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LESSON 2.5   •   Normal Distributions: Finding Areas from Values  137

                  Lesson 2.5                                                                13. (i) z = (9.2 9.12)/0.051.60; the
                                                                                                               =
                                                                                                      −
                  WHA T  DID   Y OU  LEARN ?                                                proportion of z-scores greater than
                                                                                                          =
                  LEARNING TARGET                              EXAMPLES  EXERCISES          z =1.60 is1–0.9452 0.0548.
                  Find the proportion of values to the left of a boundary in a normal distribution.  p. 131  9–12  (ii) Applet/normalcdf(lower:9.2,
                  Find the proportion of values to the right of a boundary in a normal distribution.  p. 133  13–16  upper:1000,mean: 9.12,SD: 0.05) =  Lesson 2.5
                                                                                            0.0548. The proportion of 9-ounce bags of
                  Find the proportion of values between two boundaries in a normal distribution.  p. 135  17–20
                                                                                            this brand of potato chips that weigh more
                                                                                            than 9.2 ounces is approximately 0.0548.
                  Exercises                                                                 14.  (i) z = (0.3 0.261)/0.034 1.15;
                                                                                                                 =
                                                                                                      −
                                                                                            the proportion of z-scores greater than
                  Building Concepts and Skills         of  potato  chips  weigh  less  than  the  advertised   z =1.15is1–0.8749 0.1251.
                                                                                                          =
                           (C) 2021 BFW Publishers -- for review purposes only.
                                                       9 ounces? Is this likely to pose a problem for the
                  1.   The  standard  normal  distribution  has  a  mean   company that produces these chips?
                     of          and  a  standard  deviation  of                            (ii) Applet/normalcdf(lower:0.3,
                             .                      10.  On the bench In baseball, a player’s batting average
                                                       is the proportion of times the player gets a hit out of   upper:1000,mean: 0.261,SD: 0.034) =
                  2.   True/False: The entry for each z-score in Table A is the   the total number of times at bat. The distribution of   0.1257. The proportion of Major
                     area under the standard normal curve to the left of z.  batting averages in a recent season for Major League
                  3.   What  is  the  first  step  when  finding  areas  in  any   Baseball players with at least 100 plate appearances   League Baseball players that have a
                     normal distribution?              can be modeled by a normal distribution with mean   batting average of 0.300 or higher is
                                                       µ = 0.261  and  standard  deviation  σ =  0.034.  A
                  4.   When calculating an area in a normal distribution,   player with a batting average below 0.200 is at risk of   approximately 0.1257.
                     one option is to    and then find the   sitting on the bench during important games. About
                                                                                                               =
                                                                                                       −
                     area using Table A or technology.  what percent of players are at risk?  15. (i) z = (730 538)/117 1.73; the
                  5.   In the standard normal curve, the area to the left of   11.  Lower  SATs  In  the  class  of  2019,  more  than  1.6   proportion of z-scores greater than
                                                                                                          =
                     z  = 0.53 is 0.7019. What is the area to the right of   million students took the SAT. The distribution of   z =1.73is1–0.9582 0.0418.
                     z  = 0.53? Explain your answer.   scores on the math section (out of 800) is approx-
                  6.   True/False: When finding area in a normal distribu-  imately normal with a mean of 528 and standard   (ii) Applet/normalcdf(lower: 730,
                     tion, it matters whether the boundary value is included   deviation  of  117.  About  what  proportion  of  stu-  upper:1000,mean: 528,SD:117)0.0421.
                                                                                                                   =
                                        x
                     or not—for example,  <x  120 versus  ≤ 120.  dents who took the SAT scored less than 350 on the
                                                       math section?                        About 4.2% of students who took the
                  7.   Find the proportion of observations in a standard
                     normal distribution that satisfies each of the fol-  12.  Blood pressure According to a health information   SAT math test meet the University of
                     lowing statements.                website,  the  distribution  of  adults’  diastolic  blood   Michigan requirement (at least 730).
                  (a)  z  <  2.46                      pressure (in millimeters of mercury) can be modeled
                                                       by  a  normal  distribution  with  mean  70  and  stan-
                                                                                                       −
                                                                                                             =
                  (b)  z  >− 1.66                      dard deviation 20. A healthy diastolic pressure for   16. (i) z = (120 70)/202.50; the
                  (c)  z is between 0.50 and 1.79      an adult is less than 80. About what proportion of   proportion of z-scores greater than
                                                       adults have healthy diastolic blood pressures?
                                                                                                          =
                  8.   Find the proportion of observations in a standard                    z = 2.50 is1–0.9938 0.0062.
                     normal distribution that satisfies each of the fol-  13.  Chips galore Refer to Exercise 9. What proportion
                     lowing statements.            pg   133  of 9-ounce bags of this brand of potato chips weigh   (ii) Applet/normalcdf(lower:120,
                  (a)  z  <− 1.39                      more than 9.2 ounces?               upper:1000,mean: 70,SD: 20)0.0062.
                                                                                                                  =
                  (b)  z  >  2.15                   14.  Batter up! Refer to Exercise 10. What proportion of   About 0.62% of adults have a diastolic
                            −
                                  −
                  (c)  z is between  1.11 and  0.32    players have a batting average of 0.300 or higher?
                                                    15.  SATisfactory  for  admission  Refer  to  Exercise  11.   blood pressure that is classified as
                  Mastering Concepts and Skills        The University of Michigan has a recommended SAT   hypertensive crisis (greater than 120 mm
                                                       math score of at least 730. What percent of students
                  9.   Weighing potato chips The weights of 9-ounce bags   who took the SAT math test meet this requirement?  of mercury).
                 pg   131  of a particular brand of potato chips can be mod-
                     eled by a normal distribution with mean µ =  9.12   16.  Borderline  blood  pressure  Refer  to  Exercise  12.
                     ounces  and  standard  deviation  σ = 0.05  ounce.     Diastolic blood pressures higher than 120 are clas-
                     About what percent of 9-ounce bags of this brand   sified as a hypertensive crisis and require emergency
        03_StarnesSPA4e_24432_ch02_088_153.indd   137                             07/09/20   1:57 PM
                          −
                 9.  (i) z = (9 9.12)/0.05  =−2.40: the   have batting averages less than 0.200 and
                 proportion of z-scores less than z = –2.40 is   are at risk of sitting on the bench during
                 0.0082.                              important games.
                                                                  −
                 (ii) Applet/normalcdf(lower:–1000,   11. (i) z = (350 528)/117  =−1.52; the
                                       =
                 upper: 9, mean:9.12, SD:0.05) 0.0082.   proportion of z-scores less than z = –1.52 is
                 About 0.82% of 9-ounce bags of potato chips   0.0643.
                 weigh less than 9 ounces. Because only a   (ii) Applet/normalcdf(lower:–1000,
                 small proportion of 9-ounce bags of potato   upper: 350, mean: 528,SD:117)0.0641.
                                                                             =
                 chips weigh less than the advertised 9 ounces,   The proportion of students who took the SAT
                 it is unlikely that this will be a problem for the   and scored less than 350 on the math section
                 company that produces these chips.   is about 0.0641.
                            −
                 10. (i) z = (0.2 0.261)/0.034  =−1.79; the   12. (i) z = (80 70)/200.50; the proportion
                                                                 −
                                                                       =
                 proportion of z-scores less than z =−1.79 is   of z-scores less than z = 0.50 is 0.6915.
                 0.0367.                              (ii) Tech: Applet/normalcdf(lower:–1000,
                 (ii) Tech: Applet/normalcdf(lower:–1000,  upper: 80,mean: 70,SD: 20)0.6915. The
                                                                           =
                 upper: 0.2, mean:0.261,SD: 0.034) =0.0364.   proportion of adults who have healthy
                 About 3.6% of Major League Baseball players   diastolic blood pressures is about 0.692.
                                                LESSON 2.5   •  Normal Distributions: Finding Areas from Values       137
          03_TysonTEspa4e_25177_ch02_088_153_4pp.indd   137                                                            10/11/20   7:47 PM
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