Question

Solve the following problem :

Find the expected value and variance of the r. v. X if its probability distribution is as follows.

x	– 1	0	1
P(X = x)	`(1)/(5)`	`(2)/(5)`	`(2)/(5)`

Answer 1

E(X) = \[\sum\limits_{i=1}^{3} x_i\cdot\text{P}(x_i)\]

= `-1 (1/5) + 0(2/5) + 1(2/5)`

= `(1)/(5)`

E(X²) = \[\sum\limits_{i=1}^{3} x_i^2\text{P}(x_i)\]

= `(-1)^2 (1/5) + 0^2(2/5) + 1^2(2/5)`

= `(3)/(5)`

∴ Var(X) = E(X²) – [E(X)]²

= `(3)/(5) - (1/5)^2`

= `(14)/(25)`..