Homework SourseValue x of X PXx 6 029 0 1 030 5 6 028 SolutionPx 6 P x0
| Value x of X | P(X=x) |
| -6 | 0.29 |
| 0 | |
| 1 | 0.30 |
| 5 | |
| 6 | 0.28 |
Solution
P(x = -6) + P( x=0) + P(x = 1) +P( x=5) +P(x =6) =1 Legitimate condition
Now 0.29 + P(x=0) +0.30 + P(x =5) +0.28 =1
P(x =0) +P( x =5) = 1 -0.3 -0.28 = 0.42
So, condiition : P(x =0) +P( x =5) = 0.42
we should select P(x=0) and P(x=5) any va;lues between 0 and 1 such that
above condition is satfified
So, we can have P( x =0) = 0.20 and P(x =5) =0.22
