{"id":255,"date":"2021-07-14T15:59:00","date_gmt":"2021-07-14T15:59:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-11\/"},"modified":"2023-12-05T09:17:41","modified_gmt":"2023-12-05T09:17:41","slug":"answers-to-selected-exercises-11","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-11\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"

Continuous Probability Functions - Practice<\/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

The Uniform Distribution \u2013 Practice<\/h2>\r\n17.\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\n19.\u00a01.5 \u2264 x<\/em> \u2264 4.5\r\n\r\n21.\u00a00.3333\r\n\r\n23. zero\r\n\r\n25. 0.6\r\n\r\n27.\u00a0b<\/em> is 12, and it represents the highest value of x<\/em>.\r\n\r\n29. six\r\n\r\n31.\r\n\r\n\"This\r\n\r\n33. 4.8\r\n\r\n35.\u00a0X<\/em> = The age (in years) of cars in the staff parking lot\r\n\r\n37.\u00a00.5 to 9.5\r\n\r\n39.\u00a0f<\/em>(x<\/em>) = [latex]\\frac{{1}}{{9}}[\/latex]\u00a0where x<\/em> is between 0.5 and 9.5, inclusive.\r\n\r\n41.\u00a0\u03bc<\/em> = 5\r\n\r\n43.\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\n45.\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\n

      The Exponential Distribution \u2013 Practice<\/h2>\r\n47.\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\n49. five\r\n\r\n51. [latex]f(x)={0.2}{e}^{-0.2x}[\/latex]\r\n\r\n53. 0.5350\r\n\r\n55. 6.02\r\n\r\n57.\u00a0[latex]f(x)={0.75}{e}^{-0.75x}[\/latex]\r\n\r\n59.\r\n\r\n\"This\r\n\r\n61. 0.4756\r\n\r\n63.\u00a0The mean is larger. The mean is [latex]\\frac{{1}}{{m}}=\\frac{{1}}{{0.75}}=1.33[\/latex]\r\n\r\n65.\u00a0continuous\r\n\r\n67.\u00a0m<\/em> = 0.000121\r\n\r\n69.\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\n71.\r\n
          \r\n \t
        1. Check student's solution.<\/li>\r\n \t
        2. k<\/em> = 2947.73<\/li>\r\n<\/ol>\r\n
          \r\n
          \r\n

          Continuous Probability Functions \u2013 Homework<\/h2>\r\n73.\u00a0Age is a measurement, regardless of the accuracy used.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n
          \r\n
          \r\n

          The Uniform Distribution \u2013 Homework<\/h2>\r\n<\/div>\r\n<\/section>75.\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\n77.\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. Graph the probability distribution.<\/li>\r\n \t
            4. [latex]f\\left(x\\right)=\\frac{1}{8}[\/latex]\u00a0where\u00a0\u2264 x \u2264 8<\/li>\r\n \t
            5. 4<\/li>\r\n \t
            6. 2.31<\/li>\r\n \t
            7. [latex]\\frac{{1}}{{8}}[\/latex]<\/li>\r\n \t
            8. [latex]\\frac{{1}}{{8}}[\/latex]<\/li>\r\n \t
            9. 3.2<\/li>\r\n<\/ol>\r\n79. d\r\n\r\n81. b\r\n\r\n83.\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\n85.\r\n
                    \r\n \t
                  1. [latex]P(X>650) = \\frac{700-650}{700-300} = \\frac{50}{400} = \\frac{1}{8} = 0.125[\/latex].<\/li>\r\n \t
                  2. [latex]P(400<X<650) = \\frac{650-400}{700-300} = \\frac{250}{400} = 0.625[\/latex].<\/li>\r\n \t
                  3. [latex]0.10 = \\frac{\\text{width}}{700-300}[\/latex], so width =\u00a0400(0.10) = 40. Since 700 \u2013 40 = 660, the drivers travel at least 660 miles on the furthest 10% of days.<\/li>\r\n<\/ol>\r\n

                    The Exponential Distribution \u2013 Homework<\/h2>\r\n87.\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\n89.\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\n91. a\r\n\r\n93. c\r\n\r\n95. Let T<\/em> = the life time of a light bulb.\r\n\r\na.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\nb.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
                        c. 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.\"\"<\/figure>\r\n
                        d. We want to find [latex]0.02 = P(T<t) = 1-e^{- frac{1}{8}}[\/latex].\r\nSolving for\u00a0t<\/em>, [latex]e^{- \\frac{t}{8}} = 0.98[\/latex], so [latex]- \\frac{t}{8}[\/latex]\u00a0=\u00a0ln<\/em>(0.98), and t = \u20138ln<\/em>(0.98) \u2248 0.1616 years, or roughly two months.\r\nThe warranty should cover light bulbs that last less than 2 months.\r\n<\/span>Or use [latex]\\frac{ln(\\text{area to the right})}{(-m)} = \\frac{ln(1-0.2)}{- \\frac{1}{8}} = 0.1616[\/latex].e. We must find\u00a0P<\/em>(T<\/em>\u00a0< 8|T<\/em>\u00a0> 7).\r\n<\/span>Notice that by the rule of complement events,\u00a0P<\/em>(T<\/em>\u00a0< 8|T<\/em>\u00a0> 7) = 1 \u2013\u00a0P<\/em>(T<\/em>\u00a0> 8|T<\/em>\u00a0> 7).\r\n<\/span>By the memoryless property (P<\/em>(X<\/em>\u00a0>\u00a0r<\/em>\u00a0+\u00a0t<\/em>|X<\/em>\u00a0>\u00a0r<\/em>) =\u00a0P<\/em>(X<\/em>\u00a0>\u00a0t<\/em>)).\r\nSo\u00a0P<\/em>(T<\/em>\u00a0> 8|T<\/em>\u00a0> 7) =\u00a0P<\/em>(T<\/em>\u00a0> 1) = [latex]1-(1-e^{- \\frac{1}{8}}) = e^{- \\frac{1}{8}} \\approx 0.8825[\/latex]\r\nTherefore,\u00a0P<\/em>(T<\/em>\u00a0< 8|T<\/em>\u00a0> 7) = 1 \u2013 0.8825 = 0.1175.<\/figure>\r\n
                        97.\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\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\nTherefore, m = [latex]\\frac{{1}}{{\\mu}}[\/latex] = 3 and T \u223c Exp(3).\r\n\r\na. The 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.\r\nb. 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\nc. Let 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\n99.\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]. 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.<\/li>\r\n<\/ol>\r\nNOTE\r\n\r\nWe 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.).\r\n\r\n101.\u00a0Let\u00a0<\/span>T<\/em>\u00a0= duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes,\u00a0<\/span>\u03bc<\/em>\u00a0= 7 and the decay constant is [latex]m = \\frac{1}{7}[\/latex].\u00a0 The cdf is [latex[P(T<t) = 1 - e^{\\frac{t}{7}}[\/latex].<\/span>\r\n
                            \r\n \t
                          1. [latex]P(T<2) = 1 - e^{- \\frac{2}{7}} \\approx 0.2485[\/latex].<\/li>\r\n \t
                          2. [latex]P(T>15) = 1 - P(T<15) = 1 - (1- e^{- \\frac{15}{7}}) \\approx e^{- \\frac{15}{7}} \\approx 0.1173[\/latex].<\/li>\r\n \t
                          3. [latex]P(T>15 | T>10) = P(T>5) = 1 - (1-e^{- \\frac{5}{7}}) = e^{- \\frac{5}{7}} \\approx 0.4895[\/latex].<\/li>\r\n \t
                          4. Let\u00a0X<\/em>\u00a0= # of patients arriving during a half-hour period. Then\u00a0X<\/em>\u00a0has the Poisson distribution with a mean of [latex]\\frac{30}{7}[\/latex],\u00a0X<\/em>\u00a0\u223c Poisson\u00a0[latex] (\\frac{30}{7})[\/latex].\u00a0Find\u00a0P<\/em>(X<\/em>\u00a0> 8) = 1 \u2013\u00a0P<\/em>(X<\/em>\u00a0\u2264 8) \u2248 0.0311.<\/span><\/span><\/li>\r\n<\/ol>","rendered":"

                            Continuous Probability Functions – Practice<\/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

                            The Uniform Distribution \u2013 Practice<\/h2>\n

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

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

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

                            23. zero<\/p>\n

                            25. 0.6<\/p>\n

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

                            29. six<\/p>\n

                            31.<\/p>\n

                            \"This<\/p>\n

                            33. 4.8<\/p>\n

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

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

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

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

                            43.<\/p>\n

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

                              45.<\/p>\n

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

                                The Exponential Distribution \u2013 Practice<\/h2>\n

                                47.\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.<\/p>\n

                                49. five<\/p>\n

                                51. [latex]f(x)={0.2}{e}^{-0.2x}[\/latex]<\/p>\n

                                53. 0.5350<\/p>\n

                                55. 6.02<\/p>\n

                                57.\u00a0[latex]f(x)={0.75}{e}^{-0.75x}[\/latex]<\/p>\n

                                59.<\/p>\n

                                \"This<\/p>\n

                                61. 0.4756<\/p>\n

                                63.\u00a0The mean is larger. The mean is [latex]\\frac{{1}}{{m}}=\\frac{{1}}{{0.75}}=1.33[\/latex]<\/p>\n

                                65.\u00a0continuous<\/p>\n

                                67.\u00a0m<\/em> = 0.000121<\/p>\n

                                69.<\/p>\n

                                  \n
                                1. Check student’s solution<\/li>\n
                                2. P<\/em>(x<\/em> < 5,730) = 0.5001<\/li>\n<\/ol>\n

                                  71.<\/p>\n

                                    \n
                                  1. Check student’s solution.<\/li>\n
                                  2. k<\/em> = 2947.73<\/li>\n<\/ol>\n
                                    \n
                                    \n
                                    \n

                                    Continuous Probability Functions \u2013 Homework<\/h2>\n

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

                                    \n
                                    \n

                                    The Uniform Distribution \u2013 Homework<\/h2>\n<\/div>\n<\/section>\n

                                    75.<\/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

                                      77.<\/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. Graph the probability distribution.<\/li>\n
                                      4. [latex]f\\left(x\\right)=\\frac{1}{8}[\/latex]\u00a0where\u00a0\u2264 x \u2264 8<\/li>\n
                                      5. 4<\/li>\n
                                      6. 2.31<\/li>\n
                                      7. [latex]\\frac{{1}}{{8}}[\/latex]<\/li>\n
                                      8. [latex]\\frac{{1}}{{8}}[\/latex]<\/li>\n
                                      9. 3.2<\/li>\n<\/ol>\n

                                        79. d<\/p>\n

                                        81. b<\/p>\n

                                        83.<\/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