How do you solve Let X = a random var. that takes on any of the values -1,0,1,2 with probabilities?

Question

p(X = -1) = 0.2 p(X = 0) = 0.3, p(X = 1) = 0.1 and p(X = 2) = 0.4

Find the variance.

Answer 1

mean = -1*.2 + 0* .3 + 1*.1 +2*.4 = 0.7

E(X^2) = (-1)^2 *(.2) + (0)^2*(.3) +(1)^2*(.1) +(2)^2*(.4) =1.9

variance = (1.9) - (0.7)^2 = 1.41

Answer 2

Use the formula for the Variance:

Var(X) = Exp(X^2) -( Exp(X))^2
Where Exp(X) is given by SUM [X*p(X)] over the random variable X. Where p(X) is the probability that the random variable takes on a particular value. Exp(X^2) is given analogously by SUM [(X^2)*p(X)].

By the way Exp(X) is the expected valu of X and is equivalent to the mean. I would go further but I dont do pple's h/w.

Good Luck.

Answer 3

Make table:

x | p | p.x | p.x.x
===============
-1 | 0.2 | -0.2 | 0.2
0 | 0.3 | 0 | 0
+1 | 0.1 | +0.1 | 0.1
+2 | 0.4 | +0.8 | 1.6
----------------------------
sum | 1 | +0.7 | 1.9

The mean is +0.7. The variance is

Var X = E(X^2) - (EX)^2
= 1.9 - (0.7)^2 = 1.41