Question

Using the following activity, find the expected value and variance of the r.v.X if its probability distribution is as follows.

x	1	2	3
P(X = x)	`1/5`	`2/5`	`2/5`

Solution: µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`

E(X) = `square + square + square = square`

Var(X) = `"E"("X"^2) - {"E"("X")}^2`

= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`

= `square - square`

= `square`

Answer 1

µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`

E(X) = `1(1/5)+ 2(2/5) + 3(2/5) = 11/5`

`sumx^2 "P"_"i" = 1(1/5) + 4(2/5) + 9(2/5) = 27/5`

Var(X) = `"E"("X"^2) - {"E"("X")}^2`

= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`

= `27/5 - 121/25`

= `14/25`