{"id":225,"date":"2016-04-21T22:43:43","date_gmt":"2016-04-21T22:43:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&p=225"},"modified":"2019-01-22T17:23:27","modified_gmt":"2019-01-22T17:23:27","slug":"answers-to-selected-exercises-11","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-fmcc-introstats1\/chapter\/answers-to-selected-exercises-11\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"

Continuous Probability Functions<\/h2>\r\n1.\u00a0Uniform Distribution\r\n\r\n3.\u00a0Normal Distribution\r\n\r\n5.\u00a0P<\/em>(6 < x<\/em> < 7)\r\n\r\n7. 1\r\n\r\n9. 0\r\n\r\n11. 1\r\n\r\n13.\u00a00.625\r\n\r\n15.\u00a0The probability is equal to the area from x<\/em> = [latex]\\frac{{3}}{{2}}[\/latex] to x<\/em> = 4 above the x-axis and up to f<\/em>(x<\/em>) = [latex]\\frac{{1}}{{3}}[\/latex].\r\n\r\n17.\u00a0Age is a measurement, regardless of the accuracy used.\r\n\r\n21.\u00a0It means that the value of x<\/em> is just as likely to be any number between 1.5 and 4.5.\r\n\r\n23.\u00a01.5 \u2264 x<\/em> \u2264 4.5\r\n\r\n25.\u00a00.3333\r\n\r\n27. 0\r\n\r\n29. 0.6\r\n\r\n31.\u00a0b<\/em> is 12, and it represents the highest value of x<\/em>.\r\n\r\n33. 6\r\n\r\n35.\r\n\r\n\"This\r\n\r\n37. 4.8\r\n\r\n39.\u00a0X<\/em> = The age (in years) of cars in the staff parking lot\r\n\r\n41.\u00a00.5 to 9.5\r\n\r\n43.\u00a0f<\/em>(x<\/em>) = [latex]\\frac{{1}}{{9}}[\/latex]\u00a0where x<\/em> is between 0.5 and 9.5, inclusive.\r\n\r\n45.\u00a0\u03bc<\/em> = 5\r\n\r\n47.\r\n
    \r\n \t
  1. Check student\u2019s solution.<\/li>\r\n \t
  2. [latex]\\frac{{3.5}}{{7}}[\/latex]<\/li>\r\n<\/ol>\r\n49.\r\n
      \r\n \t
    1. Check student's solution.<\/li>\r\n \t
    2. k<\/em> = 7.25<\/li>\r\n \t
    3. 7.25<\/li>\r\n<\/ol>\r\n51.\r\n
        \r\n \t
      1. X<\/em> ~ U<\/em>(1, 9)<\/li>\r\n \t
      2. Check student\u2019s solution.<\/li>\r\n \t
      3. f(x) = [latex]\\frac{{1}}{{8}}[\/latex] where\u00a01<\/span>\u2264<\/span>x<\/span>\u2264<\/span>9<\/span><\/span><\/span><\/span><\/span><\/li>\r\n \t
      4. 5<\/li>\r\n \t
      5. 2.3<\/li>\r\n \t
      6. [latex]\\frac{{15}}{{32}}[\/latex]<\/li>\r\n \t
      7. [latex]\\frac{{333}}{{800}}[\/latex]<\/li>\r\n \t
      8. [latex]\\frac{{2}}{{3}}[\/latex]<\/li>\r\n \t
      9. 8.2<\/li>\r\n<\/ol>\r\n53.\r\n
          \r\n \t
        1. X<\/em> represents the length of time a commuter must wait for a train to arrive on the Red Line.<\/li>\r\n \t
        2. X<\/em> ~ U<\/em>(0, 8)<\/li>\r\n \t
        3. [latex]f\\left(x\\right)=\\frac{1}{8}[\/latex]\u00a0where\u00a0\u2264 x \u2264 8<\/li>\r\n \t
        4. 4<\/li>\r\n \t
        5. 2.31<\/li>\r\n \t
        6. [latex]\\frac{{1}}{{8}}[\/latex]<\/li>\r\n \t
        7. [latex]\\frac{{1}}{{8}}[\/latex]<\/li>\r\n \t
        8. 3.2<\/li>\r\n<\/ol>\r\n55. 4\r\n\r\n57. 0.25\r\n\r\n59.\r\n
            \r\n \t
          1. The probability density function of X<\/em> is [latex]\\frac{{1}}{{25-16}}=\\frac{{1}}{{9}}[\/latex]\u00a0P<\/em>(X<\/em> > 19) = (25 \u2013 19) [latex]\\left(\\frac{{1}}{{9}}\\right)=\\frac{{6}}{{9}}=\\frac{{2}}{{3}}[\/latex]<\/li>\r\n<\/ol>\r\n\"\"\r\n\r\n2.P<\/em>(19 < X<\/em> < 22)\u00a0= (22 \u2013 19) [latex]\\left(\\frac{{1}}{{9}}\\right)=\\frac{{3}}{{9}}=\\frac{{1}}{{3}}[\/latex]\r\n\r\n\"\"\r\n\r\n3.The area must be 0.25, and 0.25 = (width)[latex]\\left(\\frac{{1}}{{9}}\\right)[\/latex] so width = (0.25)(9) = 2.25. Thus, the value is 25 \u2013 2.25 = 22.75.\r\n\r\n4.\u00a0This is a conditional probability question. P(x > 21| x > 18). You can do this two ways:\r\n
              \r\n \t
            • Draw the graph where a is now 18 and b is still 25. The height is [latex]\\frac{{1}}{{(25-18)}}=\\frac{{1}}{{7}}[\/latex] So, P<\/em>(x<\/em> > 21|x<\/em> > 18) = (25 \u2013 21)[latex]\\left(\\frac{{1}}{{7}}\\right)=\\frac{{4}}{{7}}[\/latex]<\/li>\r\n \t
            • Use the formula: P<\/em>(x<\/em> > 21|x<\/em> > 18) =[latex]\\frac{{P(x>21 and x>18)}}{{P(x>18)}}=\\frac{{P(x>21)}}{{P(x>18)}}=\\frac{{4}}{{7}}[\/latex]<\/li>\r\n<\/ul>\r\n63.\u00a0No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.\r\n\r\n65. 5\r\n\r\n67. [latex]f(x)={0.2}{e}^{-0.2x}[\/latex]\r\n\r\n69. 0.5350\r\n\r\n71. 6.02\r\n\r\n73.\u00a0[latex]f(x)={0.75}{e}^{-0.75x}[\/latex]\r\n\r\n75.\r\n\r\n\"This\r\n\r\n77. 0.4756\r\n\r\n79.\u00a0The mean is larger. The mean is [latex]\\frac{{1}}{{m}}=\\frac{{1}}{{0.75}}=1.33[\/latex]\r\n\r\n81.\u00a0continuous\r\n\r\n83.\u00a0m<\/em> = 0.000121\r\n\r\n85.\r\n
                \r\n \t
              1. Check student's solution<\/li>\r\n \t
              2. P<\/em>(x<\/em> < 5,730) = 0.5001<\/li>\r\n<\/ol>\r\n87.\r\n
                  \r\n \t
                1. Check student's solution.<\/li>\r\n \t
                2. k<\/em> = 2947.73<\/li>\r\n<\/ol>\r\n89.\r\n
                    \r\n \t
                  1. X<\/em> = the useful life of a particular car battery, measured in months.<\/li>\r\n \t
                  2. X<\/em> is continuous.<\/li>\r\n \t
                  3. X<\/em> ~ Exp<\/em>(0.025)<\/li>\r\n \t
                  4. 40 months<\/li>\r\n \t
                  5. 360 months<\/li>\r\n \t
                  6. 0.4066<\/li>\r\n \t
                  7. 14.27<\/li>\r\n<\/ol>\r\n91.\r\n
                      \r\n \t
                    1. X<\/em> = the time (in years) after reaching age 60 that it takes an individual to retire<\/li>\r\n \t
                    2. X<\/em> is continuous.<\/li>\r\n \t
                    3. X ~ Exp[latex]\\left(\\frac{{1}}{{5}}\\right)[\/latex]<\/li>\r\n \t
                    4. five<\/li>\r\n \t
                    5. five<\/li>\r\n \t
                    6. Check student\u2019s solution.<\/li>\r\n \t
                    7. 0.1353<\/li>\r\n \t
                    8. before<\/li>\r\n \t
                    9. 18.3<\/li>\r\n<\/ol>\r\n93. 0.3333\r\n\r\n95. 2.0794\r\n\r\n97.\r\n

                      Let T<\/em> = the life time of a light bulb.<\/p>\r\n1.The decay parameter is m<\/em> = 1\/8, and T<\/em> \u223c Exp(1\/8). The cumulative distribution function is P(T<t) = 1-[latex]{e}^{-\\frac{{t}}{{8}}}[\/latex]\u2248 0.1175.\r\n\r\n2.We want to find P<\/em>(6 < t<\/em> < 10),To do this, P<\/em>(6 < t<\/em> < 10) \u2013 P<\/em>(t<\/em> < 6)=\u00a0[latex]\\left(1-{e}^{-\\frac{{t}}{{8}}*10}\\right)-\\left(1-{e}^{-\\frac{{t}}{{8}}*6}\\right)[\/latex]\u2248 0.7135 \u2013 0.5276 = 0.1859\r\n

                      \"\"<\/span><\/figure>\r\n
                      3. We want to find 0.70 =P(T>t)=1-[latex]1-\\left(1-{e}^{-\\frac{{t}}{{8}}}\\right)={e}^{\\frac{{-t}}{{8}}}[\/latex]. [latex]{e}^{\\frac{{-t}}{{8}}}=0.70[\/latex], so [latex]\\frac{-t}{8}[\/latex]=ln(0.70)\u2248 2.85 years.\"\"99.\r\n

                      Let X<\/em> = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number<\/u> of no-hitters per season<\/u> is Poisson with mean \u03bb<\/em> = 3.<\/p>\r\n\r\n

                      <\/div>\r\nTherefore, (X<\/em> = 0) =[latex]\\frac{{{3}^{0}{e}^{-3}}}{{0!}}={e}^{-3}[\/latex]\u2248 0.0498\r\n
                      \u00a0NOTE<\/div>\r\nYou could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is 13 season. For the exponential, \u00b5 = 13.\r\n\r\nTherefore, m = [latex]\\frac{{1}}{{\\mu}}[\/latex] = 3 and T \u223c Exp(3).\r\nThe desired probability is P(T > 1) = 1 \u2013 P(T < 1) = 1 \u2013 (1 \u2013 [latex]{e}^{-3}[\/latex]) = [latex]{e}^{-3}[\/latex] \u2248 0.0498. Let T = duration of time between no-hitters. We find P(T > 2|T > 1), and by the memoryless property this is simply P(T > 1), which we found to be 0.0498 in part a.\r\nLet X = the number of no-hitters is a season. Assume that X is Poisson with mean \u03bb = 3. Then P(X > 3) = 1 \u2013 P(X \u2264 3) = 0.3528.<\/figure>\r\n101.\r\n
                        \r\n \t
                      1. [latex]\\frac{{100}}{{9}}[\/latex] = 11.11<\/li>\r\n \t
                      2. P<\/em>(X<\/em> > 10) = 1 \u2013 P<\/em>(X<\/em> \u2264 10) = 1 \u2013 Poissoncdf(11.11, 10) \u2248 0.5532.<\/li>\r\n \t
                      3. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X<\/em> who arrive between successive Type B arrivals is roughly exponential with mean \u03bc<\/em> = 9 and m<\/em> =[latex]\\frac{{1}}{{9}}[\/latex]<\/li>\r\n \t
                      4. The cumulative distribution function of X is P(X<x)=1 \u2013 [latex]{e}^{\\frac{-x}{9}}[\/latex]), thus\u00a0P(X > 20) = 1 - P(X \u2264 20) = 1 - ([latex]{e}^{\\frac{-20}{9}}[\/latex])\u22480.1084.\r\nNOTE\u00a0We could also deduce that each person arriving has a 8\/9 chance of not having Type B blood.\r\nSo the probability that none of the first 20 people arrive have Type B blood is [latex]{\\left(\\frac{8}{9}\\right)}^{20}[\/latex].\r\n(The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.).<\/li>\r\n<\/ol>","rendered":"

                        Continuous Probability Functions<\/h2>\n

                        1.\u00a0Uniform Distribution<\/p>\n

                        3.\u00a0Normal Distribution<\/p>\n

                        5.\u00a0P<\/em>(6 < x<\/em> < 7)<\/p>\n

                        7. 1<\/p>\n

                        9. 0<\/p>\n

                        11. 1<\/p>\n

                        13.\u00a00.625<\/p>\n

                        15.\u00a0The probability is equal to the area from x<\/em> = [latex]\\frac{{3}}{{2}}[\/latex] to x<\/em> = 4 above the x-axis and up to f<\/em>(x<\/em>) = [latex]\\frac{{1}}{{3}}[\/latex].<\/p>\n

                        17.\u00a0Age is a measurement, regardless of the accuracy used.<\/p>\n

                        21.\u00a0It means that the value of x<\/em> is just as likely to be any number between 1.5 and 4.5.<\/p>\n

                        23.\u00a01.5 \u2264 x<\/em> \u2264 4.5<\/p>\n

                        25.\u00a00.3333<\/p>\n

                        27. 0<\/p>\n

                        29. 0.6<\/p>\n

                        31.\u00a0b<\/em> is 12, and it represents the highest value of x<\/em>.<\/p>\n

                        33. 6<\/p>\n

                        35.<\/p>\n

                        \"This<\/p>\n

                        37. 4.8<\/p>\n

                        39.\u00a0X<\/em> = The age (in years) of cars in the staff parking lot<\/p>\n

                        41.\u00a00.5 to 9.5<\/p>\n

                        43.\u00a0f<\/em>(x<\/em>) = [latex]\\frac{{1}}{{9}}[\/latex]\u00a0where x<\/em> is between 0.5 and 9.5, inclusive.<\/p>\n

                        45.\u00a0\u03bc<\/em> = 5<\/p>\n

                        47.<\/p>\n

                          \n
                        1. Check student\u2019s solution.<\/li>\n
                        2. [latex]\\frac{{3.5}}{{7}}[\/latex]<\/li>\n<\/ol>\n

                          49.<\/p>\n

                            \n
                          1. Check student’s solution.<\/li>\n
                          2. k<\/em> = 7.25<\/li>\n
                          3. 7.25<\/li>\n<\/ol>\n

                            51.<\/p>\n

                              \n
                            1. X<\/em> ~ U<\/em>(1, 9)<\/li>\n
                            2. Check student\u2019s solution.<\/li>\n
                            3. f(x) = [latex]\\frac{{1}}{{8}}[\/latex] where\u00a01<\/span>\u2264<\/span>x<\/span>\u2264<\/span>9<\/span><\/span><\/span><\/span><\/span><\/li>\n
                            4. 5<\/li>\n
                            5. 2.3<\/li>\n
                            6. [latex]\\frac{{15}}{{32}}[\/latex]<\/li>\n
                            7. [latex]\\frac{{333}}{{800}}[\/latex]<\/li>\n
                            8. [latex]\\frac{{2}}{{3}}[\/latex]<\/li>\n
                            9. 8.2<\/li>\n<\/ol>\n

                              53.<\/p>\n

                                \n
                              1. X<\/em> represents the length of time a commuter must wait for a train to arrive on the Red Line.<\/li>\n
                              2. X<\/em> ~ U<\/em>(0, 8)<\/li>\n
                              3. [latex]f\\left(x\\right)=\\frac{1}{8}[\/latex]\u00a0where\u00a0\u2264 x \u2264 8<\/li>\n
                              4. 4<\/li>\n
                              5. 2.31<\/li>\n
                              6. [latex]\\frac{{1}}{{8}}[\/latex]<\/li>\n
                              7. [latex]\\frac{{1}}{{8}}[\/latex]<\/li>\n
                              8. 3.2<\/li>\n<\/ol>\n

                                55. 4<\/p>\n

                                57. 0.25<\/p>\n

                                59.<\/p>\n

                                  \n
                                1. The probability density function of X<\/em> is [latex]\\frac{{1}}{{25-16}}=\\frac{{1}}{{9}}[\/latex]\u00a0P<\/em>(X<\/em> > 19) = (25 \u2013 19) [latex]\\left(\\frac{{1}}{{9}}\\right)=\\frac{{6}}{{9}}=\\frac{{2}}{{3}}[\/latex]<\/li>\n<\/ol>\n

                                  \"\"<\/p>\n

                                  2.P<\/em>(19 < X<\/em> < 22)\u00a0= (22 \u2013 19) [latex]\\left(\\frac{{1}}{{9}}\\right)=\\frac{{3}}{{9}}=\\frac{{1}}{{3}}[\/latex]<\/p>\n

                                  \"\"<\/p>\n

                                  3.The area must be 0.25, and 0.25 = (width)[latex]\\left(\\frac{{1}}{{9}}\\right)[\/latex] so width = (0.25)(9) = 2.25. Thus, the value is 25 \u2013 2.25 = 22.75.<\/p>\n

                                  4.\u00a0This is a conditional probability question. P(x > 21| x > 18). You can do this two ways:<\/p>\n