SOLUTION FOR ASSIGNMENT #2                                       Spring 1999
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#1:

Plot of the following:

  x value | pdf value | CDF value
  --------+-----------+-----------
     0    | 0.017212  | 0.017212
     1    | 0.135175  | 0.152388
     2    | 0.602580  | 0.754967
     3    | 0.162015  | 0.916983
     4    | 0.077062  | 0.994044
     5    | 0.004594  | 0.998638
     6    | 0.001362  | 1.000000

#2:

E[X]   = (0*0.017212)+(1*0.135175)+(2*0.602580)+(3*0.162015)+
         (4*0.077062)+(5*0.004594)+(6*0.001362)

       = 2.1658

E[X^2] = (0^2*0.017212)+(1^2*0.135175)+(2^2*0.602580)+(3^2*0.162015)+
         (4^2*0.077062)+(5^2*0.004594)+(6^2*0.001362)

       = 5.4005

Mean = E[X] = 2.1658

Variance = E[X^2] - E[X]^2 = 0.7098

#3:

See as3_3sol.c

#4:

Using Excel, mean of the 10000 generated values is 2.1491 and the variance
is 0.695069.  These values are reasonably close to the theoretically computed
values from the distribution.  If more than 10000 values were used, one
would expect the "closeness" to be even better.

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