Homework Soursefind the expected value and standard deviation of the distru
find the expected value and standard deviation of the distrubuion by filling in the following table
E(X) = VAR(X)= SD(X)=
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | .02 | .03 | .09 | .25 | .40 | .16 | .05 |
| X*P(X) | |||||||
| (X-u) | |||||||
| (X-u)2 | |||||||
| P(X)*(X-u)2 |
Solution
E(X)= 4.63
Var(x)= 12.8669
SD(X) = 3.587046
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0.02 | 0.03 | 0.08 | .25 | .4 | .16 | .05 |
| X*P(X) | .02 | .06 | .24 | 1 | 2 | ..96 | .35 |
| (X-U) | .98 | 1.94 | 2.76 | 3 | 3 | 5.04 | 6.65 |
| (X-U)2 | 0.9604 | 3.7636 | 7.6176 | 9 | 9 | 25.4016 | 44.2225 |
| P(X)*(X-u)2 | .01921 | .11291 | .60941 | 2.25 | 3.6 | 4.06426 | 2.21113 |
