latex]P(x)=\frac{\mu^{x}e^{-\mu}{x!}[/latex] overflow in even a scientific calculator. It is recommended to use technology to do calculations for those problems.

Example

According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let [latex]X=[/latex] the number of emails an email user receives per day. The discrete random variable [latex]X[/latex] takes on the values [latex]x= 0, 1, 2, \ldots[/latex]. The random variable [latex]X[/latex] has a Poisson distribution: [latex]X \sim P(147)[/latex]. The mean is 147 emails. What is the standard deviation?

Solution

Standard Deviation = [latex]\sigma =\sqrt{\mu }=\sqrt{147}\approx 12.1244[/latex]

    The Poisson distribution can be used to approximate probabilities for a binomial distribution. The Poisson approximation to a binomial distribution was commonly used in the days before technology made both values very easy to calculate.

    Let [latex]n[/latex] represent the number of binomial trials and let [latex]p[/latex] represent the probability of a success for each trial. If [latex]n[/latex] is large enough and [latex]p[/latex] is small enough then the Poisson approximates the binomial very well. In general, [latex]n[/latex] is considered “large enough” if it is greater than or equal to 20. The probability [latex]p[/latex] 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 [latex]\mu = np[/latex].

    Example

    On a specific day in May 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 [latex]X =[/latex] the number of days with low seismic activity.

    Using the binomial distribution: [latex]P(X = x) = \frac{n!}{x!(n-x)!} p^x q^{(n-x)}[/latex]

    [latex]P(x = 10) = \frac{200!}{10!(200-10)!} 0.0102^{10} 0.9898^{190} = 0.000039[/latex]

    Using the Poisson distribution: [latex]P(x)=\frac{\mu^{x}e^{-\mu}{x!}[/latex]

    Calculate [latex]\mu = np = 200(0.0102) \approx 2.04[/latex].

    [latex]P(x = 10) = \frac{(2.04)^{10}e^{-2.04}{10!} = 0.000045[/latex]

    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.

    Videos

    Below are helpful videos for the content covered in this Section. Videos are provided from YouTube.

    Section 4.6 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).

    Formula Review

    • [latex]X \sim P(\mu)[/latex] means that [latex]X[/latex] has a Poisson probability distribution where [latex]X= \text{the number of occurrences in the interval of interest}[/latex].
    • [latex]X[/latex] takes on the values [latex]x = 0, 1, 2, 3, \ldots[/latex]
    • The mean [latex]\mu[/latex] is typically given.
    • The variance is [latex]\sigma^2 = \mu[/latex], and the standard deviation is [latex]\sigma \text{ = }\sqrt{\mu }[/latex].
    • When [latex]P(\mu)[/latex] is used to approximate a binomial distribution, [latex]\mu = np[/latex] where [latex]n[/latex] represents the number of independent trials and [latex]p[/latex] represents the probability of success in a single trial.

    Section 4.6 Practice

    On average, a clothing store gets 120 customers per day.

    1. Assume the event occurs independently on any given day. Define the random variable [latex]X[/latex].
    2. What values does [latex]X[/latex] take on?
    3. What is the probability of getting 150 customers in one day?
    4. What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day.
    5. What is the probability that the store will have more than 12 customers in the first hour?
    6. What is the probability that the store will have fewer than 12 customers in the first two hours?
    Solution
    1. [latex]X \sim P(120)[/latex]
    2. [latex]0, 1, 2, 3, 4, …[/latex]
    3. 0.0010
    4. 0.0485
    5. 0.2084
    6. 0.0214

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

    Solution

    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.

    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.

    1. Assume the event occurs independently on any given day. In words, define the random variable [latex]X[/latex].
    2. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    3. What values does [latex]X[/latex] take on?
    4. 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.
    5. 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
    1. [latex]X =[/latex] the number of U.S. teens who die from motor vehicle injuries per day.
    2. [latex]P(8)[/latex]
    3. [latex]0, 1, 2, 3, 4, ...[/latex]
    4. No
    5. No

    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 [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. Find the probability that she has no children.
    5. Find the probability that she has fewer children than the Japanese average.
    6. Find the probability that she has more children than the Japanese average.

    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 [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. How many cookies do we expect to have an extra fortune?
    5. Find the probability that none of the cookies have an extra fortune.
    6. Find the probability that more than three have an extra fortune.
    7. As [latex]n[/latex] increases, what happens involving the probabilities using the two distributions? Explain in complete sentences.
    Solution
    1. [latex]X =[/latex] the number of fortune cookies that have an extra fortune
    2. [latex]0, 1, 2, 3,... 144[/latex]
    3. [latex]X \sim B(144, 0.03)[/latex] or [latex]P(4.32)[/latex]
    4. 4.32
    5. 0.0124 or 0.0133
    6. 0.6300 or 0.6264
    7. As [latex]n[/latex] gets larger, the probabilities get closer together.

    According to the South Carolina Department of Mental Health website, 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 [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    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
    1. [latex]X =[/latex] the number of women that suffer from anorexia
    2. [latex]0, 1, 2, 3, ..., 600[/latex]
    3. [latex]X \sim P(3)[/latex]
    4. 3
    5. 0.0498
    6. 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 answer the following questions.

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. How many are expected to be audited?
    5. Find the probability that no one was audited.
    6. Find the probability that at least three were audited.
    Solution
    1. [latex]X =[/latex] the number of people audited in one year
    2. [latex]0, 1, 2, ..., 100[/latex]
    3. [latex]X \sim P(2)[/latex]
    4. 2
    5. 0.1353
    6. 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 [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. How many seniors are expected to have participated in after-school sports all four years of high school?
    5. 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.
    6. 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
    1. [latex]X =[/latex] the number of seniors who participated in after-school sports all four years of high school
    2. [latex]0, 1, 2, 3, ... 60[/latex]
    3. [latex]X \sim P(4.8)[/latex]
    4. 4.8
    5. Yes
    6. 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 [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. On average, how many pieces of egg shell do you expect to be in the cake?
    5. What is the probability that there will not be any pieces of egg shell in the cake?
    6. 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?
    7. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why?
    Solution
    1. [latex]X =[/latex] the number of shell pieces in one cake
    2. [latex]0, 1, 2, 3,...[/latex]
    3. [latex]X \sim P(1.5)[/latex]
    4. 1.5
    5. 0.2231
    6. 0.0001
    7. Yes

    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.

    1. In words, the random variable [latex]X = \underline{\hspace{2cm}}[/latex]
      • the number of times Mrs. Plum’s cats wake her up each week.
      • the number of times Mrs. Plum’s cats wake her up each hour.
      • the number of times Mrs. Plum’s cats wake her up each night.
      • the number of times Mrs. Plum’s cats wake her up.
    2. Find the probability that her cats will wake her up no more than five times next week.
      • 0.5000
      • 0.9329
      • 0.0378
      • 0.0671
    Solution
    1. the number of times Mrs. Plum’s cats wake her up each week.
    2. 0.0671

    References

    “ATL Fact Sheet,” Department of Aviation at the Hartsfield-Jackson Atlanta International Airport, 2013. Available online at http://www.atlanta-airport.com/Airport/ATL/ATL_FactSheet.aspx (accessed May 15, 2013).

    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 volume 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).

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