"

4.5 The Poisson Distribution

LEARNING OBJECTIVES

  • Recognize the Poisson probability distribution and apply it appropriately.

A Poisson experiment has the following characteristics:

  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 average, there are five words spelled incorrectly in [latex]100[/latex] pages. The interval is the [latex]100[/latex] pages.
  2. The Poisson distribution may be used to approximate the binomial distribution if the probability of success is “small” (such as [latex]0.01[/latex]) and the number of trials is “large” (such as [latex]1,000[/latex]).

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

EXAMPLE

The average number of loaves of bread put on a shelf in a bakery in a half-hour period is [latex]12[/latex]. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. We want to find the probability that three loaves are put on the shelf in any five minute interval. Why is this a Poisson experiment?

Solution

Let [latex]X[/latex] 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 [latex]12[/latex], then the average number of loaves put on the shelf in five minutes is [latex]\displaystyle{\frac{5}{30} \times 12=2}[/latex] loaves of bread.

This is a Poisson experiment because we are interested in the number of loaves happening in a fixed interval (five minutes) with an average of [latex]2[/latex] loaves in any five minutes.

Calculating Poisson Probabilities

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.

Visit the Microsoft page for more information about the Poisson.dist function.

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

Leah receives about six telephone calls every two hours.

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

Solution

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

    The probability that Leah receives [latex]4[/latex] calls in the next two hours is [latex]13.39\%[/latex].

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

    The probability that Leah receives at most [latex]6[/latex] calls in the next two hours is [latex]91.61\%[/latex].

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

    The probability that Leah receives at most [latex]6[/latex] calls in the next two hours is [latex]42.32\%[/latex].

TRY IT

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

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

EXAMPLE

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

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

Solution

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

    The probability a user receives fewer than [latex]160[/latex] emails over a six-hour period is [latex]84.86\%[/latex].

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

    The probability a user receives more than [latex]40[/latex] emails over a two-hour period is [latex]89.02\%[/latex].

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

    The probability a user receives at least [latex]600[/latex] emails over a [latex]24[/latex]-hour period is [latex]31.58\%[/latex].

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

    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 [latex]25[/latex] parts from [latex]100[/latex] metres of sheet metal.

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

Video: “The Poisson Distribution – explained with examples and illustrated using Excel – statistics Help” by Dr Nic’s Maths and Stats [7:49] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.

Exercises

  1. On average, a clothing store gets [latex]120[/latex] customers per day. Assume the store is open [latex]12[/latex] hours each day.
    1. What is the probability that the store gets [latex]150[/latex] customers in one day?
    2. What is the probability that the store gets [latex]35[/latex] customers in a 4-hour period?
    3. What is the probability that the store gets more than [latex]12[/latex] customers in a 1-hour period?
    4. What is the probability that the store gets fewer than [latex]12[/latex] customers in a 2-hour period?
    5. What is the probability that the store gets between [latex]100[/latex] and [latex]130[/latex] customers in one day?
    6. What is the probability that the store gets at most than [latex]20[/latex] customers in a 3-hour period?
    7. What is the probability that the store gets at least [latex]70[/latex] customers in a 6-hour period?
    Click to see Answer
    1. [latex]0.001[/latex]
    2. [latex]0.0485[/latex]
    3. [latex]0.2084[/latex]
    4. [latex]0.0214[/latex]
    5. [latex]0.8036[/latex]
    6. [latex]0.0353[/latex]
    7. [latex]0.1118[/latex]

     

  2. On average, [latex]8[/latex] teens in the U.S. die from motor vehicle injuries in a 24-hour period.
    1. What is the probability that at most [latex]3[/latex] teens die from motor vehicle injuries in a 24-hour period?
    2. What is the probability that between [latex]8[/latex] and [latex]12[/latex] teens die from motor vehicle injuries in a 48-hour period?
    3. What is the probability that at least [latex]3[/latex] teens die from motor vehicle injuries in a 6-hour period?
    4. What is the probability that fewer than [latex]7[/latex] teens die from motor vehicle injuries in a 12-hour period?
    5.  Is it likely that there will be no teens killed from motor vehicle injuries in any 24-hour period in the U.S.? Justify your answer numerically.
    6. Is it likely that there will be more than [latex]20[/latex] teens killed from motor vehicle injuries in any 24-hour period in the U.S.? Justify your answer numerically.
    Click to see Answer
    1. [latex]0.0424[/latex]
    2. [latex]0.1711[/latex]
    3. [latex]0.3233[/latex]
    4. [latex]0.8893[/latex]
    5. Not likely because the probability that [latex]0[/latex] teens die from motor vehicle injuries in any 24-hour period is [latex]0.0003[/latex].
    6. Not likely because the probability that more than [latex]20[/latex] teens die from motor vehicle injuries in any 24-hour period is [latex]0.000094[/latex].

     

  3. The switchboard in a law office receives an average of [latex]6[/latex] calls in an hour.
    1. What is the probability that the office receives exactly [latex]4[/latex] calls in a 30-minute period?
    2. What is the probability that the office receives at most [latex]30[/latex] calls in a 4-hour period?
    3. What is the probability that the office receives at least [latex]3[/latex] calls in a 20-minute period?
    4. What is the probability that the office receives less than [latex]45[/latex] calls in an 8-hour period?
    5. What is the probability that the office receives more than [latex]10[/latex] calls in a 1-hour period?
    6. What is the probability that the office receives between [latex]15[/latex] and [latex]20[/latex] calls in a 2-hour period?
    Click to see Answer
    1. [latex]0.168[/latex]
    2. [latex]0.9041[/latex]
    3. [latex]0.3233[/latex]
    4. [latex]0.3131[/latex]
    5. [latex]0.0426[/latex]
    6. [latex]0.2164[/latex]

     

  4. 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 [latex]60[/latex] births in a 24-hour period.
    1. What is the probability that the maternity ward will deliver [latex]3[/latex] babies in a 1-hour period?
    2. What is the probability that the maternity ward will deliver more than [latex]70[/latex] babies in a 24-hour period?
    3. What is the probability that the maternity ward will deliver at most [latex]13[/latex]0 babies in a 48-hour period?
    4. What is the probability that the maternity ward will deliver between [latex]10[/latex] and [latex]20[/latex] babies in a 6-hour period?
    5. What is the probability that the maternity ward will deliver at least [latex]200[/latex] babies in a 72-hour period?
    6. What is the probability that the maternity ward will deliver fewer than [latex]8[/latex] babies in a 4-hour period?
    Click to see Answer
    1. [latex]0.2138[/latex]
    2. [latex]0.0902[/latex]
    3. [latex]0.8315[/latex]
    4. [latex]0.8472[/latex]
    5. [latex]0.0749[/latex]
    6. [latex]0.2202[/latex]

     

  5. Fertile female cats produce an average of [latex]3[/latex] litters per year. Suppose that one fertile female cat is randomly chosen.
    1. Find the probability that the cat has no litters in one year.
    2. Find the probability that the cat has at least two litters in one year.
    3. Find the probability that the cat has exactly three litters in one year.
    Click to see Answer
    1. [latex]0.0498[/latex]
    2. [latex]0.8009[/latex]
    3. [latex]0.224[/latex]

     

  6. On average, Pierre, an amateur chef, drops [latex]3[/latex] pieces of eggshell into every two cake batters he makes.
    1. What is the probability that there will be more than [latex]5[/latex] pieces of eggshell in any two cake batters?
    2. What is the probability that there will be at most [latex]20[/latex] pieces of eggshell in a dozen cake batters?
    3. What is the probability that there will not be any pieces of eggshell in any single cake batter?
    4. What is the probability that there will be fewer than [latex]10[/latex] pieces of eggshell in any four cake batter?
    5. What is the probability that there will be at least [latex]15[/latex] pieces of eggshell in any six-cake batter?
    6. What is the probability that there will be between [latex]2[/latex] and [latex]4[/latex] pieces of eggshell in any two cake batter?
    7. Is it possible for there to be [latex]7[/latex] pieces of eggshell in any single cake batter? Why?
    Click to see Answer
    1. [latex]0.0839[/latex]
    2. [latex]0.9884[/latex]
    3. [latex]0.2231[/latex]
    4. [latex]0.9161[/latex]
    5. [latex]0.0415[/latex]
    6. [latex]0.6161[/latex]
    7. It is possible but unlikely because the probability of [latex]7[/latex] pieces of eggshell in any single cake batter is [latex]0.0008[/latex].

     

  7. On a particular nature trail in a national park, deer are spotted at a rate of [latex]1[/latex] deer every [latex]3[/latex] kilometres.
    1. What is the probability that exactly [latex]4[/latex] deer are spotted in any [latex]6[/latex] kilometre stretch of trail?
    2. What is the probability that more than [latex]2[/latex] deer are spotted in any [latex]1.5[/latex] kilometre stretch of trail?
    3. What is the probability that between [latex]6[/latex] and [latex]10[/latex] deer are spotted in any 9-kilometre stretch of trail?
    4. What is the probability that at most [latex]3[/latex] deer are spotted in any [latex]6[/latex] kilometre stretch of trail?
    5. What is the probability that at least [latex]5[/latex] deer are spotted in any [latex]3[/latex] kilometre stretch of trail?
    6. What is the probability that fewer than [latex]7[/latex] deer are spotted in any [latex]9[/latex] kilometre stretch of trail?
    7. If someone walks the entire [latex]12[/latex] kilometres of the trail, is it likely that they will not see any deer? Why?
    Click to see Answer
    1. [latex]0.0902[/latex]
    2. [latex]0.0144[/latex]
    3. [latex]0.0836[/latex]
    4. [latex]0.8571[/latex]
    5. [latex]0.0037[/latex]
    6. [latex]0.9665[/latex]
    7. It is unlikely they will not see any deer because the probability of seeing [latex]0[/latex] deer in any [latex]12[/latex] kilometres is [latex]0.0183[/latex].

     

“4.5 The Poisson Distribution” and “4.6 Exercises” from Introduction to Statistics by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Introduction to Statistics - Second Edition Copyright © 2025 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book