Question

Consider the probability distribution of a random variable X:

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

Variance of X.

Answer 1

Here, we have

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

We know that: Var(X) = E(X²) – [E(X)]²

Where E(X) = `sum_("i" = 1)^"n" x_"i""p"_"i"` and E(X²) = `sum_("i" = 1)^"n" "p"_"i"x"i"^2`

∴ E(X) = 0 × 0.1 + 1 × 0.25 + 2 × 0.3 + 3 × 0.2 + 4 × 0.15

= 0 + 0.25 + 0.6 + 0.6 + 0.6

= 2.05

E(X²) = 0 × 0.1 + 1 × 0.25 + 4 × 0.3 + 9 × 0.2 + 16 × 0.15

= 0 + 0.25 + 1.2 + 1.8 + 2.40

= 5.65

Var(X) = 1.4475