Random variables

Question

The discrete random variable X has the following probability distribution:

x -2 0 1 3 4

P(X=x) 0.2 0.1 p 0.1 q

(i) Given that E(X)=1.7 , find the values of p and q. [4]

(ii) Find Var(X) .

Answer 1

∑P(x) = 1

∑xP(x) = E(x) = 1.7

.2 + .1 + p + .1 + q = 1

-2(.2) + 0*.1 + 1p + 3 * .1 + 4q = 1.7

p + q = .6

p + 4q = 1.8

Solving, p = .2 and q = .4

var(X) = E(X - E(x))^2 = E(X^2) - (E(x))^2

E(X^2) = -2^2(.2) + 0^2*.1 + 1^2*p + 3^2 * .1 + 4^2*q =

-2^2(.2) + 0^2*.1 + 1^2*.2 + 3^2 * .1 + 4^2*.4 =

8.3

Var(X) = E(X^2) - (E(x))^2 = 8.3 - 1.7² = 5.41