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Question
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
| X | 0 | 1 | 2 | 3 | 4 | 5 |
| P(X = x) | `(1)/(32)` | `(5)/(32)` | `(10)/(32)` | `(10)/(32)` | `(5)/(32)` | `(1)/(32)` |
Solution
E(X) = \[\sum\limits_{i=1}^{6} x_i\cdot\text{P}(x_i)\]
= `0(1/32) + 1(5/32) + 2(10/32) + 3(10/32) + 4(5/32) + 5(1/32)`
= `(0 + 5 + 20 + 30 + 20 + 5)/(32)`
= `(80)/(32)`
= `(5)/(2)`
= 2.5
E(X2) = \[\sum\limits_{i=1}^{6} x_i^2\text{P}(x_i)\]
= `0^2(1/32) + 1^2(5/32) + 2^2(10/32) + 3^2(10/32) + 4^2(5/32) + 5^2(1/32)`
= `(0 + 5 + 40 + 90 + 80 + 25)/(32)`
= `(240)/(32)`
= `(15)/(2)`
∴ Var(X) = E(X2) – [E(X)]2
= `(15)/(2) - (5/2)^2`
= `(5)/(4)`
= 1.25
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