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                                                                   5-3 TWO CONTINUOUS RANDOM VARIABLES    161


                                      y




                                   2000




                                      0
                                       0      2000         x
                                   Figure 5-10 Region of integration
                                   for the probability that Y   2000  is
                                   darkly shaded and it is partitioned
                                   into two regions with x   2000 and
                                   x 
 2000.





                                   The first integral is

                                                2000   0.002y 
                    6    2000
                                                     e             0.001x   6   10     4      0.001x
                                               6
                                         6   10    °       `   ¢   e    dx            e    e     dx
                                                      0.002  2000             0.002
                                                 0                                       0
                                                                            6   10  6    1   e  2
                                                                                     e  4  a    b   0.0475
                                                                              0.002       0.001
                                   The second integral is

                                                    
     0.002y 
                   6
                                                        e           0.001x    6   10          0.003x
                                                  6
                                            6   10    °        ` ¢   e   dx               e      dx
                                                         0.002  x              0.002
                                                    2000                                2000
                                                                              6   10  6  e  6
                                                                                       a    b   0.0025
                                                                                0.002   0.003

                                   Therefore,

                                                       P 1Y 
 20002   0.0475   0.0025   0.05.

                                   Alternatively, the probability can be calculated from the marginal probability distribution of Y
                                   as follows. For y 
 0
                                                  y                                       y
                                                                                  6  0.002y
                                            f  1y2      6   10 e       dx   6   10 e          e  0.001x  dx
                                                          6  0.001x 0.002y
                                           Y

                                                  0                                       0
                                                                 e  0.001x y       6  0.002y  1   e  0.001y
                                                        6  0.002y
                                                 6   10 e        a     ` b   6   10 e       a         b
                                                                  0.001  0                    0.001
                                                 6   10  3  0.002y  11   e  0.001y 2  for y 
 0
                                                          e
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