{"id":156,"date":"2016-04-21T22:43:43","date_gmt":"2016-04-21T22:43:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&p=156"},"modified":"2019-01-22T17:23:22","modified_gmt":"2019-01-22T17:23:22","slug":"answers-to-selected-exercises-12","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-fmcc-introstats1\/chapter\/answers-to-selected-exercises-12\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"

Probability Distribution Function (PDF) for a Discrete Random Variable<\/h2>\r\n1.\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\n0.12<\/td>\r\n<\/tr>\r\n
1<\/td>\r\n0.18<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n0.30<\/td>\r\n<\/tr>\r\n
3<\/td>\r\n0.15<\/td>\r\n<\/tr>\r\n
4<\/td>\r\n0.10<\/td>\r\n<\/tr>\r\n
5<\/td>\r\n0.10<\/td>\r\n<\/tr>\r\n
6<\/td>\r\n0.05<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n3.\u00a00.10 + 0.05 = 0.15\r\n\r\n5. 1\r\n\r\n7.\u00a00.35 + 0.40 + 0.10 = 0.85\r\n\r\n9.\u00a01(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45\r\n\r\n11.\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\n0.03<\/td>\r\n<\/tr>\r\n
1<\/td>\r\n0.04<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n0.08<\/td>\r\n<\/tr>\r\n
3<\/td>\r\n0.85<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n13.\u00a0Let X<\/em> = the number of events Javier volunteers for each month.\r\n\r\n15.\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\n0.05<\/td>\r\n<\/tr>\r\n
1<\/td>\r\n0.05<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n0.10<\/td>\r\n<\/tr>\r\n
3<\/td>\r\n0.20<\/td>\r\n<\/tr>\r\n
4<\/td>\r\n0.25<\/td>\r\n<\/tr>\r\n
5<\/td>\r\n0.35<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n17.\u00a01 \u2013 0.05 = 0.95\r\n

Mean or Expected Value and Standard Deviation<\/h2>\r\n20.\u00a00.2 + 1.2 + 2.4 + 1.6 = 5.4\r\n\r\n22.The values of P<\/em>(x<\/em>) do not sum to one.\r\n\r\n24.Let X<\/em> = the number of years a physics major will spend doing post-graduate research.\r\n\r\n27.\u00a01(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years\r\n\r\n29.\u00a0X<\/em> is the number of years a student studies ballet with the teacher.\r\n\r\n31.\u00a00.10 + 0.05 + 0.10 = 0.25\r\n\r\n33.\u00a0The sum of the probabilities sum to one because it is a probability distribution.\r\n\r\n35. [latex]-2\\left(\\frac{40}{52}\\right)+30\\left(\\frac{12}{52}\\right)=-1.54+6.92=5.38[\/latex]\r\n\r\n38. The variable of interest is X<\/em>, or the gain or loss, in dollars.\r\n

The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 \u2013 12 = 40 cards that are not face cards.<\/p>\r\n

We first need to construct the probability distribution for X<\/em>. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X<\/em> to determine the expected value.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Card Event<\/th>\r\nX<\/em> net gain\/loss<\/th>\r\nP<\/em>(X<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
Face Card and Heads<\/td>\r\n6<\/td>\r\n[latex]\\left(\\frac{{12}}{{52}}\\right)\\left(\\frac{{1}}{{2}}\\right)=\\left(\\frac{{6}}{{52}}\\right)\\\\[\/latex]<\/td>\r\n<\/tr>\r\n
Face Card and Tails<\/td>\r\n2<\/td>\r\n[latex]\\left(\\frac{{12}}{{52}}\\right)\\left(\\frac{{1}}{{2}}\\right)=\\left(\\frac{{6}}{{52}}\\right)\\\\[\/latex]<\/td>\r\n<\/tr>\r\n
(Not Face Card) and (H or T)<\/td>\r\n\u20132<\/td>\r\n[latex]\\left(\\frac{{40}}{{52}}\\right)\\left(1\\right)=\\left(\\frac{{40}}{{52}}\\right)\\\\[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
    \r\n \t
  • Expected Value = (6)[latex]\\frac{{6}}{{52}}+\\left(2\\right)\\left(\\frac{{6}}{{52}}\\right)+\\left(-2\\right)\\frac{{40}}{{52}}=-\\frac{{32}}{{52}}\\\\[\/latex]<\/li>\r\n \t
  • Expected value = \u2013$0.62, rounded to the nearest cent<\/li>\r\n \t
  • If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average.<\/li>\r\n \t
  • You should not play this game to win money because the expected value indicates an expected average loss.<\/li>\r\n<\/ul>\r\n40.\r\n
      \r\n \t
    1. 0.1<\/li>\r\n \t
    2. 1.6<\/li>\r\n<\/ol>\r\n42.\r\n
        \r\n \t
      1. \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Software Company<\/th>\r\n<\/tr>\r\n
        x<\/em><\/th>\r\nP<\/em>(x<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        5,000,000<\/td>\r\n0.10<\/td>\r\n<\/tr>\r\n
        1,000,000<\/td>\r\n0.30<\/td>\r\n<\/tr>\r\n
        \u20131,000,000<\/td>\r\n0.60<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Hardware Company<\/th>\r\n<\/tr>\r\n
        x<\/em><\/th>\r\nP<\/em>(x<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        3,000,000<\/td>\r\n0.20<\/td>\r\n<\/tr>\r\n
        1,000,000<\/td>\r\n0.40<\/td>\r\n<\/tr>\r\n
        \u20131,000,00<\/td>\r\n0.40<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Biotech Firm<\/th>\r\n<\/tr>\r\n
        x<\/em><\/th>\r\nP<\/em>(x<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        6,00,000<\/td>\r\n0.10<\/td>\r\n<\/tr>\r\n
        0<\/td>\r\n0.70<\/td>\r\n<\/tr>\r\n
        \u20131,000,000<\/td>\r\n0.20<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t
      2. $200,000; $600,000; $400,000<\/li>\r\n \t
      3. third investment because it has the lowest probability of loss<\/li>\r\n \t
      4. first investment because it has the highest probability of loss<\/li>\r\n \t
      5. second investment<\/li>\r\n<\/ol>\r\n44.\u00a04.85 years\r\n\r\n46. b\r\n\r\n48.\r\n\r\n
        \r\n
        \r\n
        \r\n

        Let X<\/em> = the amount of money to be won on a ticket. The following table shows the PDF for X<\/em>.<\/p>\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\n0.969<\/td>\r\n<\/tr>\r\n
        5<\/td>\r\n[latex]\\frac{{250}}{{10000}}\\\\[\/latex]= 0.025<\/td>\r\n<\/tr>\r\n
        25<\/td>\r\n[latex]\\frac{{50}}{{10000}}\\\\[\/latex] = 0.005<\/td>\r\n<\/tr>\r\n
        100<\/td>\r\n[latex]\\frac{{10}}{{10000}}\\\\[\/latex] = 0.001<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

        Calculate the expected value of X<\/em>.<\/p>\r\n

        0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35<\/p>\r\n

        A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money.<\/p>\r\n\r\n

        Binomial Distribution<\/h2>\r\n49.\u00a0X<\/em> = the number that reply \u201cyes\u201d\r\n\r\n50.\u00a00, 1, 2, 3, 4, 5, 6, 7, 8\r\n\r\n52.\u00a05.7\r\n\r\n54.\u00a00.4151\r\n\r\n57.\r\n

        X<\/em> = the number of patients calling in claiming to have the flu, who actually have the flu.<\/p>\r\n

        X<\/em> = 0, 1, 2, ...25<\/p>\r\n61.\r\n

          \r\n \t
        1. X<\/em> = the number of DVDs a Video to Go customer rents<\/li>\r\n \t
        2. 0.12<\/li>\r\n \t
        3. 0.11<\/li>\r\n \t
        4. 0.77<\/li>\r\n<\/ol>\r\n63. 4.43\r\n\r\n<\/section>65. 0.4734\r\n\r\n67.\r\n
            \r\n \t
          • X<\/em> = number of questions answered correctly<\/li>\r\n \t
          • X<\/em> ~ B<\/em>(<\/span>32,\u00a0<\/span>1<\/span>3<\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/li>\r\n \t
          • We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find P<\/em>(x<\/em> > 24). The event \"more than 24\" is the complement of \"less than or equal to 24.\"<\/li>\r\n \t
          • Using your calculator's distribution menu: 1 \u2013 binomcdf(<\/span>32,\u00a0<\/span>1<\/span>3<\/span><\/span>,<\/span>\u00a024<\/span>)<\/span><\/span><\/span><\/span><\/span><\/li>\r\n \t
          • P<\/em>(x<\/em> > 24) = 0<\/li>\r\n \t
          • The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero.<\/li>\r\n<\/ul>\r\n69.\r\n
              \r\n \t
            1. X<\/em> = the number of college and universities that offer online offerings.<\/li>\r\n \t
            2. 0, 1, 2, \u2026, 13<\/li>\r\n \t
            3. X<\/em> ~ B<\/em>(13, 0.96)<\/li>\r\n \t
            4. 12.48<\/li>\r\n \t
            5. 0.0135<\/li>\r\n \t
            6. P<\/em>(x<\/em> = 12) = 0.3186 P<\/em>(x<\/em> = 13) = 0.5882 More likely to get 13.<\/li>\r\n<\/ol>\r\n71.\r\n
                \r\n \t
              1. X<\/em> = the number of fencers who do not<\/strong> use the foil as their main weapon<\/li>\r\n \t
              2. 0, 1, 2, 3,... 25<\/li>\r\n \t
              3. X<\/em> ~ B<\/em>(25,0.40)<\/li>\r\n \t
              4. 10<\/li>\r\n \t
              5. 0.0442<\/li>\r\n \t
              6. The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.<\/li>\r\n<\/ol>\r\n73.\r\n
                  \r\n \t
                1. X<\/em> = the number of audits in a 20-year period<\/li>\r\n \t
                2. 0, 1, 2, \u2026, 20<\/li>\r\n \t
                3. X<\/em> ~ B<\/em>(20, 0.02)<\/li>\r\n \t
                4. 0.4<\/li>\r\n \t
                5. 0.6676<\/li>\r\n \t
                6. 0.0071<\/li>\r\n<\/ol>\r\n75.\r\n
                    \r\n \t
                  1. X<\/em> = the number of matches<\/li>\r\n \t
                  2. 0, 1, 2, 3<\/li>\r\n \t
                  3. X<\/em> ~ B<\/em>(<\/span>3<\/span>,<\/span>1<\/span>6<\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/li>\r\n \t
                  4. In dollars: \u22121, 1, 2, 3<\/li>\r\n \t
                  5. 1<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/li>\r\n \t
                  6. Multiply each Y<\/em> value by the corresponding X<\/em> probability from the PDF table. The answer is \u22120.0787. You lose about eight cents, on average, per game.<\/li>\r\n \t
                  7. The house has the advantage.<\/li>\r\n<\/ol>\r\n77.\r\n\r\n
                    \r\n
                    \r\n
                    \r\n
                      \r\n \t
                    1. X<\/em> ~ B<\/em>(15, 0.281)\r\n
                      \"This<\/span><\/figure>\r\n
                        \r\n \t
                      1. Mean = \u03bc<\/em> = np<\/em> = 15(0.281) = 4.215<\/li>\r\n \t
                      2. Standard Deviation = \u03c3<\/em> =\u00a0[latex]\\sqrt{npq}=\\sqrt{15\\left(0.281\\right)\\left(0.719\\right)}=1.7409[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                      3. P<\/em>(x<\/em> > 5) = 1 \u2013 P<\/em>(x<\/em> \u2264 5) = 1 \u2013 binomcdf(15, 0.281, 5) = 1 \u2013 0.7754 = 0.2246\r\n
                        <\/div>\r\nP<\/em>(x<\/em> = 3) = binompdf(15, 0.281, 3) = 0.1927\r\n
                        <\/div>\r\nP<\/em>(x<\/em> = 4) = binompdf(15, 0.281, 4) = 0.2259\r\n
                        <\/div>\r\nIt is more likely that four people are literate that three people are.<\/li>\r\n<\/ol>\r\n

                        Geometric Distribution<\/h2>\r\n<\/section>78.\u00a0X<\/em> = the number of freshmen selected from the study until one replied \"yes\" that same-sex couples should have the right to legal marital status.\r\n\r\n80.\u00a01,2,\u2026\r\n\r\n82. 1.4\r\n\r\n85.\r\n
                          \r\n \t
                        1. X<\/em> = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.<\/li>\r\n \t
                        2. X<\/em> ~ G<\/em>(0.40)<\/li>\r\n \t
                        3. 2.5<\/li>\r\n \t
                        4. 0.0187<\/li>\r\n \t
                        5. 0.2304<\/li>\r\n<\/ol>\r\n87.\r\n
                          X<\/em> = the number of pages that advertise footwear<\/div>\r\n
                          X<\/em> takes on the values 0, 1, 2, ..., 20<\/div>\r\n
                          X<\/em> ~ B<\/em>(20, [latex]\\frac{{29}}{{192}}\\\\[\/latex])<\/div>\r\n
                          3.02<\/div>\r\n
                          No<\/div>\r\n
                          0.9997<\/div>\r\n
                          X<\/em> = the number of pages we must survey until we find one that advertises footwear. X<\/em> ~ G<\/em>([latex]\\frac{{29}}{{192}}\\\\[\/latex])<\/div>\r\n
                          0.3881<\/div>\r\n
                          6.6207 pages<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>\r\n
                          <\/div>\r\n
                          89.0, 1, 2, and 3<\/div>\r\n
                          <\/div>\r\n
                          \r\n\r\n91.\r\n
                            \r\n \t
                          1. X<\/em> ~ G<\/em>(0.25)<\/li>\r\n \t
                          2. Mean = [latex]\\mu=\\frac{{1}}{{p}}={{1}}{{0.25}}=4\\\\[\/latex], Standard Deviation = [latex]\\sigma=\\sqrt{\\frac{{1-p}}{{{p^{2}}}}}\\\\[\/latex]\u2248 3.4641<\/li>\r\n \t
                          3. P<\/em>(x<\/em> = 10) = geometpdf(0.25, 10) = 0.0188<\/li>\r\n \t
                          4. P<\/em>(x<\/em> = 20) = geometpdf(0.25, 20) = 0.0011<\/li>\r\n \t
                          5. P<\/em>(x<\/em> \u2264 5) = geometcdf(0.25, 5) = 0.7627<\/li>\r\n<\/ol>\r\n92.\u00a0X<\/em> = the number of business majors in the sample.\r\n\r\n94.\u00a02, 3, 4, 5, 6, 7, 8, 9\r\n\r\n96.\u00a06.26\r\n\r\n98.\r\n
                              \r\n \t
                            1. X<\/em> = the number of pages that advertise footwear<\/li>\r\n \t
                            2. 0, 1, 2, 3, ..., 20<\/li>\r\n \t
                            3. X<\/em> ~ H<\/em>(29, 163, 20); r<\/em> = 29, b<\/em> = 163, n<\/em> = 20<\/li>\r\n \t
                            4. 3.03<\/li>\r\n \t
                            5. 1.5197<\/li>\r\n<\/ol>\r\n100.\r\n
                                \r\n \t
                              1. X<\/em> = the number of Patriots picked<\/li>\r\n \t
                              2. 0, 1, 2, 3, 4<\/li>\r\n \t
                              3. X<\/em> ~ H<\/em>(4, 8, 9)<\/li>\r\n \t
                              4. Without replacement<\/li>\r\n<\/ol>\r\n103.\u00a00, 1, 2, 3, 4, \u2026\r\n\r\n105.\u00a00.0485\r\n\r\n107.\u00a00.0214\r\n\r\n109.\u00a0X<\/em> = the number of U.S. teens who die from motor vehicle injuries per day.\r\n\r\n111.0, 1, 2, 3, 4, \u2026\r\n\r\n113. No\r\n\r\n115.\r\n
                                  \r\n \t
                                1. X<\/em> ~ P<\/em>(5.5); \u03bc<\/em> = 5.5;\u00a0[latex]=\\sqrt{5.5}{\\approx}2.3452[\/latex]<\/li>\r\n \t
                                2. P<\/em>(x<\/em> \u2264 6) = poissoncdf(5.5, 6) \u2248 0.6860<\/li>\r\n \t
                                3. There is a 15.7% probability that the law staff will receive more calls than they can handle.<\/li>\r\n \t
                                4. P<\/em>(x<\/em> > 8) = 1 \u2013 P<\/em>(x<\/em> \u2264 8) = 1 \u2013 poissoncdf(5.5, 8) \u2248 1 \u2013 0.8944 = 0.1056<\/li>\r\n<\/ol>\r\n117.\r\n

                                  Let X<\/em> = the number of defective bulbs in a string.<\/p>\r\n

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

                                  \r\n
                                  \u03bc<\/em> = np<\/em> = 100(0.03) = 3<\/div>\r\n
                                  X<\/em> ~ P<\/em>(3)<\/div>\r\n
                                  P<\/em>(x<\/em> \u2264 4) = poissoncdf(3, 4) \u2248 0.8153<\/div>\r\n<\/div>\r\n

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

                                  \r\n
                                  X<\/em> ~ B<\/em>(100, 0.03)<\/div>\r\n
                                  P<\/em>(x<\/em> \u2264 4) = binomcdf(100, 0.03, 4) \u2248 0.8179<\/div>\r\n<\/div>\r\n

                                  The Poisson approximation is very good\u2014the difference between the probabilities is only 0.0026.<\/p>\r\n\r\n<\/div>\r\n119.\r\n

                                    \r\n \t
                                  1. X<\/em> = the number of children for a Spanish woman<\/li>\r\n \t
                                  2. 0, 1, 2, 3,...<\/li>\r\n \t
                                  3. X<\/em> ~ P<\/em>(1.47)<\/li>\r\n \t
                                  4. 0.2299<\/li>\r\n \t
                                  5. 0.5679<\/li>\r\n \t
                                  6. 0.4321<\/li>\r\n<\/ol>\r\n121.\r\n
                                      \r\n \t
                                    1. X<\/em> = the number of fortune cookies that have an extra fortune<\/li>\r\n \t
                                    2. 0, 1, 2, 3,... 144<\/li>\r\n \t
                                    3. X<\/em> ~ B<\/em>(144, 0.03) or P<\/em>(4.32)<\/li>\r\n \t
                                    4. 4.32<\/li>\r\n \t
                                    5. 0.0124 or 0.0133<\/li>\r\n \t
                                    6. 0.6300 or 0.6264<\/li>\r\n \t
                                    7. As n<\/em> gets larger, the probabilities get closer together.<\/li>\r\n<\/ol>\r\n123.\r\n
                                        \r\n \t
                                      1. X<\/em> = the number of people audited in one year<\/li>\r\n \t
                                      2. 0, 1, 2, ..., 100<\/li>\r\n \t
                                      3. X<\/em> ~ P<\/em>(2)<\/li>\r\n \t
                                      4. 2<\/li>\r\n \t
                                      5. 0.1353<\/li>\r\n \t
                                      6. 0.3233<\/li>\r\n<\/ol>\r\n125.\r\n
                                          \r\n \t
                                        1. X<\/em> = the number of shell pieces in one cake<\/li>\r\n \t
                                        2. 0, 1, 2, 3,...<\/li>\r\n \t
                                        3. X<\/em> ~ P<\/em>(1.5)<\/li>\r\n \t
                                        4. 1.5<\/li>\r\n \t
                                        5. 0.2231<\/li>\r\n \t
                                        6. 0.0001<\/li>\r\n \t
                                        7. Yes<\/li>\r\n<\/ol>\r\n127. 0.0671\r\n
                                          <\/div>\r\n
                                          \r\n

                                          <\/h2>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section>","rendered":"

                                          Probability Distribution Function (PDF) for a Discrete Random Variable<\/h2>\n

                                          1.<\/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>\n0.12<\/td>\n<\/tr>\n
                                          1<\/td>\n0.18<\/td>\n<\/tr>\n
                                          2<\/td>\n0.30<\/td>\n<\/tr>\n
                                          3<\/td>\n0.15<\/td>\n<\/tr>\n
                                          4<\/td>\n0.10<\/td>\n<\/tr>\n
                                          5<\/td>\n0.10<\/td>\n<\/tr>\n
                                          6<\/td>\n0.05<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                          3.\u00a00.10 + 0.05 = 0.15<\/p>\n

                                          5. 1<\/p>\n

                                          7.\u00a00.35 + 0.40 + 0.10 = 0.85<\/p>\n

                                          9.\u00a01(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45<\/p>\n

                                          11.<\/p>\n\n\n\n\n\n\n\n\n
                                          x<\/em><\/th>\nP<\/em>(x<\/em>)<\/th>\n<\/tr>\n<\/thead>\n
                                          0<\/td>\n0.03<\/td>\n<\/tr>\n
                                          1<\/td>\n0.04<\/td>\n<\/tr>\n
                                          2<\/td>\n0.08<\/td>\n<\/tr>\n
                                          3<\/td>\n0.85<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                          13.\u00a0Let X<\/em> = the number of events Javier volunteers for each month.<\/p>\n

                                          15.<\/p>\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>\n0.05<\/td>\n<\/tr>\n
                                          1<\/td>\n0.05<\/td>\n<\/tr>\n
                                          2<\/td>\n0.10<\/td>\n<\/tr>\n
                                          3<\/td>\n0.20<\/td>\n<\/tr>\n
                                          4<\/td>\n0.25<\/td>\n<\/tr>\n
                                          5<\/td>\n0.35<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                          17.\u00a01 \u2013 0.05 = 0.95<\/p>\n

                                          Mean or Expected Value and Standard Deviation<\/h2>\n

                                          20.\u00a00.2 + 1.2 + 2.4 + 1.6 = 5.4<\/p>\n

                                          22.The values of P<\/em>(x<\/em>) do not sum to one.<\/p>\n

                                          24.Let X<\/em> = the number of years a physics major will spend doing post-graduate research.<\/p>\n

                                          27.\u00a01(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years<\/p>\n

                                          29.\u00a0X<\/em> is the number of years a student studies ballet with the teacher.<\/p>\n

                                          31.\u00a00.10 + 0.05 + 0.10 = 0.25<\/p>\n

                                          33.\u00a0The sum of the probabilities sum to one because it is a probability distribution.<\/p>\n

                                          35. [latex]-2\\left(\\frac{40}{52}\\right)+30\\left(\\frac{12}{52}\\right)=-1.54+6.92=5.38[\/latex]<\/p>\n

                                          38. The variable of interest is X<\/em>, or the gain or loss, in dollars.<\/p>\n

                                          The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 \u2013 12 = 40 cards that are not face cards.<\/p>\n

                                          We first need to construct the probability distribution for X<\/em>. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X<\/em> to determine the expected value.<\/p>\n\n\n\n\n\n\n\n
                                          Card Event<\/th>\nX<\/em> net gain\/loss<\/th>\nP<\/em>(X<\/em>)<\/th>\n<\/tr>\n<\/thead>\n
                                          Face Card and Heads<\/td>\n6<\/td>\n[latex]\\left(\\frac{{12}}{{52}}\\right)\\left(\\frac{{1}}{{2}}\\right)=\\left(\\frac{{6}}{{52}}\\right)\\\\[\/latex]<\/td>\n<\/tr>\n
                                          Face Card and Tails<\/td>\n2<\/td>\n[latex]\\left(\\frac{{12}}{{52}}\\right)\\left(\\frac{{1}}{{2}}\\right)=\\left(\\frac{{6}}{{52}}\\right)\\\\[\/latex]<\/td>\n<\/tr>\n
                                          (Not Face Card) and (H or T)<\/td>\n\u20132<\/td>\n[latex]\\left(\\frac{{40}}{{52}}\\right)\\left(1\\right)=\\left(\\frac{{40}}{{52}}\\right)\\\\[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n