Chapter 4: Discrete Random Variables

4.10 Using Technology for the Poisson Distribution

Learning Objectives

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

  • use technology to solve problems involving the Poisson Distribution

This is another section to show technology tools that can be used to help with the statistics concepts taught in this chapter. This section will cover tools to help solve problems involving the Poisson Distribution.

Graphing Calculator

Similar to all the other technology sections, the main tool introduced will be the TI-83, 83+, 84, or 84+ Graphing Calculator. The graphing calculator can help to find the probabilities for the poisson probability distribution.

 

Using the TI-83, 83+, 84, 84+ Calculator for Poisson Distributions

  1. Access [DISTR] by pressing 2nd key, vars key. (Note: DISTR stands for "Distributions")
  2. The functions for the Poisson Distribution are:
    • <poissonpdf()> corresponds to P(X = x)
    • <poissoncdf()> corresponds to P(X ≤ x)

The syntax for the instructions is as follows:

  • poissonpdf(λ, x) corresponds to P(X = x)
  • poissoncdf(λ, x) corresponds to P(X ≤ x)

Legend

  • A blank calculator button represents a button press
  • [ ] represents yellow command or green letter behind a key
  • < > represents items on the screen

 

Let's use these calculator functions with some of the same examples we did in 4.6 Poisson Distribution.

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].

Find the probability that Leah receives more than one email in the next 15 minutes

For this problem, we are asking to find [latex]P(X > 1)[/latex]. Note, the TI calculators use λ (lambda) for the mean. We can input [latex]1 - \text{poissoncdf}(0.75, 1)[/latex]. The result is [latex]P(X > 1) = 0.1734.[/latex]. Let's see another example.

Your Turn!

Atlanta’s Hartsfield-Jackson International Airport is the busiest airport 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 [latex]X =[/latex] the number of airplanes arriving and departing from Hartsfield-Jackson in one hour. The average number of arrivals and departures per hour is [latex]\frac{2,500}{24} \approx 104.1667[/latex].
  2. [latex]X \sim P(104.1667)[\latex] so [latex]P(X = 100)=\text{poissonpdf}(104.1667, 100) ≈ 0.0366[/latex]
  3. [latex]P(X \le 100)= \text{poissoncdf}(104.1667,100) \approx 0.3651[/latex]

 

Note

To compute poisson probabilities on a graphing calculator, the syntax is as follows:

  • To calculate [latex]P(X = x)[/latex]: poissonpdf([latex]\lambda[/latex], x).
  • To calculate [latex]P(X \leq x)[/latex]: poissoncdf([latex]\lambda[/latex], x).
  • If you wanted to instead find [latex]P(X>x)[/latex], use 1 - poissoncdf([latex]\lambda[/latex], x).

Remember, the TI calculators use [latex]\lambda[/latex] (lambda) for the mean.

 

Here's some more examples, with solution, for you to try, from the problems covered in the Section.

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.

  1. What is the probability that an email user receives exactly 160 emails per day?
  2. What is the probability that an email user receives at most 160 emails per day?
Solution
  1. [latex]P(x = 160) = \text{poissonpdf}(147, 160) ≈ 0.0180[/latex]
  2. [latex]P(x \leq 160) = \text{poissoncdf}(147, 160) ≈ 0.8666[/latex]

Your Turn!

According to a recent poll by the Pew Internet Project, girls between the ages of 14 and 17 send an average of 187 text messages each day. Let [latex]X =[/latex]the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable [latex]X[/latex] takes on the values [latex]x = 0, 1, 2 …[/latex]. The random variable [latex]X[/latex] has a Poisson distribution: [latex]X \sim P(187)[/latex]. The mean is 187 text messages.

  1. What is the probability that a teen girl sends exactly 175 texts per day?
  2. What is the probability that a teen girl sends at most 150 texts per day?
Solution
  1. [latex]P(X=175)=\text{poissonpdf}(187,175) \approx 0.0203[/latex]
  2. [latex]P(X\le150)= \text{poissoncdf} (187,150) \approx 0.0030[/latex]

    Poisson Distribution and Binomial Distribution

    Remember, we saw in 4.6 Poisson Distribution that the Poisson distribution can be used to approximate probabilities for a Binomial distribution. It was mentioned that the Poisson approximation to a Binomial distribution was commonly used in the days before technology made both values very easy to calculate. Well, now we have the technology! But let's see how we can use it for these problems. Remember the set-up:

    • Let [latex]n[/latex] represent the number of binomial trials
    • 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].

    Here's some examples:

    Example

    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 similar?

    Your Turn!

    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?

     

    Helpful Videos for the Graphing Calculator

    Below are links to helpful videos for using the graphing calculator for the concepts covered on this page:

     

    Additional Technology Tools

    In addition to the graphing calculator, there are some additional technology tools that can be used for the concepts covered on this page. Below are links to helpful videos for those tools:

     

    Microsoft Excel

    In Microsoft Excel, both Poisson probabilities are computed using POISSON.DIST(x, [latex]\lambda[/latex], True/False), where False computes [latex]P(X=x)[/latex] and is equivalent to poissonpdf(), and True computes [latex]P(X \leq x)[/latex] and is equivalent to poissoncdf().

     

    Section Practice

    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 [latex]X =[/latex] the number of calls received at noon.

    1. Find the mean and standard deviation of [latex]X[/latex].
    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?
    Solution
    1. [latex]X \sim P(5.5); \mu = 5.5; \sigma = \sqrt{5.5} \approx 2.3452[/latex]
    2. [latex]P(x \le 6) = \text{poissoncdf} (5.5, 6) \approx 0.6860[/latex]
    3. There is a 15.7% probability that the law staff will receive more calls than they can handle.
    4. [latex]P(x > 8) = 1 – P(x \le 8) = 1 – \text{poissoncdf} (5.5, 8) \approx 1 – 0.8944 = 0.1056[/latex]

    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 [latex]X =[/latex] the number of births in an hour.

    1. Find the mean and standard deviation of [latex]X[/latex].
    2. Sketch a graph of the probability distribution of [latex]X[/latex].
    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?

    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 [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 Spanish average.
    6. Find the probability that she has more children than the Spanish average.
    Solution
    1. [latex]X =[/latex] the number of children for a Spanish woman
    2. [latex]0, 1, 2, 3,...[/latex]
    3. [latex]X \sim P(1.47)[/latex]
    4. 0.2299
    5. 0.5679
    6. 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 [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}}[/latex]
    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
    1. [latex]X =[/latex] the number of litters a fertile, female cat produces per year
    2. [latex]0, 1, 2, 3, ...[/latex]
    3. [latex]X \sim P(3)[/latex]
    4. 0.0498
    5. 0.8009
    6. 0.2240

    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.

    Solution

    Let [latex]X =[/latex]the number of defective bulbs in a string.

    1. Using the Poisson distribution:
      • [latex]\mu = np = 100(0.03) = 3[/latex]
      • [latex]X \sim P(3)[/latex]
      • [latex]P(x \le 4) = \text{poissoncdf} (3, 4) \approx 0.8153[/latex]
    2. Using the binomial distribution:
      • [latex]X \sim B(100, 0.03)[/latex]
      • [latex]P(x \le 4) = \text{binomcdf} (100, 0.03, 4) \approx 0.8179[/latex]
      • The Poisson approximation is very good—the difference between the probabilities is only 0.0026.

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