A random variable X has the probability function as follows:X-101P(X)0

1.	A random variable X has the probability function as follows:X-101P(X)0.20.30.5Find E(3X + 1), E(X2 ) and Var(X).
Answer» E(X) = (-1) (0.2) + 0(0.3) + 1(0.5) = -0.2 + 0.5 = 0.3 So E(3X + 1) = 3 E(X) + 1 = 3(0.3) + 1 = 1.9 E(X² ) = (-1)² (0.2) + 0² (0.3) + 1² (0.5) = 0.2 + 0.5 = 0.7 Var(X) = E(X² ) – [E(X)]² = 0.7 – (0.3)² = 0.61

A random variable X has the probability function as follows:X-101P(X)0.20.30.5Find E(3X + 1), E(X2 ) and Var(X).

Answer»

E(X) = (-1) (0.2) + 0(0.3) + 1(0.5) = -0.2 + 0.5 = 0.3

So E(3X + 1) = 3 E(X) + 1 = 3(0.3) + 1 = 1.9

E(X² ) = (-1)² (0.2) + 0² (0.3) + 1² (0.5) = 0.2 + 0.5 = 0.7

Var(X) = E(X² ) – [E(X)]² = 0.7 – (0.3)² = 0.61