{"id":276,"date":"2016-04-21T22:43:41","date_gmt":"2016-04-21T22:43:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&p=276"},"modified":"2019-01-22T17:23:34","modified_gmt":"2019-01-22T17:23:34","slug":"answers-to-selected-exercises-8","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-fmcc-introstats1\/chapter\/answers-to-selected-exercises-8\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"1.\u00a0mean = 4 hours; standard deviation = 1.2 hours; sample size = 16\r\n\r\n3.\r\n\r\n1. Check student's solution.\r\n\r\n2. 3.5, 4.25, 0.2441\r\n\r\n5.\u00a0The fact that the two distributions are different accounts for the different probabilities.\r\n\r\n7.\r\n
    \r\n \t
  1. \u03a7<\/em> = amount of change students carry<\/li>\r\n \t
  2. \u03a7<\/em> ~ E<\/em>(0.88, 0.88)<\/li>\r\n \t
  3. X<\/span>\u23af\u23af\u23af<\/span><\/span><\/span><\/span><\/span><\/span><\/span> = average amount of change carried by a sample of 25 sstudents.<\/li>\r\n \t
  4. X<\/span>\u23af\u23af\u23af<\/span><\/span><\/span><\/span><\/span><\/span><\/span> ~ N<\/em>(0.88, 0.176)<\/li>\r\n \t
  5. 0.0819<\/li>\r\n \t
  6. 0.1882<\/li>\r\n \t
  7. The distributions are different. Part a is exponential and part b is normal.<\/li>\r\n<\/ol>\r\n9.\r\n
      \r\n \t
    1. length of time for an individual to complete IRS form 1040, in hours.<\/li>\r\n \t
    2. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours.<\/li>\r\n \t
    3. N<\/em>(<\/span>10<\/span>.53,\u00a0<\/span>1<\/span>3<\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/li>\r\n \t
    4. Yes. I would be surprised, because the probability is almost 0.<\/li>\r\n \t
    5. No. I would not be totally surprised because the probability is 0.2312<\/li>\r\n<\/ol>\r\n11.\r\n
      \r\n
      \r\n
        \r\n \t
      1. the length of a song, in minutes, in the collection<\/li>\r\n \t
      2. U<\/em>(2, 3.5)<\/li>\r\n \t
      3. the average length, in minutes, of the songs from a sample of five albums from the collection<\/li>\r\n \t
      4. N<\/em>(2.75, 0.0220)<\/li>\r\n \t
      5. 2.74 minutes<\/li>\r\n \t
      6. 0.03 minutes<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
        <\/div>\r\n
        <\/div>\r\n
        \r\n\r\n13.\r\n
          \r\n \t
        1. True. The mean of a sampling distribution of the means is approximately the mean of the data distribution.<\/li>\r\n \t
        2. True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal.<\/li>\r\n \t
        3. The standard deviation of the sampling distribution of the means will decrease making it approximately the same as the standard deviation of X as the sample size increases.<\/li>\r\n<\/ol>\r\n15.\r\n
            \r\n \t
          1. X<\/em> = the yearly income of someone in a third world country<\/li>\r\n \t
          2. the average salary from samples of 1,000 residents of a third world country<\/li>\r\n \t
          3. [latex]\\overline{X}[\/latex]\u00a0\u223c N<\/em>[latex]\\left(2000,\\frac{{8000}}{{\\sqrt{1000}}}\\right)[\/latex]<\/li>\r\n \t
          4. Very wide differences in data values can have averages smaller than standard deviations.<\/li>\r\n \t
          5. The distribution of the sample mean will have higher probabilities closer to the population mean.\r\n
            P<\/em>(2000 < [latex]\\overline{X}[\/latex]< 2100) = 0.1537<\/div>\r\n
            P<\/em>(2100 < [latex]\\overline{X}[\/latex]< 2200) = 0.1317<\/div><\/li>\r\n<\/ol>\r\n17.\u00a0[latex]\\displaystyle\\overline{{X}}[\/latex] ~ N<\/em>(4.59,\u00a0[latex]\\frac{{0.10}}{{\\sqrt{16}}}[\/latex])\r\n\r\n<\/div>\r\n

            The Central Limit Theorem for Sums<\/h2>\r\n18.\u00a00.3345\r\n\r\n20.\u00a07,833.92\r\n\r\n22.\u00a00.0089\r\n\r\n24.\u00a07,326.49\r\n\r\n26.\u00a077.45%\r\n\r\n28.\u00a00.4207\r\n\r\n30.\u00a03,888.5\r\n\r\n32.\u00a00.8186\r\n\r\n34. 5\r\n\r\n36.\u00a00.9772\r\n\r\n38.\u00a0The sample size, n<\/em>, gets larger.\r\n\r\n40. 49\r\n\r\n42.\u00a026.00\r\n\r\n44.\u00a00.1587\r\n\r\n46.\u00a01,000\r\n\r\n49.\r\n
              \r\n \t
            1. the total length of time for nine criminal trials<\/li>\r\n \t
            2. N<\/em>(189, 21)<\/li>\r\n \t
            3. 0.0432<\/li>\r\n \t
            4. 162.09; ninety percent of the total nine trials of this type will last 162 days or more.<\/li>\r\n<\/ol>\r\n51.\r\n
                \r\n \t
              1. X<\/em> = the salary of one elementary school teacher in the district<\/li>\r\n \t
              2. X<\/em> ~ N<\/em>(44,000, 6,500)<\/li>\r\n \t
              3. \u03a3X<\/em> ~ sum of the salaries of ten elementary school teachers in the sample<\/li>\r\n \t
              4. \u03a3X<\/em> ~ N<\/em>(44000, 20554.80)<\/li>\r\n \t
              5. 0.9742<\/li>\r\n \t
              6. $52,330.09<\/li>\r\n \t
              7. 466,342.04<\/li>\r\n \t
              8. Sampling 70 teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.<\/li>\r\n \t
              9. If every teacher received a $3,000 raise, the distribution of X<\/em> would shift to the right by $3,000. In other words, it would have a mean of $47,000.<\/li>\r\n<\/ol>\r\n

                Using the Central Limit Theorem<\/h2>\r\n53.\u00a0N<\/em>(25, 0.0577)\r\n\r\n55.\u00a00.0003\r\n\r\n57.\u00a025.07\r\n\r\n59.\r\n
                  \r\n \t
                1. N<\/em>(2,500, 5.7735)<\/li>\r\n \t
                2. 0<\/li>\r\n<\/ol>\r\n61. 10\r\n\r\n63. N[latex]\\left(10,\\frac{{10}}{{8}}\\right)[\/latex]\r\n\r\n65.\u00a00.7799\r\n\r\n67.\u00a01.69\r\n\r\n69.\u00a00.0072\r\n\r\n74.\r\n
                    \r\n \t
                  1. \r\n
                      \r\n \t
                    1. X<\/em> = the closing stock prices for U.S. semiconductor manufacturers<\/li>\r\n \t
                    2. i. $20.71; ii. $17.31; iii. 35<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n \r\n
                        \r\n \t
                      1. Exponential distribution, \u03a7 ~ Exp<\/em>[latex]\\left(\\frac{{1}}{{20.71}}\\right)[\/latex]<\/li>\r\n \t
                      2. Answers will vary.<\/li>\r\n \t
                      3. i. $20.71; ii. $11.14<\/li>\r\n \t
                      4. Answers will vary.<\/li>\r\n \t
                      5. Answers will vary.<\/li>\r\n \t
                      6. Answers will vary.<\/li>\r\n \t
                      7. N[latex\\left(20.71,\\frac{{17.31}}{{\\sqrt{5}}}\\right)<\/li>\r\n<\/ol>\r\n76.\u00a0two hours.\r\n\r\n78. 40.3\r\n\r\n80. almost zero\r\n\r\n82.\r\n
                          \r\n \t
                        1. 0<\/li>\r\n \t
                        2. 0.1123<\/li>\r\n \t
                        3. 0.0162<\/li>\r\n \t
                        4. 0.0003<\/li>\r\n \t
                        5. 0.0268<\/li>\r\n<\/ol>\r\n85.\r\n
                            \r\n \t
                          1. Check student\u2019s solution.<\/li>\r\n \t
                          2. [latex]\\overline{X}~\\left(60,\\frac{{9}}{{\\sqrt{25}}}\\right)[\/latex]<\/li>\r\n \t
                          3. 0.5000<\/li>\r\n \t
                          4. 59.06<\/li>\r\n \t
                          5. 0.8536<\/li>\r\n \t
                          6. 0.1333<\/li>\r\n \t
                          7. N<\/em>(1500, 45)<\/li>\r\n \t
                          8. 1530.35<\/li>\r\n \t
                          9. 0.6877<\/li>\r\n<\/ol>\r\n87.\r\n
                              \r\n \t
                            1. $52,330<\/li>\r\n \t
                            2. $46,634<\/li>\r\n<\/ol>\r\n89.\u00a0We have\r\n