{"id":48,"date":"2021-01-12T22:19:31","date_gmt":"2021-01-12T22:19:31","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/probability-distribution-function-pdf-for-a-discrete-random-variable\/"},"modified":"2021-05-11T20:09:06","modified_gmt":"2021-05-11T20:09:06","slug":"probability-distribution-function-pdf-for-a-discrete-random-variable","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/probability-distribution-function-pdf-for-a-discrete-random-variable\/","title":{"rendered":"Random Variable – Discrete"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
\r\n

A discrete probability distribution function<\/span> has two characteristics:<\/p>\r\n\r\n

    \r\n \t
  1. Each probability is between zero and one, inclusive.<\/li>\r\n \t
  2. The sum of the probabilities is one.<\/li>\r\n<\/ol>\r\n
    \r\n

    Example <\/span>4.1<\/span><\/h3>\r\n<\/header>
    \r\n
    \r\n

    A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X<\/em> = the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, x<\/em> = 0, 1, 2, 3, 4, 5.<\/p>\r\n

    P<\/em>(x<\/em>) = probability that X<\/em> takes on a value x<\/em>.<\/p>\r\n\r\n

    \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    x<\/em><\/th>\r\nP<\/em>(x<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    0<\/td>\r\n$P(x=0) = \\frac{2}{50}$<\/td>\r\n<\/tr>\r\n
    1<\/td>\r\n$P(x=1) = \\frac{11}{50}$<\/td>\r\n<\/tr>\r\n
    2<\/td>\r\n$P(x=2) = \\frac{23}{50}$<\/td>\r\n<\/tr>\r\n
    3<\/td>\r\n$P(x=3) = \\frac{9}{50}$<\/td>\r\n<\/tr>\r\n
    4<\/td>\r\n$P(x=4) = \\frac{4}{50}$<\/td>\r\n<\/tr>\r\n
    5<\/td>\r\n$P(x=5) = \\frac{1}{50}$<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
    Table <\/span>4.2<\/span><\/div>\r\n<\/div>\r\n

    X<\/em> takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because:<\/p>\r\n\r\n

      \r\n \t
    1. Each P<\/em>(x<\/em>) is between zero and one, inclusive.<\/li>\r\n \t
    2. The sum of the probabilities is one, that is,\r\n$\\frac{2}{50}+\\frac{11}{50}+\\frac{23}{50}+\\frac{9}{50}+\\frac{4}{50}+\\frac{1}{50}=1$<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section><\/div>\r\n
      \r\n

      Try It <\/span>4.1<\/span><\/h3>\r\n<\/header>
      \r\n
      <\/header>
      \r\n
      \r\n
      \r\n

      A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let X<\/em> = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x<\/em> = 0, 1, 2, 3, 4, 5. P<\/em>(x<\/em>) = the probability that X<\/em> takes on value x<\/em>. Why is this a discrete probability distribution function (two reasons)?<\/p>\r\n\r\n

      \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      X<\/em><\/th>\r\nP<\/em>(x<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      0<\/td>\r\n$P(x=0) = \\frac{4}{50}$<\/td>\r\n<\/tr>\r\n
      1<\/td>\r\n$P(x=1) = \\frac{8}{50}$<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n$P(x=2) = \\frac{16}{50}$<\/td>\r\n<\/tr>\r\n
      3<\/td>\r\n$P(x=3) = \\frac{14}{50}$<\/td>\r\n<\/tr>\r\n
      4<\/td>\r\n$P(x=4) = \\frac{6}{50}$<\/td>\r\n<\/tr>\r\n
      5<\/td>\r\n$P(x=5) = \\frac{2}{50}$<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
      Table <\/span>4.3<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
      \r\n

      Example <\/span>4.2<\/span><\/h3>\r\n<\/header>
      \r\n
      \r\n

      Suppose Nancy has classes three days<\/strong> a week. She attends classes three days a week 80%<\/strong> of the time, two days 15%<\/strong> of the time, one day 4%<\/strong> of the time, and no days 1%<\/strong> of the time. Suppose one week is randomly selected.<\/p>\r\n \r\n

      <\/header>
      \r\n
      \r\n
      \r\n
        \r\n \t
      1. Let X<\/em> = the number of days Nancy ____________________.<\/li>\r\n \t
      2. X<\/em> takes on what values?<\/li>\r\n \t
      3. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example 4.1<\/a>. The table should have two columns labeled x<\/em> and P<\/em>(x<\/em>). What does the P<\/em>(x<\/em>) column sum to?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
        \r\n
        \r\n
        \r\n

        Solution<\/h4>\r\n
          \r\n \t
        1. Let X<\/em> = the number of days Nancy attends class per week<\/span>.<\/li>\r\n \t
        2. 0, 1, 2, and 3\r\n<\/span><\/li>\r\n \t
        3. \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          x<\/em><\/th>\r\nP<\/em>(x<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
          0<\/td>\r\n0.01<\/td>\r\n<\/tr>\r\n
          1<\/td>\r\n0.04<\/td>\r\n<\/tr>\r\n
          2<\/td>\r\n0.15<\/td>\r\n<\/tr>\r\n
          3<\/td>\r\n0.80<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
          Table <\/span>4.4<\/span><\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
          \r\n
          <\/div>\r\n
          \r\n

          Solution <\/span>4.2<\/span><\/h4>\r\n
          \r\n

          c.<\/p>\r\n\r\n

          \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          x<\/em><\/th>\r\nP<\/em>(x<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
          0<\/td>\r\n0.01<\/td>\r\n<\/tr>\r\n
          1<\/td>\r\n0.04<\/td>\r\n<\/tr>\r\n
          2<\/td>\r\n0.15<\/td>\r\n<\/tr>\r\n
          3<\/td>\r\n0.80<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
          Table <\/span>4.4<\/span><\/div>\r\n<\/div>\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>4.2<\/span><\/h3>\r\n<\/header>
          \r\n
          <\/header>
          \r\n
          \r\n
          \r\n

          Jeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X<\/em> and what values does it take on?<\/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

          A discrete probability distribution function<\/span> has two characteristics:<\/p>\n

            \n
          1. Each probability is between zero and one, inclusive.<\/li>\n
          2. The sum of the probabilities is one.<\/li>\n<\/ol>\n
            \n
            \n

            Example <\/span>4.1<\/span><\/h3>\n<\/header>\n
            \n
            \n

            A child psychologist is interested in the number of times a newborn baby’s crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X<\/em> = the number of times per week a newborn baby’s crying wakes its mother after midnight. For this example, x<\/em> = 0, 1, 2, 3, 4, 5.<\/p>\n

            P<\/em>(x<\/em>) = probability that X<\/em> takes on a value x<\/em>.<\/p>\n

            \n\n\n\n\n\n\n\n\n\n\n
            x<\/em><\/th>\nP<\/em>(x<\/em>)<\/th>\n<\/tr>\n<\/thead>\n
            0<\/td>\n$P(x=0) = \\frac{2}{50}$<\/td>\n<\/tr>\n
            1<\/td>\n$P(x=1) = \\frac{11}{50}$<\/td>\n<\/tr>\n
            2<\/td>\n$P(x=2) = \\frac{23}{50}$<\/td>\n<\/tr>\n
            3<\/td>\n$P(x=3) = \\frac{9}{50}$<\/td>\n<\/tr>\n
            4<\/td>\n$P(x=4) = \\frac{4}{50}$<\/td>\n<\/tr>\n
            5<\/td>\n$P(x=5) = \\frac{1}{50}$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
            Table <\/span>4.2<\/span><\/div>\n<\/div>\n

            X<\/em> takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because:<\/p>\n

              \n
            1. Each P<\/em>(x<\/em>) is between zero and one, inclusive.<\/li>\n
            2. The sum of the probabilities is one, that is,
              \n$\\frac{2}{50}+\\frac{11}{50}+\\frac{23}{50}+\\frac{9}{50}+\\frac{4}{50}+\\frac{1}{50}=1$<\/li>\n<\/ol>\n<\/div>\n<\/section>\n<\/div>\n
              \n
              \n

              Try It <\/span>4.1<\/span><\/h3>\n<\/header>\n
              \n
              \n
              <\/header>\n
              \n
              \n
              \n

              A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let X<\/em> = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x<\/em> = 0, 1, 2, 3, 4, 5. P<\/em>(x<\/em>) = the probability that X<\/em> takes on value x<\/em>. Why is this a discrete probability distribution function (two reasons)?<\/p>\n

              \n\n\n\n\n\n\n\n\n\n\n
              X<\/em><\/th>\nP<\/em>(x<\/em>)<\/th>\n<\/tr>\n<\/thead>\n
              0<\/td>\n$P(x=0) = \\frac{4}{50}$<\/td>\n<\/tr>\n
              1<\/td>\n$P(x=1) = \\frac{8}{50}$<\/td>\n<\/tr>\n
              2<\/td>\n$P(x=2) = \\frac{16}{50}$<\/td>\n<\/tr>\n
              3<\/td>\n$P(x=3) = \\frac{14}{50}$<\/td>\n<\/tr>\n
              4<\/td>\n$P(x=4) = \\frac{6}{50}$<\/td>\n<\/tr>\n
              5<\/td>\n$P(x=5) = \\frac{2}{50}$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
              Table <\/span>4.3<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n
              \n
              \n

              Example <\/span>4.2<\/span><\/h3>\n<\/header>\n
              \n
              \n

              Suppose Nancy has classes three days<\/strong> a week. She attends classes three days a week 80%<\/strong> of the time, two days 15%<\/strong> of the time, one day 4%<\/strong> of the time, and no days 1%<\/strong> of the time. Suppose one week is randomly selected.<\/p>\n

               <\/p>\n

              \n
              <\/header>\n
              \n
              \n
              \n
                \n
              1. Let X<\/em> = the number of days Nancy ____________________.<\/li>\n
              2. X<\/em> takes on what values?<\/li>\n
              3. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example 4.1<\/a>. The table should have two columns labeled x<\/em> and P<\/em>(x<\/em>). What does the P<\/em>(x<\/em>) column sum to?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n
                \n
                \n
                \n
                \n

                Solution<\/h4>\n
                  \n
                1. Let X<\/em> = the number of days Nancy attends class per week<\/span>.<\/li>\n
                2. 0, 1, 2, and 3
                  \n<\/span><\/li>\n
                3. \n\n\n\n\n\n\n\n\n
                  x<\/em><\/th>\nP<\/em>(x<\/em>)<\/th>\n<\/tr>\n<\/thead>\n
                  0<\/td>\n0.01<\/td>\n<\/tr>\n
                  1<\/td>\n0.04<\/td>\n<\/tr>\n
                  2<\/td>\n0.15<\/td>\n<\/tr>\n
                  3<\/td>\n0.80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                  Table <\/span>4.4<\/span><\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                  \n
                  <\/div>\n
                  \n

                  Solution <\/span>4.2<\/span><\/h4>\n
                  \n

                  c.<\/p>\n

                  \n\n\n\n\n\n\n\n\n
                  x<\/em><\/th>\nP<\/em>(x<\/em>)<\/th>\n<\/tr>\n<\/thead>\n
                  0<\/td>\n0.01<\/td>\n<\/tr>\n
                  1<\/td>\n0.04<\/td>\n<\/tr>\n
                  2<\/td>\n0.15<\/td>\n<\/tr>\n
                  3<\/td>\n0.80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                  Table <\/span>4.4<\/span><\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n
                  \n
                  \n

                  Try It <\/span>4.2<\/span><\/h3>\n<\/header>\n
                  \n
                  \n
                  <\/header>\n
                  \n
                  \n
                  \n

                  Jeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X<\/em> and what values does it take on?<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n","protected":false},"author":1,"menu_order":1,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[47],"contributor":[],"license":[],"class_list":["post-48","chapter","type-chapter","status-publish","hentry","chapter-type-standard"],"part":47,"_links":{"self":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/48","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":4,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/48\/revisions"}],"predecessor-version":[{"id":369,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/48\/revisions\/369"}],"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\/48\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/media?parent=48"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=48"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/contributor?post=48"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/license?post=48"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}