{"id":50,"date":"2021-01-12T22:19:32","date_gmt":"2021-01-12T22:19:32","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/binomial-distribution\/"},"modified":"2023-06-25T11:45:38","modified_gmt":"2023-06-25T11:45:38","slug":"binomial-distribution","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/binomial-distribution\/","title":{"rendered":"Binomial Distribution"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
\r\n

There are three characteristics of a binomial experiment.<\/p>\r\n\r\n

    \r\n \t
  1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n<\/em> denotes the number of trials.<\/li>\r\n \t
  2. There are only two possible outcomes, called \"success\" and \"failure,\" for each trial. The letter p<\/em> denotes the probability of a success on one trial, and q<\/em> denotes the probability of a failure on one trial. p<\/em> + q<\/em> = 1.<\/li>\r\n \t
  3. The n<\/em> trials are independent and are repeated using identical conditions. Because the n<\/em> trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, p<\/em>, of a success and probability, q<\/em>, of a failure remain the same. For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with probability p<\/em> = 0.6. Then, q<\/em> = 0.4. This means that for every true-false statistics question Joe answers, his probability of success (p<\/em> = 0.6) and his probability of failure (q<\/em> = 0.4) remain the same.<\/li>\r\n<\/ol>\r\n

    The outcomes of a binomial experiment fit a binomial probability distribution<\/span>. The random variable X<\/em> = the number of successes obtained in the n<\/em> independent trials.<\/p>\r\n

    The mean, \u03bc<\/em>, and variance, \u03c3<\/em>2<\/sup>, for the binomial probability distribution are \u03bc<\/em> = np<\/em> and \u03c3<\/em>2<\/sup> = npq<\/em>. The standard deviation, \u03c3<\/em>, is then $\\sigma = \\sqrt{npq}$<\/p>\r\nAny experiment that has characteristics two and three and where n<\/em> = 1 is called a Bernoulli Trial<\/span> (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.\r\n

    \r\n

    Example <\/span>4.9<\/span><\/h3>\r\n<\/header>
    \r\n
    \r\n

    At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A \"success\" could be defined as an individual who withdrew. The random variable X<\/em> = the number of students who withdraw from the randomly selected elementary physics class.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

    \r\n

    Try It <\/span>4.9<\/span><\/h3>\r\n<\/header>
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    <\/header>
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    \r\n

    The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a \"success\" be in this case?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

    \r\n

    Example <\/span>4.10<\/span><\/h3>\r\n<\/header>
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    \r\n

    Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X<\/em> as the number of wins, then X<\/em> takes on the values 0, 1, 2, 3, ..., 20. The probability of a success is p<\/em> = 0.55. The probability of a failure is q<\/em> = 0.45. The number of trials is n<\/em> = 20. The probability question can be stated mathematically as P<\/em>(x<\/em> = 15) or more simply, $P(15)$.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

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    Try It <\/span>4.10<\/span><\/h3>\r\n<\/header>
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    <\/header>
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    \r\n
    \r\n

    A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the trick is 35%, and the probability that the dolphin does not successfully perform the trick is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

    \r\n

    Example <\/span>4.11<\/span><\/h3>\r\n<\/header>
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    \r\n
    <\/header>
    \r\n
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    \r\n

    A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more than ten heads? Let X<\/em> = the number of heads in 15 flips of the fair coin. X<\/em> takes on the values 0, 1, 2, 3, ..., 15. Since the coin is fair, p<\/em> = 0.5 and q<\/em> = 0.5. The number of trials is n<\/em> = 15. The probability question can be stated mathematically as $P(x>10)$ where $x$ is the number of heads flipped.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n

    \r\n

    Try It <\/span>4.11<\/span><\/h3>\r\n<\/header>
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    <\/header>
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    \r\n

    A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

    \r\n

    Example <\/span>4.12<\/span><\/h3>\r\n<\/header>
    \r\n
    \r\n

    Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly.<\/p>\r\n \r\n

    <\/header>
    \r\n
    \r\n
    \r\n
      \r\n \t
    1. a. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial.<\/li>\r\n \t
    2. If we are interested in the number of students who do their homework on time, then how do we define X<\/em>?<\/li>\r\n \t
    3. What values does x<\/em> take on?<\/li>\r\n \t
    4. What is a \"failure,\" in words?<\/li>\r\n \t
    5. If p<\/em> + q<\/em> = 1, then what is q<\/em>?<\/li>\r\n \t
    6. The words \"at least\" translate as what kind of inequality for the probability question P<\/em>(x<\/em> ____ 40).<\/li>\r\n<\/ol>\r\n

      Solution 4.11<\/h4>\r\n
        \r\n \t
      1. failure<\/li>\r\n \t
      2. X<\/em> = the number of statistics students who do their homework on time\r\n<\/span><\/li>\r\n \t
      3. 0, 1, 2, \u2026, 50\r\n<\/span><\/li>\r\n \t
      4. \r\n

        Failure is defined as a student who does not complete his or her homework on time.<\/p>\r\n

        The probability of a success is p<\/em> = 0.70. The number of trials is n<\/em> = 50.<\/p>\r\n<\/li>\r\n \t

      5. q<\/em> = 0.30\r\n<\/span><\/li>\r\n \t
      6. greater than or equal to (\u2265) \r\n<\/span>The probability question is P<\/em>(x<\/em> \u2265 40).<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
        \r\n
        \r\n
        <\/div>\r\n
        \r\n

        Solution <\/span>4.12<\/span><\/h4>\r\n
        \r\n

        f. greater than or equal to (\u2265)\r\n\r\n<\/span>The probability question is P<\/em>(x<\/em> \u2265 40).<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n

        \r\n

        Try It <\/span>4.12<\/span><\/h3>\r\n<\/header>
        \r\n
        <\/header>
        \r\n
        \r\n
        \r\n

        Sixty-five percent of people pass the state driver\u2019s exam on the first try. A group of 50 individuals who have taken the driver\u2019s exam is randomly selected. Give two reasons why this is a binomial problem.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

        \r\n

        Notation for the Binomial: B<\/em> = Binomial Probability Distribution Function<\/h3>\r\n

        X<\/em> ~ B<\/em>(n<\/em>, p<\/em>)<\/p>\r\n

        Read this as \"X<\/em> is a random variable with a binomial distribution.\" The parameters are n<\/em> and p<\/em>; n<\/em> = number of trials, p<\/em> = probability of a success on each trial.<\/p>\r\n\r\n<\/section>\r\n

        \r\n

        Example <\/span>4.13<\/span><\/h3>\r\n<\/header>
        \r\n
        \r\n

        It 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?<\/p>\r\n

        Let X<\/em> = the number of workers who have a high school diploma but do not pursue any further education.<\/p>\r\n

        X<\/em> takes on the values 0, 1, 2, ..., 20 where n<\/em> = 20, p<\/em> = 0.41, and q<\/em> = 1 \u2013 0.41 = 0.59. X<\/em> ~ B<\/em>(20, 0.41)<\/p>\r\n

        Find P<\/em>(x<\/em> \u2264 12).<\/p>\r\n

        P<\/em>(x<\/em> \u2264 12) = 0.9738 can be found using Google Sheets using the formula =BINOM.DIST(12, 20, 0.41, TRUE)<\/strong>. See the Sheets section below for more information.<\/p>\r\n\r\n

        \r\n

        NOTE<\/span><\/h3>\r\n<\/header>
        \r\n
        \r\n

        If you want to find P<\/em>(x<\/em> = 12), use =BINOM.DIST(12, n, p, FALSE)<\/strong> (See the note about using Google Sheets below.).<\/p>\r\nIf you want to find P<\/em>(x<\/em> > 12), use 1 - BINOM.DIST(12, 20, 0.41, TRUE)<\/strong>.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

        The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.<\/p>\r\n

        The graph of X<\/em> ~ B<\/em>(20, 0.41) is as follows:<\/p>\r\n\r\n

        \r\n
        \"This<\/span><\/figure>\r\n
        Figure <\/span>4.2<\/span><\/div>\r\n<\/div>\r\n

        The y<\/em>-axis contains the probability of x<\/em>, where X<\/em> = the number of workers who have only a high school diploma.<\/p>\r\n

        The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, \u03bc<\/em> = np<\/em> = (20)(0.41) = 8.2.<\/p>\r\n

        The formula for the variance is \u03c32<\/sup> = npq<\/em>. The standard deviation is $\\sigma = \\sqrt{npq} = \\sqrt{(20)(0.41)(0.59)}=2.20$<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

        \r\n

        Google Sheets: Binomial Distribution<\/h3>\r\n<\/header>\r\n
        \r\n

        BINOMDIST<\/h4>\r\n
        \r\n
        \r\n
        \r\n

        Calculates the probability of drawing a certain number of successes (or a maximum number of successes) in a certain number of tries given a population of a certain size containing a certain number of successes, with replacement of draws.<\/p>\r\n\r\n

        Sample Usage<\/h4>\r\nBINOMDIST(4,100,0.005,FALSE)<\/code>\r\n\r\nBINOMDIST(A2,A3,A4,TRUE)<\/code>\r\n

        Syntax<\/h4>\r\nBINOMDIST(num_successes, num_trials, prob_success, cumulative)<\/code>\r\n
          \r\n \t
        • \r\n

          num_successes<\/code> - The number of successes for which to calculate the probability in num_trials<\/code> trials.<\/p>\r\n\r\n

            \r\n \t
          • If cumulative<\/code> is TRUE<\/code> then BINOMDIST<\/code> returns the probability of num_successes<\/code> or fewer successes, otherwise the probability of exactly num_successes<\/code> successes.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
          • num_trials<\/code> - The number of independent trials.<\/li>\r\n \t
          • prob_success<\/code> - The probability of success in any given trial.<\/li>\r\n \t
          • cumulative<\/code> - [<\/strong> FALSE<\/code> by default ]<\/strong> - Whether to use the binomial cumulative distribution.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
            \r\n

            Try It <\/span>4.13<\/span><\/h3>\r\n<\/header>
            \r\n
            <\/header>
            \r\n
            \r\n
            \r\n

            About 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 Google Sheets to find the answer.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

            \r\n

            Example <\/span>4.14<\/span><\/h3>\r\n<\/header>
            \r\n
            \r\n
            <\/header>
            \r\n
            \r\n
            \r\n

            In the 2013 Jerry\u2019s Artarama<\/em> art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X<\/em> = the number of pages that feature signature artists.<\/p>\r\n\r\n

              \r\n \t
            1. What values does x<\/em> take on?<\/li>\r\n \t
            2. What is the probability distribution? Find the following probabilities:\r\n
                \r\n \t
              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 (i) mean and (ii) standard deviation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
                \r\n
                <\/div>\r\n
                \r\n

                Solution <\/span>4.14<\/span><\/h4>\r\nNote: in our solutions below, we use the BINOMDIST formula in a spreadsheet.\r\n
                \r\n
                  \r\n \t
                1. x<\/em> = 0, 1, 2, 3, 4, 5, 6, 7, 8<\/li>\r\n \t
                2. $X\\approx B(100,~ \\frac{8}{560})$\r\n
                    \r\n \t
                  1. \r\n
                      \r\n \t
                    1. $P(x=2) =$ BINOMDIST(2,100,8\/560,FALSE)<\/code>=0.2466<\/li>\r\n \t
                    2. $P(x\\leq6) =$ BINOMDIST(6,100,8\/560,TRUE)<\/code> = 0.9994<\/li>\r\n \t
                    3. $P(x>3) = 1-P(x\\leq 3) =$ 1 - BINOMDIST(3,100,8\/560,TRUE)<\/code> = 1-0.9443 = 0.0557<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                    4. Mean: $np=(100)\\left( \\frac{8}{560} \\right) = \\frac{800}{560}=1.4268$\r\nStandard Deviation: $\\sqrt{npq} = \\sqrt{(100)\\left(\\frac{8}{560} \\right)\\left(\\frac{552}{560} \\right)}=1.1867$<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
                      \r\n

                      Try It <\/span>4.14<\/span><\/h3>\r\n<\/header>
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                      <\/header>
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                      \r\n
                      \r\n

                      According to a Gallup poll, 60% of American adults prefer saving over spending. Let X<\/em> = the number of American adults out of a random sample of 50 who prefer saving to spending.<\/p>\r\n\r\n

                        \r\n \t
                      1. What is the probability distribution for X<\/em>?<\/li>\r\n \t
                      2. Use your calculator to find the following probabilities:\r\n
                          \r\n \t
                        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 X<\/em>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
                          \r\n

                          Example <\/span>4.15<\/span><\/h3>\r\n<\/header>
                          \r\n
                          \r\n

                          The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let X<\/em> = the number of people who will develop pancreatic cancer.<\/p>\r\n\r\n

                          <\/header>
                          \r\n
                          \r\n
                          \r\n
                            \r\n \t
                          1. What is the probability distribution for X<\/em>?<\/li>\r\n \t
                          2. Using the formulas for this distribution, calculate the (i) mean and (ii) standard deviation of X<\/em>.<\/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<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
                            \r\n
                            <\/div>\r\n
                            \r\n

                            Solution <\/span>4.15<\/span><\/h4>\r\nNote: in our solutions below, we use the BINOMDIST formula in a spreadsheet.\r\n
                            \r\n
                              \r\n \t
                            1. The probability distribution is Binomial since there are:\r\n
                                \r\n \t
                              1. a fixed number of trials (200 people sampled),<\/li>\r\n \t
                              2. the trials are independent (one person getting pancreatic cancer does not affect the probability someone else will),<\/li>\r\n \t
                              3. the probability of success stays the same on each trial (each person has a 0.0128 probability of getting pancreatic cancer), and<\/li>\r\n \t
                              4. each trial can result in only one of two outcomes (either a person gets pancreatic cancer or the do not)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                              5. Since this is a binomial distribution, we can use the following \"shortcut\" formulas to calculate the mean and standard deviation:\r\n
                                  \r\n \t
                                1. mean = $n \\cdot p = 200 \\cdot 0.0128 = 2.56$ (ie we would expect about 2.56 people out of the 200 to get pancreatic cancer)<\/li>\r\n \t
                                2. standard deviation = $\\sqrt{npq} = \\sqrt{200 \\cdot 0.0128 \\cdot (1-0.0128)} = 1.588$<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                3. We can use a spreadsheet to calculate P(0) + P(1) + P(2) + ... +P(7) + P(8)\r\nTo calculate P(0) when n=200 and p=0.0178, use the formula =BINOMDIST(0, 200, 0.0178, FALSE)<\/code>\r\nTo calculate P(1) when n=200 and p=0.0178, use the formula =BINOMDIST(1, 200, 0.0178, FALSE)<\/code>\r\nDo this for all probabilities up to 8, and then add those probabilities up.\r\nThe probability that at most 8 people will get pancreatic cancer out of these 200 is 0.9988 which means it will be very unlikely that more than 8 people would get pancreatic cancer.<\/li>\r\n \t
                                4. From our spreadsheet calculations in part c, we see that P(5) = 0.0707 and P(6) = 0.0298, so it's more likely that 5 people out of the 200 would get pancreatic cancer.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
                                  \r\n

                                  Try It <\/span>4.15<\/span><\/h3>\r\n<\/header>
                                  \r\n
                                  <\/header>
                                  \r\n
                                  \r\n
                                  \r\n

                                  During 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 X<\/em> = the number of shots that scored points.<\/p>\r\n\r\n

                                    \r\n \t
                                  1. What is the probability distribution for X<\/em>?<\/li>\r\n \t
                                  2. Using the formulas, calculate the (i) mean and (ii) standard deviation of X<\/em>.<\/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<\/div>\r\n<\/div>\r\n
                                    \r\n
                                    <\/div>\r\n
                                    \r\n

                                    Solution <\/span>4.14<\/span><\/h4>\r\nNote: in our solutions below, we use the BINOMDIST formula in a spreadsheet.\r\n
                                    <\/div>\r\n\r\n\r\n<\/section><\/div>\r\n\r\n\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
                                    \r\n

                                    Example <\/span>4.16<\/span><\/h3>\r\n<\/header>
                                    \r\n
                                    \r\n\r\nThe following example illustrates a problem that is not<\/strong> 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? The names of all committee members are put into a box, and two names are drawn without replacement<\/strong>. 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 $\\frac{6}{15}$. The probability of a student on the second draw is $\\frac{5}{15}$, when the first draw selects a student. The probability is $\\frac{6}{15}$, 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.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n
                                    \r\n

                                    Try It <\/span>4.16<\/span><\/h3>\r\n<\/header>
                                    \r\n
                                    <\/header>
                                    \r\n
                                    \r\n
                                    \r\n

                                    A 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.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>","rendered":"


                                    \n[latexpage]
                                    \n<\/span><\/p>\n

                                    \n

                                    There are three characteristics of a binomial experiment.<\/p>\n

                                      \n
                                    1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n<\/em> denotes the number of trials.<\/li>\n
                                    2. There are only two possible outcomes, called “success” and “failure,” for each trial. The letter p<\/em> denotes the probability of a success on one trial, and q<\/em> denotes the probability of a failure on one trial. p<\/em> + q<\/em> = 1.<\/li>\n
                                    3. The n<\/em> trials are independent and are repeated using identical conditions. Because the n<\/em> trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, p<\/em>, of a success and probability, q<\/em>, of a failure remain the same. For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with probability p<\/em> = 0.6. Then, q<\/em> = 0.4. This means that for every true-false statistics question Joe answers, his probability of success (p<\/em> = 0.6) and his probability of failure (q<\/em> = 0.4) remain the same.<\/li>\n<\/ol>\n

                                      The outcomes of a binomial experiment fit a binomial probability distribution<\/span>. The random variable X<\/em> = the number of successes obtained in the n<\/em> independent trials.<\/p>\n

                                      The mean, \u03bc<\/em>, and variance, \u03c3<\/em>2<\/sup>, for the binomial probability distribution are \u03bc<\/em> = np<\/em> and \u03c3<\/em>2<\/sup> = npq<\/em>. The standard deviation, \u03c3<\/em>, is then $\\sigma = \\sqrt{npq}$<\/p>\n

                                      Any experiment that has characteristics two and three and where n<\/em> = 1 is called a Bernoulli Trial<\/span> (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.<\/p>\n

                                      \n
                                      \n

                                      Example <\/span>4.9<\/span><\/h3>\n<\/header>\n
                                      \n
                                      \n

                                      At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A “success” could be defined as an individual who withdrew. The random variable X<\/em> = the number of students who withdraw from the randomly selected elementary physics class.<\/p>\n<\/div>\n<\/section>\n<\/div>\n

                                      \n
                                      \n

                                      Try It <\/span>4.9<\/span><\/h3>\n<\/header>\n
                                      \n
                                      \n
                                      <\/header>\n
                                      \n
                                      \n
                                      \n

                                      The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a “success” be in this case?<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

                                      \n
                                      \n

                                      Example <\/span>4.10<\/span><\/h3>\n<\/header>\n
                                      \n
                                      \n

                                      Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X<\/em> as the number of wins, then X<\/em> takes on the values 0, 1, 2, 3, …, 20. The probability of a success is p<\/em> = 0.55. The probability of a failure is q<\/em> = 0.45. The number of trials is n<\/em> = 20. The probability question can be stated mathematically as P<\/em>(x<\/em> = 15) or more simply, $P(15)$.<\/p>\n<\/div>\n<\/section>\n<\/div>\n

                                      \n
                                      \n

                                      Try It <\/span>4.10<\/span><\/h3>\n<\/header>\n
                                      \n
                                      \n
                                      <\/header>\n
                                      \n
                                      \n
                                      \n

                                      A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the trick is 35%, and the probability that the dolphin does not successfully perform the trick is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

                                      \n
                                      \n

                                      Example <\/span>4.11<\/span><\/h3>\n<\/header>\n
                                      \n
                                      \n
                                      \n
                                      <\/header>\n
                                      \n
                                      \n
                                      \n

                                      A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more than ten heads? Let X<\/em> = the number of heads in 15 flips of the fair coin. X<\/em> takes on the values 0, 1, 2, 3, …, 15. Since the coin is fair, p<\/em> = 0.5 and q<\/em> = 0.5. The number of trials is n<\/em> = 15. The probability question can be stated mathematically as $P(x>10)$ where $x$ is the number of heads flipped.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n

                                      \n
                                      \n

                                      Try It <\/span>4.11<\/span><\/h3>\n<\/header>\n
                                      \n
                                      \n
                                      <\/header>\n
                                      \n
                                      \n
                                      \n

                                      A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

                                      \n
                                      \n

                                      Example <\/span>4.12<\/span><\/h3>\n<\/header>\n
                                      \n
                                      \n

                                      Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly.<\/p>\n

                                       <\/p>\n

                                      \n
                                      <\/header>\n
                                      \n
                                      \n
                                      \n
                                        \n
                                      1. a. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial.<\/li>\n
                                      2. If we are interested in the number of students who do their homework on time, then how do we define X<\/em>?<\/li>\n
                                      3. What values does x<\/em> take on?<\/li>\n
                                      4. What is a “failure,” in words?<\/li>\n
                                      5. If p<\/em> + q<\/em> = 1, then what is q<\/em>?<\/li>\n
                                      6. The words “at least” translate as what kind of inequality for the probability question P<\/em>(x<\/em> ____ 40).<\/li>\n<\/ol>\n

                                        Solution 4.11<\/h4>\n
                                          \n
                                        1. failure<\/li>\n
                                        2. X<\/em> = the number of statistics students who do their homework on time
                                          \n<\/span><\/li>\n
                                        3. 0, 1, 2, \u2026, 50
                                          \n<\/span><\/li>\n
                                        4. \n

                                          Failure is defined as a student who does not complete his or her homework on time.<\/p>\n

                                          The probability of a success is p<\/em> = 0.70. The number of trials is n<\/em> = 50.<\/p>\n<\/li>\n

                                        5. q<\/em> = 0.30
                                          \n<\/span><\/li>\n
                                        6. greater than or equal to (\u2265)
                                          \n<\/span>The probability question is P<\/em>(x<\/em> \u2265 40).<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n
                                          \n
                                          \n
                                          \n
                                          <\/div>\n
                                          \n

                                          Solution <\/span>4.12<\/span><\/h4>\n
                                          \n

                                          f. greater than or equal to (\u2265)
                                          \n
                                          \n<\/span>The probability question is P<\/em>(x<\/em> \u2265 40).<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n

                                          \n
                                          \n

                                          Try It <\/span>4.12<\/span><\/h3>\n<\/header>\n
                                          \n
                                          \n
                                          <\/header>\n
                                          \n
                                          \n
                                          \n

                                          Sixty-five percent of people pass the state driver\u2019s exam on the first try. A group of 50 individuals who have taken the driver\u2019s exam is randomly selected. Give two reasons why this is a binomial problem.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

                                          \n

                                          Notation for the Binomial: B<\/em> = Binomial Probability Distribution Function<\/h3>\n

                                          X<\/em> ~ B<\/em>(n<\/em>, p<\/em>)<\/p>\n

                                          Read this as “X<\/em> is a random variable with a binomial distribution.” The parameters are n<\/em> and p<\/em>; n<\/em> = number of trials, p<\/em> = probability of a success on each trial.<\/p>\n<\/section>\n

                                          \n
                                          \n

                                          Example <\/span>4.13<\/span><\/h3>\n<\/header>\n
                                          \n
                                          \n

                                          It 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?<\/p>\n

                                          Let X<\/em> = the number of workers who have a high school diploma but do not pursue any further education.<\/p>\n

                                          X<\/em> takes on the values 0, 1, 2, …, 20 where n<\/em> = 20, p<\/em> = 0.41, and q<\/em> = 1 \u2013 0.41 = 0.59. X<\/em> ~ B<\/em>(20, 0.41)<\/p>\n

                                          Find P<\/em>(x<\/em> \u2264 12).<\/p>\n

                                          P<\/em>(x<\/em> \u2264 12) = 0.9738 can be found using Google Sheets using the formula =BINOM.DIST(12, 20, 0.41, TRUE)<\/strong>. See the Sheets section below for more information.<\/p>\n

                                          \n
                                          \n

                                          NOTE<\/span><\/h3>\n<\/header>\n
                                          \n
                                          \n

                                          If you want to find P<\/em>(x<\/em> = 12), use =BINOM.DIST(12, n, p, FALSE)<\/strong> (See the note about using Google Sheets below.).<\/p>\n

                                          If you want to find P<\/em>(x<\/em> > 12), use 1 – BINOM.DIST(12, 20, 0.41, TRUE)<\/strong>.<\/p>\n<\/div>\n<\/section>\n<\/div>\n

                                          The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.<\/p>\n

                                          The graph of X<\/em> ~ B<\/em>(20, 0.41) is as follows:<\/p>\n

                                          \n
                                          \"This<\/span><\/figure>\n
                                          Figure <\/span>4.2<\/span><\/div>\n<\/div>\n

                                          The y<\/em>-axis contains the probability of x<\/em>, where X<\/em> = the number of workers who have only a high school diploma.<\/p>\n

                                          The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, \u03bc<\/em> = np<\/em> = (20)(0.41) = 8.2.<\/p>\n

                                          The formula for the variance is \u03c32<\/sup> = npq<\/em>. The standard deviation is $\\sigma = \\sqrt{npq} = \\sqrt{(20)(0.41)(0.59)}=2.20$<\/p>\n<\/div>\n<\/section>\n<\/div>\n

                                          \n
                                          \n

                                          Google Sheets: Binomial Distribution<\/h3>\n<\/header>\n
                                          \n

                                          BINOMDIST<\/h4>\n
                                          \n
                                          \n
                                          \n

                                          Calculates the probability of drawing a certain number of successes (or a maximum number of successes) in a certain number of tries given a population of a certain size containing a certain number of successes, with replacement of draws.<\/p>\n

                                          Sample Usage<\/h4>\n

                                          BINOMDIST(4,100,0.005,FALSE)<\/code><\/p>\n

                                          BINOMDIST(A2,A3,A4,TRUE)<\/code><\/p>\n

                                          Syntax<\/h4>\n

                                          BINOMDIST(num_successes, num_trials, prob_success, cumulative)<\/code><\/p>\n

                                            \n
                                          • \n

                                            num_successes<\/code> – The number of successes for which to calculate the probability in num_trials<\/code> trials.<\/p>\n

                                              \n
                                            • If cumulative<\/code> is TRUE<\/code> then BINOMDIST<\/code> returns the probability of num_successes<\/code> or fewer successes, otherwise the probability of exactly num_successes<\/code> successes.<\/li>\n<\/ul>\n<\/li>\n
                                            • num_trials<\/code> – The number of independent trials.<\/li>\n
                                            • prob_success<\/code> – The probability of success in any given trial.<\/li>\n
                                            • cumulative<\/code> – [<\/strong> FALSE<\/code> by default ]<\/strong> – Whether to use the binomial cumulative distribution.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
                                              \n
                                              \n

                                              Try It <\/span>4.13<\/span><\/h3>\n<\/header>\n
                                              \n
                                              \n
                                              <\/header>\n
                                              \n
                                              \n
                                              \n

                                              About 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 Google Sheets to find the answer.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

                                              \n
                                              \n

                                              Example <\/span>4.14<\/span><\/h3>\n<\/header>\n
                                              \n
                                              \n
                                              \n
                                              <\/header>\n
                                              \n
                                              \n
                                              \n

                                              In the 2013 Jerry\u2019s Artarama<\/em> art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X<\/em> = the number of pages that feature signature artists.<\/p>\n

                                                \n
                                              1. What values does x<\/em> take on?<\/li>\n
                                              2. What is the probability distribution? Find the following probabilities:\n
                                                  \n
                                                1. the probability that two pages feature signature artists<\/li>\n
                                                2. the probability that at most six pages feature signature artists<\/li>\n
                                                3. the probability that more than three pages feature signature artists.<\/li>\n<\/ol>\n<\/li>\n
                                                4. Using the formulas, calculate the (i) mean and (ii) standard deviation.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                                                  \n
                                                  <\/div>\n
                                                  \n

                                                  Solution <\/span>4.14<\/span><\/h4>\n

                                                  Note: in our solutions below, we use the BINOMDIST formula in a spreadsheet.<\/p>\n

                                                  \n
                                                    \n
                                                  1. x<\/em> = 0, 1, 2, 3, 4, 5, 6, 7, 8<\/li>\n
                                                  2. $X\\approx B(100,~ \\frac{8}{560})$\n
                                                      \n
                                                    1. \n
                                                        \n
                                                      1. $P(x=2) =$ BINOMDIST(2,100,8\/560,FALSE)<\/code>=0.2466<\/li>\n
                                                      2. $P(x\\leq6) =$ BINOMDIST(6,100,8\/560,TRUE)<\/code> = 0.9994<\/li>\n
                                                      3. $P(x>3) = 1-P(x\\leq 3) =$ 1 - BINOMDIST(3,100,8\/560,TRUE)<\/code> = 1-0.9443 = 0.0557<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n
                                                      4. Mean: $np=(100)\\left( \\frac{8}{560} \\right) = \\frac{800}{560}=1.4268$
                                                        \nStandard Deviation: $\\sqrt{npq} = \\sqrt{(100)\\left(\\frac{8}{560} \\right)\\left(\\frac{552}{560} \\right)}=1.1867$<\/li>\n<\/ol>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n
                                                        \n
                                                        \n

                                                        Try It <\/span>4.14<\/span><\/h3>\n<\/header>\n
                                                        \n
                                                        \n
                                                        <\/header>\n
                                                        \n
                                                        \n
                                                        \n

                                                        According to a Gallup poll, 60% of American adults prefer saving over spending. Let X<\/em> = the number of American adults out of a random sample of 50 who prefer saving to spending.<\/p>\n

                                                          \n
                                                        1. What is the probability distribution for X<\/em>?<\/li>\n
                                                        2. Use your calculator to find the following probabilities:\n
                                                            \n
                                                          1. the probability that 25 adults in the sample prefer saving over spending<\/li>\n
                                                          2. the probability that at most 20 adults prefer saving<\/li>\n
                                                          3. the probability that more than 30 adults prefer saving<\/li>\n<\/ol>\n<\/li>\n
                                                          4. Using the formulas, calculate the (i) mean and (ii) standard deviation of X<\/em>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n
                                                            \n
                                                            \n

                                                            Example <\/span>4.15<\/span><\/h3>\n<\/header>\n
                                                            \n
                                                            \n

                                                            The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let X<\/em> = the number of people who will develop pancreatic cancer.<\/p>\n

                                                            \n
                                                            <\/header>\n
                                                            \n
                                                            \n
                                                            \n
                                                              \n
                                                            1. What is the probability distribution for X<\/em>?<\/li>\n
                                                            2. Using the formulas for this distribution, calculate the (i) mean and (ii) standard deviation of X<\/em>.<\/li>\n
                                                            3. Use your calculator to find the probability that at most eight people develop pancreatic cancer<\/li>\n
                                                            4. Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n
                                                              \n
                                                              <\/div>\n
                                                              \n

                                                              Solution <\/span>4.15<\/span><\/h4>\n

                                                              Note: in our solutions below, we use the BINOMDIST formula in a spreadsheet.<\/p>\n

                                                              \n
                                                                \n
                                                              1. The probability distribution is Binomial since there are:\n
                                                                  \n
                                                                1. a fixed number of trials (200 people sampled),<\/li>\n
                                                                2. the trials are independent (one person getting pancreatic cancer does not affect the probability someone else will),<\/li>\n
                                                                3. the probability of success stays the same on each trial (each person has a 0.0128 probability of getting pancreatic cancer), and<\/li>\n
                                                                4. each trial can result in only one of two outcomes (either a person gets pancreatic cancer or the do not)<\/li>\n<\/ol>\n<\/li>\n
                                                                5. Since this is a binomial distribution, we can use the following “shortcut” formulas to calculate the mean and standard deviation:\n
                                                                    \n
                                                                  1. mean = $n \\cdot p = 200 \\cdot 0.0128 = 2.56$ (ie we would expect about 2.56 people out of the 200 to get pancreatic cancer)<\/li>\n
                                                                  2. standard deviation = $\\sqrt{npq} = \\sqrt{200 \\cdot 0.0128 \\cdot (1-0.0128)} = 1.588$<\/li>\n<\/ol>\n<\/li>\n
                                                                  3. We can use a spreadsheet to calculate P(0) + P(1) + P(2) + … +P(7) + P(8)
                                                                    \nTo calculate P(0) when n=200 and p=0.0178, use the formula =BINOMDIST(0, 200, 0.0178, FALSE)<\/code>
                                                                    \nTo calculate P(1) when n=200 and p=0.0178, use the formula =BINOMDIST(1, 200, 0.0178, FALSE)<\/code>
                                                                    \nDo this for all probabilities up to 8, and then add those probabilities up.
                                                                    \nThe probability that at most 8 people will get pancreatic cancer out of these 200 is 0.9988 which means it will be very unlikely that more than 8 people would get pancreatic cancer.<\/li>\n
                                                                  4. From our spreadsheet calculations in part c, we see that P(5) = 0.0707 and P(6) = 0.0298, so it’s more likely that 5 people out of the 200 would get pancreatic cancer.<\/li>\n<\/ol>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n
                                                                    \n
                                                                    \n

                                                                    Try It <\/span>4.15<\/span><\/h3>\n<\/header>\n
                                                                    \n
                                                                    \n
                                                                    <\/header>\n
                                                                    \n
                                                                    \n
                                                                    \n

                                                                    During 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 X<\/em> = the number of shots that scored points.<\/p>\n

                                                                      \n
                                                                    1. What is the probability distribution for X<\/em>?<\/li>\n
                                                                    2. Using the formulas, calculate the (i) mean and (ii) standard deviation of X<\/em>.<\/li>\n
                                                                    3. Use your calculator to find the probability that DeAndre scored with 60 of these shots.<\/li>\n
                                                                    4. Find the probability that DeAndre scored with more than 50 of these shots.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                                                                      \n
                                                                      <\/div>\n
                                                                      \n

                                                                      Solution <\/span>4.14<\/span><\/h4>\n

                                                                      Note: in our solutions below, we use the BINOMDIST formula in a spreadsheet.<\/p>\n

                                                                      <\/div>\n

                                                                      <\/p>\n<\/section>\n<\/div>\n

                                                                      <\/p>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

                                                                      \n
                                                                      \n

                                                                      Example <\/span>4.16<\/span><\/h3>\n<\/header>\n
                                                                      \n
                                                                      \n

                                                                      The following example illustrates a problem that is not<\/strong> 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? The names of all committee members are put into a box, and two names are drawn without replacement<\/strong>. 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 $\\frac{6}{15}$. The probability of a student on the second draw is $\\frac{5}{15}$, when the first draw selects a student. The probability is $\\frac{6}{15}$, 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.<\/p>\n<\/div>\n<\/section>\n<\/div>\n

                                                                      \n
                                                                      \n

                                                                      Try It <\/span>4.16<\/span><\/h3>\n<\/header>\n
                                                                      \n
                                                                      \n
                                                                      <\/header>\n
                                                                      \n
                                                                      \n
                                                                      \n

                                                                      A 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.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n","protected":false},"author":1,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-50","chapter","type-chapter","status-publish","hentry"],"part":47,"_links":{"self":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/50","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":13,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/50\/revisions"}],"predecessor-version":[{"id":636,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/50\/revisions\/636"}],"part":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/parts\/47"}],"metadata":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/50\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/media?parent=50"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=50"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/contributor?post=50"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/license?post=50"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}