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)`

Answer 1

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