Expectation and Variance

Statistics — Third Secondary

Expectation and Variance

Worked examples

Example 1→Two fair coins are tossed, and X is the number of heads. The distribution is f(0) = 1 ÷ 4, f(1) = 1 ÷ 2, f(2) = 1 ÷ 4. Find the expectation, the variance and the standard deviation of X.

μ = 0 × 1 ÷ 4 + 1 × 1 ÷ 2 + 2 × 1 ÷ 4 = 0 + 0.5 + 0.5 = 1.
Σ x² f(x) = 0 × 1 ÷ 4 + 1 × 1 ÷ 2 + 4 × 1 ÷ 4 = 0 + 0.5 + 1 = 1.5.
σ² = 1.5 − 1² = 0.5.
σ = √0.5 ≈ 0.71.
Example 2 A random variable X takes the values 1, 2, 3, 4 with probabilities f(x) = x ÷ 10. Find the expectation and the variance.

Comparison aspect	Main idea	Keywords
Column 3	x × f(x) = 0.1 0.4 0.9 1.6 with sum μ = 3
Column 4	x² × f(x) = 0.1 0.8 2.7 6.4 with sum = 10
•	σ² = 10 − 3² = 1
•	σ = √1 = 1
Example 3	In a game a player pays 10 pounds to play once. If the player wins which happens with probability 0.1 the player receives 50 pounds otherwise nothing is received. Let X be the net gain of the player. Find the expectation and the variance and decide whether the game favours the player	gameplayerpayspounds

If the player wins, the net gain = 50 − 10 = 40 pounds with probability 0.1. If the player does not win, the net gain = −10 pounds with probability 0.9.
μ = 40 × 0.1 + (−10) × 0.9 = 4 − 9 = −5.
Σ x² f(x) = 1600 × 0.1 + 100 × 0.9 = 160 + 90 = 250.
σ² = 250 − (−5)² = 250 − 25 = 225, and σ = 15 pounds.
The expectation is negative, so the game favours the organiser. The player loses 5 pounds on average in the long run for each game played.
Example 4 If E(X) = 4 and Var(X) = 5, and Y = 3X + 2, find E(Y), Var(Y) and σ_Y.

E(Y) = 3 × 4 + 2 = 14.
Var(Y) = 3² × 5 = 45.
σ_Y = √45 ≈ 6.71.
Notice that the constant 2 affected the expectation only, and did not change the variance.

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