{"id":1245,"date":"2021-08-20T18:04:21","date_gmt":"2021-08-20T18:04:21","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/?post_type=chapter&p=1245"},"modified":"2023-12-05T09:12:03","modified_gmt":"2023-12-05T09:12:03","slug":"poisson-distribution-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/poisson-distribution-2\/","title":{"raw":"Poisson Probability Distribution Function","rendered":"Poisson Probability Distribution Function"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • State Poisson probabilities using mathematical notation<\/li>\r\n \t
  • Calculate the mean and standard deviation of a Poisson random variable<\/li>\r\n \t
  • Calculate a Poisson probability using technology<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n

    Notation for the Poisson: [latex]P=[\/latex] Poisson Probability Distribution Function<\/h2>\r\n

    [latex]X{\\sim}P(\\mu)[\/latex]<\/p>\r\nRead this as \"[latex]X[\/latex]\u00a0is a random variable with a Poisson distribution.\" The parameter is [latex]\\mu[\/latex]\u00a0(or [latex]\\lambda[\/latex]); [latex]\\mu[\/latex]\u00a0(or [latex]\\lambda[\/latex])[latex]=[\/latex] the mean for the interval of interest.\u00a0The standard deviation of the Poisson distribution with mean\u00a0\u00b5<\/em>\u00a0is\u00a0\u03a3<\/em>=\u221a\u03bc<\/em>\r\n

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    Example<\/h3>\r\nLeah'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?\r\n\r\n[reveal-answer q=\"877222\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"877222\"]\r\n\r\nLet [latex]X=[\/latex]\u00a0the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or [latex]\\frac{1}{4}[\/latex] hour.)\r\n

    [latex]x=0,1,2,3,...[\/latex]<\/p>\r\nIf Leah receives, on the average, six telephone calls in two hours, and there are eight 15-minute intervals in two hours, then Leah receives [latex]\\left(\\frac{1}{8}\\right)\\left(6\\right)=0.75[\/latex] calls in 15 minutes, on average. So, [latex]\\mu=0.75[\/latex] for this problem.\r\n

    [latex]X{\\sim}P(0.75)[\/latex]<\/p>\r\nFind [latex]P(x>1)[\/latex]. [latex]P(x>1)=0.1734[\/latex] (calculator or computer)\r\n

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    • Press 1 \u2013 and then press 2nd<\/sup> DISTR.<\/li>\r\n \t
    • Arrow down to poissoncdf. Press ENTER.<\/li>\r\n \t
    • Enter (.75,1).<\/li>\r\n \t
    • The result is [latex]P(x>1)=0.1734[\/latex].<\/li>\r\n<\/ul>\r\nNote:<\/strong>\u00a0The TI calculators use [latex]\\lambda[\/latex]\u00a0(lambda) for the mean.\r\n\r\nThe probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:\r\n

      [latex]P(x>1)=1\u2212\\text{poissoncdf}(0.75,1)[\/latex]<\/p>\r\nThe graph of [latex]X{\\sim}P(0.75)[\/latex] is:\r\n\r\n\"This\r\n\r\nThe [latex]y=[\/latex]-axis contains the probability of [latex]x=[\/latex] where [latex]X=[\/latex] the number of calls in 15 minutes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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      Try It<\/h3>\r\nA customer service center receives about ten emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? Use the TI-83+ or TI-84 calculator to find the answer.\r\n\r\n<\/div>\r\n
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      Example<\/h3>\r\nAccording 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=[\/latex]0, 1, 2 \u2026. The random variable\u00a0[latex]X[\/latex]\u00a0<\/em>has a Poisson distribution: [latex]X{\\sim}P(147)[\/latex]. The mean is 147 emails.\r\n
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      1. What is the probability that an email user receives exactly 160 emails per day?<\/li>\r\n \t
      2. What is the probability that an email user receives at most 160 emails per day?<\/li>\r\n \t
      3. What is the standard deviation?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"372382\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"372382\"]\r\n
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        1. [latex]P(x=160)=\\text{poissonpdf}(147, 160){\\approx}0.0180[\/latex]<\/li>\r\n \t
        2. [latex]P(x\\leq160)=\\text{poissoncdf}(147, 160){\\approx}0.8666[\/latex]<\/li>\r\n \t
        3. Standard Deviation[latex]=\\sigma=\\sqrt{\\mu}=\\sqrt{144}\\approx12.1244[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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          Try It<\/h3>\r\n

          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\u00a0[latex]X=[\/latex]\u00a0the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable\u00a0[latex]X[\/latex]\u00a0takes on the values\u00a0[latex]x[\/latex]\u00a0= 0, 1, 2 \u2026. The random variable\u00a0[latex]X[\/latex]\u00a0has a Poisson distribution: [latex]X{\\sim}P(187)[\/latex]. The mean is 187 text messages.<\/p>\r\n\r\n

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          1. What is the probability that a teen girl sends exactly 175 texts per day?<\/li>\r\n \t
          2. What is the probability that a teen girl sends at most 150 texts per day?<\/li>\r\n \t
          3. What is the standard deviation?<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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            Example<\/h3>\r\n

            Text message users receive or send an average of 41.5 text messages per day.<\/p>\r\n\r\n

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            1. How many text messages does a text message user receive or send per hour?<\/li>\r\n \t
            2. What is the probability that a text message user receives or sends two messages per hour?<\/li>\r\n \t
            3. What is the probability that a text message user receives or sends more than two messages per hour?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"518574\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"518574\"]\r\n
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              1. Let\u00a0[latex]X[\/latex]\u00a0= the number of texts that a user sends or receives in one hour. The average number of texts received per hour is [latex]\\frac{41.5}{24} \\approx 1.7292[\/latex]<\/span>.<\/li>\r\n \t
              2. [latex]X{\\sim}P(1.7292)[\/latex], so [latex]P(x=2)[\/latex] = poissonpdf(1.7292, 2) \u2248 0.2653<\/li>\r\n \t
              3. [latex]P(x>2) = 1\u2013P(x\u22642) = 1[\/latex] \u2013 poissoncdf(1.7292, 2) \u2248 1 \u2013 0.7495 = 0.2505<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n
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                Try It<\/h3>\r\n

                Atlanta\u2019s Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,500 arrivals and departures each day.<\/p>\r\n\r\n

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                1. How many airplanes arrive and depart the airport per hour?<\/li>\r\n \t
                2. What is the probability that there are exactly 100 arrivals and departures in one hour?<\/li>\r\n \t
                3. What is the probability that there are at most 100 arrivals and departures in one hour?<\/li>\r\n<\/ol>\r\n<\/div>\r\n
                  \r\n

                  Example<\/h3>\r\n

                  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?<\/p>\r\n[reveal-answer q=\"401654\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"401654\"]\r\n

                  Let [latex]X[\/latex]\u00a0= the number of days with low seismic activity.<\/p>\r\n

                  Using the binomial distribution:<\/p>\r\n\r\n

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                  • [latex]P(x=10)[\/latex] = binompdf(200, .0102, 10) \u2248 0.000039<\/li>\r\n<\/ul>\r\n

                    Using the Poisson distribution:<\/p>\r\n\r\n

                      \r\n \t
                    • Calculate\u00a0\u03bc<\/em>\u00a0=\u00a0np<\/em>\u00a0= 200(0.0102) \u2248 2.04<\/li>\r\n \t
                    • [latex]P(x=10)[\/latex] = poissonpdf(2.04, 10) \u2248 0.000045<\/li>\r\n<\/ul>\r\n

                      We expect the approximation to be good because [latex]n[\/latex]\u00a0is large (greater than 20) and\u00a0[latex]p[\/latex]\u00a0is small (less than 0.05). The results are close\u2014both probabilities reported are almost 0.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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                      Try It<\/h3>\r\n

                      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?<\/p>\r\n\r\n<\/div>\r\n