{"id":1223,"date":"2021-08-20T17:00:23","date_gmt":"2021-08-20T17:00:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/?post_type=chapter&p=1223"},"modified":"2023-12-05T09:09:16","modified_gmt":"2023-12-05T09:09:16","slug":"binomial-distribution-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/binomial-distribution-2\/","title":{"raw":"Binomial Probability Distribution Function","rendered":"Binomial Probability Distribution Function"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • State binomial probabilities using mathematical notation<\/li>\r\n \t
  • Calculate the mean and standard deviation of a binomial random variable<\/li>\r\n \t
  • Calculate a binomial probability using technology<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n

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

    [latex]X\\sim{B}(n,p)[\/latex]<\/p>\r\nRead this as \"[latex]X[\/latex] is a random variable with a binomial distribution.\" The parameters are [latex]n[\/latex] and [latex]p[\/latex]; [latex]n=[\/latex]\u00a0number of trials, [latex]p=[\/latex] probability of a success on each trial.\r\n

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    Example<\/h3>\r\nIt has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?\r\n\r\nLet [latex]X=[\/latex] the number of workers who have a high school diploma but do not pursue any further education.\r\n\r\n[latex]X[\/latex] takes on the values 0, 1, 2, ..., 20 where [latex]n=20[\/latex], [latex]p=0.41[\/latex], and [latex]q=1\u20130.41=0.59[\/latex]. [latex]X\\sim{B}(20,0.41)[\/latex]\r\n\r\nFind [latex]P(x\\leq12)[\/latex]. [latex]P(x\\leq12)=0.9738[\/latex]. (calculator or computer)\r\n\r\n[reveal-answer q=\"156496\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"156496\"]\r\n
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    • Go into 2nd DISTR. The syntax for the instructions are as follows:<\/li>\r\n \t
    • To calculate ([latex]x=[\/latex] value): binompdf(<\/strong>n<\/em>, p<\/em><\/strong>, number)<\/strong> if \"number\" is left out, the result is the binomial probability table.<\/li>\r\n \t
    • To calculate P<\/em>([latex]x\\leq[\/latex] value): binomcdf(<\/strong>n<\/em>, p<\/em><\/strong>, number)<\/strong> if \"number\" is left out, the result is the cumulative binomial probability table.<\/li>\r\n \t
    • For this problem: After you are in 2nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,0.41,12). The result is\u00a0[latex]P(x\\leq12)=0.9738[\/latex].<\/li>\r\n<\/ul>\r\n

      Note<\/h4>\r\nIf you want to find [latex]P(x=12)[\/latex], use the pdf (binompdf). If you want to find [latex]P(x>12)[\/latex], use [latex]1-\\text{binomcdf}(20,0.41,12)[\/latex].\r\n\r\nThe probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.\r\n\r\nThe graph of [latex]X\\sim{B}(20,0.41)[\/latex] is as follows:\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 workers who have only a high school diploma.\r\n\r\nThe number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, [latex]\\mu=np=(20)(0.41)=8.2[\/latex].\r\n\r\nThe formula for the variance is [latex]\\sigma^{2}=npq[\/latex]. The standard deviation is [latex]\\sqrt{{{n}{p}{q}}}[\/latex].\r\n

      [latex]\\sigma=\\sqrt{(20)(0.41)(0.59)}={2.20}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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      Try It<\/h3>\r\nAbout 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer.\r\n\r\n[reveal-answer q=\"226275\"]Show Solution[\/reveal-answer][hidden-answer a=\"226275\"][latex]P(x\\leq14)=0.9695[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n
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      Example<\/h3>\r\nIn the 2013 Jerry's Artarama<\/em> art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let [latex]X=[\/latex]\u00a0the number of pages that feature signature artists.\r\n
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      1. What values does [latex]x[\/latex] take on?<\/li>\r\n \t
      2. What is the probability distribution? Find the following probabilities:\r\n
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        1. the probability that two pages feature signature artists<\/li>\r\n \t
        2. the probability that at most six pages feature signature artists<\/li>\r\n \t
        3. the probability that more than three pages feature signature artists.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        4. Using the formulas, calculate the (a) mean and (b) standard deviation.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"81853\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"81853\"]\r\n
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          1. [latex]x=0,1,2,3,4,5,6,7,8[\/latex]<\/li>\r\n \t
          2. [latex]{X}\\sim{B}{({100},\\frac{{8}}{{560}})}[\/latex]\r\n
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            1. [latex]{P}{({x}={2})}=\\text{binompdf}{({100},\\frac{{8}}{{560}},{2})}={0.2466}[\/latex]<\/li>\r\n \t
            2. [latex]{P}{({x}\\leq{6})}=\\text{binomcdf}{({100},\\frac{{8}}{{560}},{6})}={0.9994}[\/latex]<\/li>\r\n \t
            3. [latex]{P}{({x}{>}{3})}={1}-{P}{({x}\\leq{3})}={1}-\\text{binomcdf}{({100},\\frac{{8}}{{560}},{3})}={1}-{0.9443}={0.0557}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
            4. Using the formulas, calculate the (a) mean and (b) standard deviation.\r\n
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              1. [latex]\\text{Mean}={n}{p}={({100})}{(\\frac{{8}}{{560}})}=\\frac{{800}}{{560}}\\approx{1.4286}[\/latex]<\/li>\r\n \t
              2. [latex]\\text{Standard Deviation}=\\sqrt{{{n}{p}{q}}}=\\sqrt{{{({100})}{({8560})}{({552560})}}}\\approx{1.1867}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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                Try It<\/h3>\r\nAccording to a Gallup poll, 60% of American adults prefer saving over spending. Let [latex]X=[\/latex] the number of American adults out of a random sample of 50 who prefer saving to spending.\r\n
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                1. What is the probability distribution for [latex]X[\/latex]?<\/li>\r\n \t
                2. Use your calculator to find the following probabilities:\r\n
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                  1. the probability that 25 adults in the sample prefer saving over spending<\/li>\r\n \t
                  2. the probability that at most 20 adults prefer saving<\/li>\r\n \t
                  3. the probability that more than 30 adults prefer saving<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                  4. Using the formulas, calculate the (i) mean and (ii) standard deviation of [latex]X[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"457993\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457993\"]\r\n
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                    1. [latex]X\u223cB(50,0.6)[\/latex]<\/li>\r\n \t
                    2. Using the TI-83, 83+, 84 calculator with instructions as provided earlier:\r\n
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                      1. [latex]P(x=25)=\\text{binompdf}(50,0.6,25)=0.0405[\/latex]<\/li>\r\n \t
                      2. [latex]P(x\\leq20)=\\text{binomcdf}(50,0.6,20)=0.0034[\/latex]<\/li>\r\n \t
                      3. [latex]P(x>30)=1-\\text{binomcdf}(50,0.6,30)=1\u20130.5535=0.4465[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                      4. Using the formulas, calculate the (a) mean and (b) standard deviation of [latex]X[\/latex].\r\n
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                        1. [latex]\\text{Mean}=np=50(0.6)=30[\/latex]<\/li>\r\n \t
                        2. Standard Deviation[latex]=\\sqrt{{{n}{p}{q}}}=\\sqrt{{{50}{({0.6})}{({0.4})}}}\\approx{3.4641}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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                          Example<\/h3>\r\nThe lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let [latex]X=[\/latex]\u00a0the number of people who will develop pancreatic cancer.\r\n
                            \r\n \t
                          1. What is the probability distribution for [latex]X[\/latex]?<\/li>\r\n \t
                          2. Using the formulas, calculate the (i) mean and (ii) standard deviation\u00a0of [latex]X[\/latex].<\/li>\r\n \t
                          3. Use your calculator to find the probability that at most eight people develop pancreatic cancer.<\/li>\r\n \t
                          4. Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"586098\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"586098\"]\r\n
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                            1. [latex]X\\sim{B}(200,0.0128)[\/latex]<\/li>\r\n \t
                            2. Using the formulas, calculate the (a) mean and (b) standard deviation of [latex]X[\/latex].\r\n
                                \r\n \t
                              1. Mean[latex]=np=200(0.0128)=2.56[\/latex]<\/li>\r\n \t
                              2. [latex]\\text{Standard Deviation}=\\sqrt{{{n}{p}{q}}}=\\sqrt{{{({200})}{({0.0128})}{({0.9872})}}}\\approx{1.5897}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                              3. Using the TI-83, 83+, 84 calculator:\r\n[latex]P(x\\leq8)=\\text{binomcdf}(200,0.0128,8)=0.9988[\/latex]<\/li>\r\n \t
                              4. [latex]P(x=5)=\\text{binompdf}(200,0.0128,5)=0.0707[\/latex]\r\n[latex]P(x=6)=\\text{binompdf}(200,0.0128,6)=0.0298[\/latex] So [latex]P(x=5)>P(x=6)[\/latex]; it is more likely that five people will develop cancer than six.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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                                Try It<\/h3>\r\nDuring the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let [latex]X=[\/latex] the number of shots that scored points.\r\n
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                                1. What is the probability distribution for [latex]X[\/latex]?<\/li>\r\n \t
                                2. Using the formulas, calculate the (i) mean and (ii) standard deviation of [latex]X[\/latex].<\/li>\r\n \t
                                3. Use your calculator to find the probability that DeAndre scored with 60 of these shots.<\/li>\r\n \t
                                4. Find the probability that DeAndre scored with more than 50 of these shots.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"522994\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522994\"]\r\n
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                                  1. [latex]X\\sim{B}(80,0.613)[\/latex]<\/li>\r\n \t
                                  2. Using the formulas, calculate the (a) mean and (b) standard deviation of [latex]X[\/latex].\r\n
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                                    1. Mean = np<\/em> = 80(0.613) = 49.04<\/li>\r\n \t
                                    2. Standard Deviation = [latex]\\displaystyle\\sqrt{{{n}{p}{q}}}=\\sqrt{{{80}{({0.613})}{({0.387})}}}\\approx{4.3564}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                    3. Using the TI-83, 83+, 84 calculator:\r\n[latex]P(x=60)=\\text{binompdf}(80,0.613,60)=0.0036[\/latex]<\/li>\r\n \t
                                    4. [latex]P(x>50)=1\u2013P(x\u226450)=1\u2013\\text{binomcdf}(80,0.613,50)=1\u20130.6282=0.3718[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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                                      Example<\/h3>\r\nThe following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students?\r\n\r\n[reveal-answer q=\"52898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"52898\"]The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is [latex]\\frac{6}{16}[\/latex], when the first draw selects a staff member. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence.[\/hidden-answer]\r\n\r\n<\/div>\r\n
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                                      Try It<\/h3>\r\nA lacrosse team is selecting a captain. The names of all the seniors are put into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this is binomial or not and state why.\r\n\r\n[reveal-answer q=\"395022\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"395022\"]This is not binomial because the names are not replaced, which means the probability changes for each time a name is drawn. This violates the condition of independence.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n