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PQ220 6234F.Ch 03  13/04/2002  03:19 PM  Page 67






                                                      3-4 MEAN AND VARIANCE OF A DISCRETE RANDOM VARIABLE  67









                                   0     2    4    6     8   10       0    2    4     6    8    10
                                                (a)                               (b)
                                   Figure 3-5  A probability distribution can be viewed as a loading with the mean equal
                                   to the balance point. Parts (a) and (b) illustrate equal means, but Part (a) illustrates a
                                   larger variance.


                                       The mean of a discrete random variable X is a weighted average of the possible values of
                                   X, with weights equal to the probabilities. If  f 1x2  is the probability mass function of a loading
                                   on a long, thin beam,  E1X2  is the point at which the beam balances. Consequently,  E1X2
                                   describes the “center’’ of the distribution of X in a manner similar to the balance point of a
                                   loading. See Fig. 3-5.
                                       The variance of a random variable X is a measure of dispersion or scatter in the possible
                                   values for X. The variance of X uses weight  f 1x2  as the multiplier of each possible squared
                                   deviation 1x   2 2 . Figure 3-5 illustrates probability distributions with equal means but dif-
                                   ferent variances. Properties of summations and the definition of   can be used to show the
                                   equality of the formulas for variance.

                                                                        2
                                                              2
                                                V1X2    a  1x   2 f 1x2    a  x f 1x2   2  a  xf 1x2 	  2 a  f 1x2
                                                      x               x           x           x
                                                        2
                                                                 2
                                                                              2
                                                                      2
                                                      a  x f 1x2   2  	     a  x f 1x2    2
                                                      x                    x
                                   Either formula for V1x2  can be used. Figure 3-6 illustrates that two probability distributions
                                   can differ even though they have identical means and variances.
                 EXAMPLE 3-9       In Example 3-4, there is a chance that a bit transmitted through a digital transmission channel
                                   is received in error. Let X equal the number of bits in error in the next four bits transmitted.
                                   The possible values for X are 50, 1, 2, 3, 46 . Based on a model for the errors that is presented
                                   in the following section, probabilities for these values will be determined. Suppose that the
                                   probabilities are

                                              P1X   02   0.6561   P1X   22   0.0486   P1X   42   0.0001
                                              P1X   12   0.2916   P1X   32   0.0036











                                   0     2    4    6     8    10     0     2    4    6     8    10
                                                (a)                               (b)
                                   Figure 3-6 The probability distributions illustrated in Parts (a) and (b) differ even
                                   though they have equal means and equal variances.
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