Question

Find the expected value and variance of X using the following p.m.f.

x	–2	–1	0	1	2
P(x)	0.2	0.3	0.1	0.15	0.25

Answer 1

Expected value of X = E(X) = `sum_(i = 1)^5 x_i * P(x_i)`

= (– 2) × (0.2) + (–1) × (0.3) + 0 × (0.1) + 1 × (0.15) + 2 × (0.25)

= – 0.4 – 0.3 + 0 + 0.15 + 0.5

= – 0.05

E(X²) = `sum_(i = 1)^5 x_i^2 * P(x_i)`

= (– 2)² × (0.2) + (–1)² × (0.3) + 0² × (0.1) + 1² × (0.15) + 2² × (0.25)

= 0.8 + 0.3 + 0 + 0.15 + 1

= 2.25

∴ Variance of X = Var(X) = E(X²) – [E(X)]²

= 2.25 – (– 0.05)²

= 2.2475