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48                     Fundamentals of Probability and Statistics for Engineers

                p X (x)
                                          f X (x)

                                            1
                                            —
                                            3
                1                           1
               —                            —
                                            3e
                2e                          1
                                            —
                                            6e
               (a)         3          x       (b)       3              x

            Figure 3.9 (a) Partial probability mass function, p (x), and (b) partial probability


                                                    X
                      density function, f (x), of X, as described in Example 3.4
                                    X
             For part (b),
                    P…2 < X   6†ˆ F X …6†  F X …2†
                                       e            2=3     2=3  e
                                         2                        2
                                ˆ   1        …1   e   †ˆ e         :
                                        2                        2
             Figure 3.9 shows p (x) for the discrete portion and f (x) for the continuous
                             X                            X
           portion of X. They are given by:
                                       8
                                          1
                                       <    ;  at x ˆ 3;
                                p …x†ˆ
                                 X       2e
                                         0;  elsewhere;
                                       :
           and
                                        8
                                          0;  for x < 0;
                                        >
                                        >
                                          1
                                        >
                               dF X …x†  <  e  x=3 ;  for 0   x <  ; 3
                                        >
                       f …x†ˆ         ˆ
                        X                 3
                                 dx     >
                                        >  1
                                        >     x=3
                                        >       ;  for x   3:
                                        :   e
                                          6
           Note again that the area under f (x) is no longer one but is
                                       X
                                                  1
                                    1   p …3†ˆ 1    :
                                        X
                                                  2e
                         >
           To  obtain  P(X     2)  and  P(2    X    6),  both  the  discrete  and  continuous
                                      <
           portions come into play, and we have, for part (a),
                                  Z  1
                       P…X > 2†ˆ      f …x† dx ‡ p …3†
                                                 X
                                       X
                                   2
                                  1  Z  3   x=3  1  Z  1   x=3  1
                                ˆ      e   dx ‡       e   dx ‡
                                  3  2          6  3           2e
                                ˆ e  2=3
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