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136    CHAPTER 2   •  Modeling One-Variable Quantitative Data

                                                                 SOLUTION:
                 LESSON   APP  2.5    Answers
                       100 −151.6
                1. (i) z =     =−2.06
                          25
                Table A: The proportion of z-scores less                                             1.  Draw a normal distribution.
                than z = –2.06 is 0.0197.
                Tech: Applet/normalcdf(lower:–1000,           280  288  296  304 305  312  320  328
                                                                                325
                upper: –2.06, mean:0,SD:1) = 0.0197                   Distance traveled (yards)
                (ii) Tech: Applet/normalcdf(lower:–1000,       305 −304                              2.  Perform calculations—show your
                                                             z
                                       =
                upper:100,mean:151.6, SD:25) 0.0195.          (i)      =  8  =0.13               work!
                About 2% of teen boys have cholesterol         325 −304                             (i)    Standardize and use  Table A  or
                           (C) 2021 BFW Publishers -- for review purposes only.
                                                              =
                levels less than 100 mg/dl.                   z       8  =2.63                   technology; or
                                                                                                   (ii)    Use technology without standardizing.
                       200 −151.6                            Using  Table A :   0.9957 −0.5517 0.4440
                                                                          =
                2. (i) z =      =1.94
                          25                                 Using technology:     Applet/normalcdf(lower:0.13,upper:2.63,
                Table A: The proportion of z-scores less   mean:0,SD:1)0.4440
                                                                  =
                than z =1.94 is 0.9738, so the proportion     (ii)    Applet/normalcdf(lower:305,upper:325,mean:304,SD:8) 0.4459          Be sure to answer the question that was
                                                                                          =

                of z-scores greater than z =1.94 is                                              asked.

                1–0.9738  = 0.0262.                         About   45%  of McIlroy’s drives travel between 305 and 325 yards.
                                                                                                       FOR PRACTICE     TRY EXERCISE 17.
                Tech: Applet/normalcdf(lower:1.94,
                                    =
                upper:1000,mean: 0, SD:1)0.0262.
                (ii) Tech: Applet/normalcdf(lower: 200,
                                        =
                upper:1000,mean:151.6, SD:25) 0.0264.                               LESSON   APP     2 . 5
                About 2.6% of teen boys have cholesterol
                levels greater than 200 mg/dl.           What cholesterol levels are unhealthy for teen boys?
                       170 −151.6
                3.  (i) z =    = 0.74 and                          High levels of cholesterol in the blood increase the
                          25                           risk of heart disease. For teenage boys, the distribu-
                   200  −151.6                         tion of blood cholesterol is approximately normal


                z =        =1.94                       with mean   µ = 151.6  milligrams of cholesterol per


                      25                               deciliter of blood  (mg/dl)  and standard deviation
                                                              19
                                                          σ =  25 mg/dl  .
                Table A: The proportion of z-scores       1.   About what proportion of teen boys have

                less than z = 0.74 is 0.7704 and the        cholesterol levels less than  100 mg/dl  ?

                proportion of z-scores less than         2.   Cholesterol levels of 200 or higher are consid-

                z =1.94 is 0.9738, so the proportion      ered high for teenagers. What percent of teen             Tim Macpherson/Cultura Creative/Alamy
                of z-scores between z = 0.74 and          boys have high cholesterol?
                                   =
                z =1.94 is 0.9738–0.7704 0.2034.          3.    Cholesterol levels between   170 mg/dl   and

                Tech: Applet/normalcdf(lower:0.74,           200 mg/dl  are considered borderline high for
                                                          teenagers. What percent of teen boys have
                                   =
                upper:1.94,mean: 0, SD:1)0.2035.           borderline high cholesterol levels?
                (ii) Tech: Applet/normalcdf(lower:170,
                                       =
                upper: 200,mean:151.6, SD:25) 0.2044.
                About 20.4% of teen boys have
                cholesterol levels between 170 and
                200 mg/dl.
                                                  03_StarnesSPA4e_24432_ch02_088_153.indd   136  Answers to Lesson 2.5 Exercises  07/09/20   1:57 PM
                                                         CHAPTER 2 ACTIVITY: NORMAL
                                                  DISTRIBUTIONS CANDY AND PIRANHA      1.  0; 1
                                                  This optional activity can be used any time   2.  True
                                                  after Lesson 2.5. Access it by clicking on the   3.  draw a normal distribution
                                                  link in the TE-book or by logging into the   4.  standardize
                                                  teachers’ resources on our digital platform.
                                                                                       5.  The area to the right of
                                                                                                     =
                                                                                       z = 0.53is1–0.7019 0.2981.
                                                         FULL SOLUTIONS TO LESSON 2.5
                                                  EXERCISES                            6.  False. Because a point on the number line
                                                                                       has no width, so there is no area directly above
                                                  You can find the full solutions for this lesson   the point x =120, so the areas under the
                                                  by clicking on the link in the TE-book or by   curve with x <120 and  x ≤120 are the same.
                                                  logging into the teachers’ resources on our   7.  (a) 0.9931 (b) 1–0.0485 0.9515
                                                                                                          =
                                                  digital platform.
                                                                                                    =
                                                                                       (c) 0.9633–0.69150.2718
                                                                                                          =
                                                                                       8.  (a) 0.0823 (b) 1–0.9842 0.0158
                                                                                                    =
                                                                                       (c) 0.3745–0.1335 0.2410
                136       CHAPTER 2   •   Modeling One-Variable Quantitative Data
          03_TysonTEspa4e_25177_ch02_088_153_4pp.indd   136                                                            10/11/20   7:47 PM
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