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84   Chapter 3/Discrete Random Variables and Probability Distributions


                                                               E X i ( ) = 1 p+ ( 0 1 −  p) =  p
                                   and
                                                                            2
                                                                                     p −
                                                                 p p +(0
                                                       V X i ( ) = (1 − ) 2  −  p) (1 −  p) = (1  p)
                                   Sums of random variables are discussed in Chapter 5, and there the intuitively reasonable result that
                                                                             +
                                                                   (
                                                           E X) =  E X ) + E X ) … + E X n )
                                                                                   (
                                                                         (
                                                            (
                                                                     1
                                                                           2
                                   is derived. Furthermore, for the independent trials of a binomial experiment, Chapter 5 also
                                   shows that
                                                                   (
                                                                                   (
                                                           V X) = V X ) + V X ) … + V X n )
                                                                             +
                                                                         (
                                                            (
                                                                     1
                                                                           2
                                                              p − )
                                           (
                                   Because  E X i ) =  p  and V X i ( ) = (1  p ,  we obtain the solution E X ( ) =  np  and V X ( ) =
                                     (
                                   np 1−  p).
                        Mean and
                         Variance     If X is a binomial random variable with parameters p and n,
                                                                          2
                                                        X
                                                                                    np
                                                      E
                                                                             V X
                                                  μ = ( ) = np    and    σ = ( ) = (1   − ) p        (3-8)
               Example 3-19    Mean and Variance  For the number of transmitted bits received in error in Example 3-16, n = 4
                               and p = 0.1, so
                                                                               9
                                                                                   .
                                                                          .
                                                                             0
                                                                           1
                                                                       4
                                                                              .
                                                                         0
                                              0
                                            4
                                      E X ( ) = ( ) = 0 4  and  V X ( ) = ( )( ) = 0 36
                                                    .
                                                1
                                               .
               and these results match those obtained from a direct calculation in Example 3-9.
               Exercises            FOR SECTION 3-6
                  Problem available in WileyPLUS at instructor’s discretion.
                           Tutoring problem available in WileyPLUS at instructor’s discretion
               3-91.  For each scenario (a)–(j), state whether or not the bino-  (g)   Reconsider the situation in part (f). Now suppose that the
               mial distribution is a reasonable model for the random variable   sample of 40 chips consists of chips with 1 and with 0
               and why. State any assumptions you make.            defects.

               (a)   A production process produces thousands of temperature trans-  (h)  A illing operation attempts to ill detergent packages to the

                  ducers. Let X denote the number of nonconforming transduc-  advertised weight. Let X denote the number of detergent
                  ers in a sample of size 30 selected at random from the process.  packages that are underi lled.
               (b)   From a batch of 50 temperature transducers, a sample of   (i)  Errors in a digital communication channel occur in bursts that
                  size 30 is selected without replacement. Let X denote the   affect several consecutive bits. Let X denote the number of
                  number of nonconforming transducers in the sample.  bits in error in a transmission of 100,000 bits.
               (c)   Four identical electronic components are wired to a con-  (j) Let X denote the number of surface l aws in a large coil of
                  troller that can switch from a failed component to one of   galvanized steel.
                                                                                                            .
                  the remaining spares. Let X denote the number of compo-  3-92.   Let X be a binomial random variable with p = 0 2

                  nents that have failed after a speciied period of operation.  and n = 20. Use the binomial table in Appendix A to determine
               (d)  Let X denote the number of accidents that occur along the   the following probabilities.
                                                                    (
                                                                                         (
                  federal highways in Arizona during a one-month period.  (a) P X ≤ ) 3      (b) P X > )
                                                                                             10
                                                                    (
                                                                                         (
               (e)  Let X denote the number of correct answers by a student   (c) P X = ) 6      (d) P 6 ≤ X 11)
                                                                                              ≤
                  taking a multiple-choice exam in which a student can elim-  3-93.   Let X  be a binomial random variable with p = 0 1.
                  inate some of the choices as being incorrect in some ques-  and n = 10. Calculate the following probabilities from the bino-
                  tions and all of the incorrect choices in other questions.  mial probability mass function and from the binomial table in
               (f)   Defects occur randomly over the surface of a semiconductor   Appendix A and compare results.
                  chip. However, only 80% of defects can be found by testing.   (        (
                  A sample of 40 chips with one defect each is tested. Let X   (a) P X ≤ ) 2      (b) P X > ) 8
                                                                    (
                                                                                         (
                  denote the number of chips in which the test i  nds a defect.  (c) P X = ) 4      (d) P 5 ≤  X ≤ 7)
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