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Worked Examples · Example 3

Q.A random variable XX has the following probability distribution:
XX: 1, 2, 3, 4; P(X)P(X): 0.1, 0.3, 0.4, 0.2. Find E(X)E(X) and Var(X)Var(X).

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Step 1 — Compute E(X)E(X).

E(X)=∑xip(xi)=1(0.1)+2(0.3)+3(0.4)+4(0.2)E(X) = \sum x_i p(x_i) = 1(0.1)+2(0.3)+3(0.4)+4(0.2)

=0.1+0.6+1.2+0.8=2.7= 0.1+0.6+1.2+0.8 = 2.7

Step 2 — Compute E(X2)E(X^2).

E(X2)=∑xi2p(xi)=12(0.1)+22(0.3)+32(0.4)+42(0.2)E(X^2) = \sum x_i^2 p(x_i) = 1^2(0.1)+2^2(0.3)+3^2(0.4)+4^2(0.2)

=0.1+1.2+3.6+3.2=8.1= 0.1+1.2+3.6+3.2 = 8.1

Step 3 — Compute Var(X)Var(X).

Var(X)=E(X2)−[E(X)]2=8.1−(2.7)2=8.1−7.29=0.81Var(X) = E(X^2) - [E(X)]^2 = 8.1 - (2.7)^2 = 8.1 - 7.29 = 0.81

Dual-solve check (independent recomputation). Recomputing E(X)E(X) by adding the products in a different order: 0.8+1.2=2.00.8+1.2 = 2.0, then 2.0+0.6=2.62.0+0.6=2.6, then 2.6+0.1=2.72.6+0.1=2.7 — same value. Recomputing Var(X)Var(X) via the definition E[(X−μ)2]E[(X-\mu)^2] with μ=2.7\mu=2.7: …

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