{"id":255,"date":"2016-04-21T22:43:42","date_gmt":"2016-04-21T22:43:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&p=255"},"modified":"2021-06-15T17:54:37","modified_gmt":"2021-06-15T17:54:37","slug":"answers-to-selected-exercises-10","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/chapter\/answers-to-selected-exercises-10\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"

THE STANDARD NORMAL DISTRIBUTION<\/h2>\n1 ounces of water in a bottle\n\n3 2\n\n5 \u20134\n\n7 \u20132\n\n9 The mean becomes zero.\n\n11 z = 2\n\n13 z = 2.78\n\n15 x = 20\n\n17 x = 6.5 368\n\n19 x = 1 21 x = 1.97\n\n23 z = \u20131.67\n\n25 z \u2248 \u20130.33\n\n27 0.67, right\n\n29 3.14, left\n\n31 about 68%\n\n33 about 4%\n\n35 between \u20135 and \u20131\n\n37 about 50%\n\n39 about 27%\n\n41. The lifetime of a Sunshine CD player measured in years.\n\n44. c\n\n46. a. Use the z-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. b. Use the z-score formula. z = 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. c. 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>\n

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

50. \u00a0Let X = an SAT math score and Y = an ACT math score. a. 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. b. 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. c. X \u2013 \u00b5 \u03c3 = 700 \u2013 514 117 \u2248 1.59, the z-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>\n\n

Using the Normal Distribution<\/h2>\n

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

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

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

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

59.\u00a00.0013<\/p>\n

61.\u00a056.03<\/p>\n

63.\u00a00.1186<\/p>\n

65.<\/p>\n\n

  1. Check student\u2019s solution.<\/li>\n\t
  2. 3, 0.1979<\/li>\n<\/ol>\n67.\n
    1. Check student\u2019s solution.<\/li>\n\t
    2. 0.70, 4.78 years<\/li>\n<\/ol>\n69.\u00a07.99\n\n71.\u00a00.0668\n\n73.\n
      1. X<\/em> ~ N<\/em>(66, 2.5)<\/li>\n\t
      2. 0.5404<\/li>\n\t
      3. No, the probability that an Asian male is over 72 inches tall is 0.0082<\/li>\n<\/ol>\n75.\n
        1. X<\/em> ~ N<\/em>(36, 10)<\/li>\n\t
        2. The probability that a person consumes more than 40% of their calories as fat is 0.3446.<\/li>\n\t
        3. Approximately 25% of people consume less than 29.26% of their calories as fat.<\/li>\n<\/ol>\n77.\n
          1. X<\/em> = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.<\/li>\n\t
          2. X<\/em> ~ N<\/em>(3, 1.5)<\/li>\n\t
          3. The probability that the child spends less than one hour a day unsupervised is 0.0918.<\/li>\n\t
          4. The probability that a child spends over ten hours a day unsupervised is less than 0.0001.<\/li>\n\t
          5. 2.21 hours<\/li>\n<\/ol>\n\u00a0\n\n79.\n
            1. X<\/em> = the distribution of the number of days a particular type of criminal trial will take<\/li>\n\t
            2. X<\/em> ~ N<\/em>(21, 7)<\/li>\n\t
            3. The probability that a randomly selected trial will last more than 24 days is 0.3336.<\/li>\n\t
            4. 22.77<\/li>\n<\/ol>\n81.\n
              1. mean = 5.51, s<\/em> = 2.15<\/li>\n\t
              2. Check student's solution.<\/li>\n\t
              3. Check student's solution.<\/li>\n\t
              4. Check student's solution.<\/li>\n\t
              5. X<\/em> ~ N<\/em>(5.51, 2.15)<\/li>\n\t
              6. 0.6029<\/li>\n\t
              7. The cumulative frequency for less than 6.1 minutes is 0.64.<\/li>\n\t
              8. 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.<\/li>\n\t
              9. 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.<\/li>\n\t
              10. The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well.<\/li>\n<\/ol>\n83.\n
                1. mean = 60,136\n
                  \ns<\/em> = 10,468<\/li>\n\t
                2. Answers will vary.<\/li>\n\t
                3. Answers will vary.<\/li>\n\t
                4. Answers will vary.<\/li>\n\t
                5. X<\/em> ~ N<\/em>(60136, 10468)<\/li>\n\t
                6. 0.7440<\/li>\n\t
                7. The cumulative relative frequency is 43\/60 = 0.717.<\/li>\n\t
                8. The answers for part f and part g are not the same, because the normal distribution is only an approximation.<\/li>\n<\/ol>

                  85.<\/p>\n\n