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Course: Statistics — Third Secondary
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Expectation and Variance

Statistics — Third Secondary

Expectation and Variance

Worked examples

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