{"id":248,"date":"2021-07-14T15:58:58","date_gmt":"2021-07-14T15:58:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-12\/"},"modified":"2023-12-05T09:13:39","modified_gmt":"2023-12-05T09:13:39","slug":"answers-to-selected-exercises-12","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/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 \u2014 Practice<\/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 \u2014 Practice<\/h2>\r\n19.\u00a00.2 + 1.2 + 2.4 + 1.6 = 5.4\r\n\r\n21.The values of P<\/em>(x<\/em>) do not sum to one.\r\n\r\n23.Let X<\/em> = the number of years a physics major will spend doing post-graduate research.\r\n\r\n25.\u00a01 \u2013 0.35 \u2013 0.20 \u2013 0.15 \u2013 0.10 \u2013 0.05 = 0.15\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

Binomial Distribution \u2014 Practice<\/h2>\r\n37.\u00a0X<\/em>\u00a0= the number that reply \u201cyes\u201d\r\n\r\n39.\u00a00, 1, 2, 3, 4, 5, 6, 7, 8\r\n\r\n41. 5.7\r\n\r\n43. 0.4151\r\n

Geometric Distribution \u2014 Practice<\/h2>\r\n45.\u00a0X<\/em>\u00a0= 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\n47. 1,2,...\r\n\r\n49. 1.4\r\n

Hypergeometric Distribution \u2014 Practice<\/h2>\r\n51.\u00a0X<\/em>\u00a0= the number of business majors in the sample.\r\n\r\n53.\u00a02, 3, 4, 5, 6, 7, 8, 9\r\n\r\n55. 6.26\r\n\r\n
\r\n
\r\n
\r\n

Poisson Distribution \u2014 Practice<\/h2>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>57. 0, 1, 2, 3, 4, ...\r\n\r\n59. 0.0485\r\n\r\n61. 0.0214\r\n\r\n63.\u00a0X<\/em>\u00a0= the number of U.S. teens who die from motor vehicle injuries per day.\r\n\r\n65. 0, 1, 2, 3, 4, ...\r\n\r\n67. No\r\n

Mean or Expected Value and Standard Deviation \u2014 Homework<\/h2>\r\n71. The variable of interest is X<\/em>, or the gain or loss, in dollars. The face cards are jack, queen, and king. There are (3)(4) = 12 face cards and 52 \u2013 12 = 40 cards that are not face cards.\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\n73.\r\n
      \r\n \t
    1. 0.1<\/li>\r\n \t
    2. 1.6<\/li>\r\n<\/ol>\r\n75.\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\n77.\u00a04.85 years\r\n\r\n79. b\r\n\r\n81.\u00a0Let <\/span>X<\/em> = the amount of money to be won on a ticket. The following table shows the PDF for <\/span>X<\/em>.<\/span>\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\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>. 0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35. 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

        \r\n
        \r\n
        \r\n

        Binomial Distribution \u2014 Homework<\/span><\/h2>\r\n
        83.\u00a0X<\/em> = the number of patients calling in claiming to have the flu, who actually have the flu.\u00a0X<\/em> = 0, 1, 2, ...2585. 0.016587.\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\n89. d. 4.43\r\n\r\n<\/section>91. c\r\n\r\n93.\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\n95.\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\n97.\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\n99.\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\n101.\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\n103.\r\n\r\n
                    \r\n
                    \r\n
                    a. X<\/em> ~ B<\/em>(15, 0.281)\r\n
                    \"This<\/span><\/figure>\r\nb. Mean = \u03bc<\/em> = np<\/em> = 15(0.281) = 4.215\r\n\r\nStandard Deviation =\u00a0\u03c3\u00a0<\/em>=\u00a0[latex]\\sqrt{npq}=\\sqrt{15\\left(0.281\\right)\\left(0.719\\right)}=1.7409[\/latex]\r\n\r\nc. 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\r\nP<\/em>(x<\/em> = 3) = binompdf(15, 0.281, 3) = 0.1927\r\n\r\nP<\/em>(x<\/em> = 4) = binompdf(15, 0.281, 4) = 0.2259\r\n\r\nIt is more likely that four people are literate than three people are.\r\n\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section>\r\n

                    Geometric Distribution \u2014 Homework<\/h2>\r\n
                    \r\n
                    \r\n
                    \r\n
                    \r\n
                    \r\n\r\n105.\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\n107.\r\n
                        \r\n \t
                      1. X<\/em> = the number of pages that advertise footwear<\/li>\r\n \t
                      2. X<\/em> takes on the values 0, 1, 2, ..., 20<\/li>\r\n \t
                      3. X<\/em> ~ B<\/em>(20, [latex]\\frac{{29}}{{192}}[\/latex])<\/li>\r\n \t
                      4. 3.02<\/li>\r\n \t
                      5. No<\/li>\r\n \t
                      6. 0.9997<\/li>\r\n \t
                      7. 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])<\/li>\r\n \t
                      8. 0.3881<\/li>\r\n \t
                      9. 6.6207 pages<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>\r\n
                        <\/div>\r\n
                        109. 0, 1, 2, and 3<\/div>\r\n
                        <\/div>\r\n
                        \r\n\r\n111.\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\n
                          \r\n
                          \r\n
                          \r\n

                          Hypergeometric Distribution \u2014 Homework<\/h2>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n113.\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\n115.\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\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>
                              \r\n
                              \r\n
                              \r\n

                              Poisson Distribution \u2013 Homework<\/h2>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>\r\n
                              \r\n
                              \r\n
                              \r\n
                              \r\n\r\n117.\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\n119.\r\n

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

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

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

                                  \r\n
                                    \r\n \t
                                  • X<\/em> ~ B<\/em>(100, 0.03)<\/li>\r\n \t
                                  • P<\/em>(x<\/em> \u2264 4) = binomcdf(100, 0.03, 4) \u2248 0.8179<\/li>\r\n<\/ul>\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\n121.\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\n123.\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\n125.\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\n127.\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\n129. d\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/section><\/div>\r\n<\/section><\/div>\r\n<\/section>","rendered":"

                                            Probability Distribution Function (PDF) for a Discrete Random Variable \u2014 Practice<\/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 \u2014 Practice<\/h2>\n

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

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

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

                                            25.\u00a01 \u2013 0.35 \u2013 0.20 \u2013 0.15 \u2013 0.10 \u2013 0.05 = 0.15<\/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

                                            Binomial Distribution \u2014 Practice<\/h2>\n

                                            37.\u00a0X<\/em>\u00a0= the number that reply \u201cyes\u201d<\/p>\n

                                            39.\u00a00, 1, 2, 3, 4, 5, 6, 7, 8<\/p>\n

                                            41. 5.7<\/p>\n

                                            43. 0.4151<\/p>\n

                                            Geometric Distribution \u2014 Practice<\/h2>\n

                                            45.\u00a0X<\/em>\u00a0= the number of freshmen selected from the study until one replied “yes” that same-sex couples should have the right to legal marital status.<\/p>\n

                                            47. 1,2,…<\/p>\n

                                            49. 1.4<\/p>\n

                                            Hypergeometric Distribution \u2014 Practice<\/h2>\n

                                            51.\u00a0X<\/em>\u00a0= the number of business majors in the sample.<\/p>\n

                                            53.\u00a02, 3, 4, 5, 6, 7, 8, 9<\/p>\n

                                            55. 6.26<\/p>\n

                                            \n
                                            \n
                                            \n
                                            \n

                                            Poisson Distribution \u2014 Practice<\/h2>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n

                                            57. 0, 1, 2, 3, 4, …<\/p>\n

                                            59. 0.0485<\/p>\n

                                            61. 0.0214<\/p>\n

                                            63.\u00a0X<\/em>\u00a0= the number of U.S. teens who die from motor vehicle injuries per day.<\/p>\n

                                            65. 0, 1, 2, 3, 4, …<\/p>\n

                                            67. No<\/p>\n

                                            Mean or Expected Value and Standard Deviation \u2014 Homework<\/h2>\n

                                            71. The variable of interest is X<\/em>, or the gain or loss, in dollars. The face cards are 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