{"id":261,"date":"2021-07-14T15:59:01","date_gmt":"2021-07-14T15:59:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-10\/"},"modified":"2023-12-05T09:21:19","modified_gmt":"2023-12-05T09:21:19","slug":"answers-to-selected-exercises-10","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-10\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"

The Standard Normal Distribution \u2014 Practice<\/h2>\r\n1. ounces of water in a bottle\r\n\r\n3. 2\r\n\r\n5. \u20134\r\n\r\n7. \u20132\r\n\r\n9. The mean becomes zero.\r\n\r\n11. z = 2\r\n\r\n13. z = 2.78\r\n\r\n15. x = 20\r\n\r\n17. x = 6.5 368\r\n\r\n19. x = 1 21 x = 1.97\r\n\r\n23. z = \u20131.67\r\n\r\n25. z \u2248 \u20130.33\r\n\r\n27. 0.67, right\r\n\r\n29. 3.14, left\r\n\r\n31. about 68%\r\n\r\n33. about 4%\r\n\r\n35. between \u20135 and \u20131\r\n\r\n37. about 50%\r\n\r\n39. about 27%\r\n\r\n41. The lifetime of a Sunshine CD player measured in years.\r\n

Using the Normal Distribution \u2014 Practice<\/h2>\r\n

43.\u00a0P<\/em>(x<\/em> < 1)<\/p>\r\n

45.\u00a0Yes, because they are the same in a continuous distribution:P<\/em>(x<\/em> = 1) = 0<\/p>\r\n

47.\u00a01 \u2013 P<\/em>(x<\/em> < 3) or P<\/em>(x<\/em> > 3)<\/p>\r\n

49.\u00a01 \u2013 0.543 = 0.457<\/p>\r\n

51.\u00a00.0013<\/p>\r\n

53.\u00a056.03<\/p>\r\n

55.\u00a00.1186<\/p>\r\n

57.<\/p>\r\n\r\n

    \r\n \t
  1. Check student\u2019s solution.<\/li>\r\n \t
  2. 3, 0.1979<\/li>\r\n<\/ol>\r\n59.\r\n
      \r\n \t
    1. Check student\u2019s solution.<\/li>\r\n \t
    2. 0.70, 4.78 years<\/li>\r\n<\/ol>\r\n
      \r\n
      \r\n
      \r\n
      \r\n

      The Standard Normal Distribution \u2014 Homework<\/h2>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section>61. c\r\n\r\n63. 1. Use the z<\/em>-score formula. z = \u20130.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average. 2. Use the z-score formula. z<\/em> = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average. 3. Height = 79 + 3.5(3.89) = 90.67 inches, which is over 7.7 feet tall. There are very few NBA players this tall so the answer is no, not likely.<\/span>\r\n

      65. <\/span><\/p>\r\n

      a. iv <\/span><\/p>\r\n

      b. Kyle\u2019s blood pressure is equal to 125 + (1.75)(14) = 149.5. 369<\/span><\/p>\r\n

      67. \u00a0Let X = an SAT math score and Y = an ACT math score. 1. X = 720 [latex]\\frac{{720-52}}{{15}}[\/latex] = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520. 2. z = 1.5 The math SAT score is 520 + 1.5(115) \u2248 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520. 3. X \u2013 \u00b5 \u03c3 = 700 \u2013 514 117 \u2248 1.59, the z<\/em>-score for the SAT. Y \u2013 \u00b5 \u03c3 = 30 \u2013 21 5.3 \u2248 1.70, the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z-score).<\/span><\/p>\r\n\r\n

      Using the Normal Distribution \u2014 Homework<\/h2>\r\n69.\u00a07.99\r\n\r\n71.\u00a00.0668\r\n\r\n73.\r\n
        \r\n \t
      1. X<\/em> ~ N<\/em>(66, 2.5)<\/li>\r\n \t
      2. 0.5404<\/li>\r\n \t
      3. No, the probability that an Asian male is over 72 inches tall is 0.0082.<\/li>\r\n<\/ol>\r\n75.\r\n
          \r\n \t
        1. X<\/em> ~ N<\/em>(36, 10)<\/li>\r\n \t
        2. The probability that a person consumes more than 40% of their calories as fat is 0.3446.<\/li>\r\n \t
        3. Approximately 25% of people consume less than 29.26% of their calories as fat.<\/li>\r\n<\/ol>\r\n77.\r\n
            \r\n \t
          1. X<\/em> = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.<\/li>\r\n \t
          2. X<\/em> ~ N<\/em>(3, 1.5)<\/li>\r\n \t
          3. The probability that the child spends less than one hour a day unsupervised is 0.0918.<\/li>\r\n \t
          4. The probability that a child spends over ten hours a day unsupervised is less than 0.0001.<\/li>\r\n \t
          5. 2.21 hours<\/li>\r\n<\/ol>\r\n79.\r\n
              \r\n \t
            1. X<\/em> = the distribution of the number of days a particular type of criminal trial will take<\/li>\r\n \t
            2. X<\/em> ~ N<\/em>(21, 7)<\/li>\r\n \t
            3. The probability that a randomly selected trial will last more than 24 days is 0.3336.<\/li>\r\n \t
            4. 22.77<\/li>\r\n<\/ol>\r\n81.\r\n\r\na) mean = 5.51, s<\/em> = 2.15\r\n\r\nb) Check student's solution.\r\n\r\nc) Check student's solution.\r\n\r\nd) Check student's solution.\r\n\r\ne) X ~ N<\/em>(5.51, 2.15)\r\n\r\nf) 0.6029\r\n\r\ng) The cumulative frequency for less than 6.1 minutes is 0.64.\r\n\r\nh) The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one.\r\n\r\ni) The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.\r\n\r\nj) The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well.\r\n\r\n83.\r\n\r\na) mean = 60,136;\u00a0s<\/em> = 10,468\r\n\r\nb) Answers will vary.\r\n\r\nc) Answers will vary.\r\n\r\nd) Answers will vary.\r\n\r\ne) X ~ N<\/em>(60136, 10468)\r\n\r\nf) 0.7440\r\n\r\ng) The cumulative relative frequency is 43\/60 = 0.717.\r\n\r\nh) The answers for part f and part g are not the same, because the normal distribution is only an approximation.\r\n

              85.<\/p>\r\n\r\n