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PQ220 6234F.Ch 03  13/04/2002  03:19 PM  Page 68






               68     CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

                                 Now


                                            E1X2   0f 102 	 1f 112 	 2f 122 	 3f 132 	 4f 142
                                                  010.65612 	 110.29162 	 210.04862 	 310.00362 	 410.00012
                                                  0.4

                                 Although X never assumes the value 0.4, the weighted average of the possible values is 0.4.
                                    To calculate V1X2,  a table is convenient.




                                             x     x   0.4    1x   0.42 2   f 1x2    f 1x21x   0.42 2
                                             0       0.4        0.16      0.6561      0.104976
                                             1       0.6        0.36      0.2916      0.104976
                                             2       1.6        2.56      0.0486      0.124416
                                             3       2.6        6.76      0.0036      0.024336
                                             4       3.6       12.96      0.0001      0.001296



                                                                  5
                                                             2
                                                                                2
                                                     V1X2        a   f 1x 21x   0.42   0.36
                                                                         i
                                                                       i
                                                                 i 1
                                 The alternative formula for variance could also be used to obtain the same result.
               EXAMPLE 3-10      Two new product designs are to be compared on the basis of revenue potential. Marketing
                                 feels that the revenue from design A can be predicted quite accurately to be $3 million. The
                                 revenue potential of design B is more difficult to assess. Marketing concludes that there is a
                                 probability of 0.3 that the revenue from design B will be $7 million, but there is a 0.7 proba-
                                 bility that the revenue will be only $2 million. Which design do you prefer?
                                    Let X denote the revenue from design A. Because there is no uncertainty in the revenue
                                 from design A, we can model the distribution of the random variable X as $3 million with
                                 probability 1. Therefore, E1X2   $3  million.
                                    Let Y denote the revenue from design B. The expected value of Y in millions of dollars is

                                                        E1Y2   $710.32 	 $210.72   $3.5

                                 Because E(Y) exceeds E(X), we might prefer design B. However, the variability of the result
                                 from design B is larger. That is,

                                                       2           2              2
                                                            17   3.52 10.32 	 12   3.52 10.72
                                                           5.25 millions of dollars squared

                                 Because the units of the variables in this example are millions of dollars, and because the vari-
                                 ance of a random variable squares the deviations from the mean, the units of   2  are millions
                                 of dollars squared. These units make interpretation difficult.
                                    Because the units of standard deviation are the same as the units of the random variable,
                                 the standard deviation  is easier to interpret. In this example, we can summarize our results

                                 as “the average deviation of Y from its mean is $2.29 million.’’
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