Expected value and variance of a discrete variable
The expected value is the probability-weighted sum of the values; the variance measures the spread around it.
Worked example
The random variable X has this distribution. Work out the expected value and the variance.
| x | P(X = x) |
|---|---|
| 1 | 0.10 |
| 2 | 0.20 |
| 4 | 0.50 |
| 5 | 0.20 |
Worked solution
- 1.E(X) = Σ xᵢ · pᵢ = 1 · 0.10 + 2 · 0.20 + 4 · 0.50 + 5 · 0.20 = 3.50
- 2.Var(X) = Σ (xᵢ − E(X))² · pᵢ = 1.65
- 3.Or: Var(X) = E(X²) − E(X)² = 13.9000 − 3.5000² = 1.65
Typical mistake: Confusing E(X²) with E(X)².
Your turn
Three problems of exactly this type, each time with new numbers. Comma or point; fractions like 1/6 work too.
Task 1 of 3Expected value
Task
The random variable X has this distribution. Work out the expected value and the variance.
| x | P(X = x) |
|---|---|
| 0 | 0.30 |
| 1 | 0.40 |
| 3 | 0.20 |
| 4 | 0.10 |
E(X) (2 decimal places)
Var(X) (2 decimal places)
Comma or point; fractions like 1/6 work too.
Formulas for this topic
- E(X) = Σ xᵢ · P(X = xᵢ)
- Var(X) = Σ (xᵢ − E(X))² · P(X = xᵢ)
- Shortcut: Var(X) = E(X²) − E(X)²
Typical mistake: Confusing E(X²) with E(X)².
Frequently asked questions
What is the typical mistake in "Expected value and variance of a discrete variable"?
Confusing E(X²) with E(X)².
Must the probabilities add up to 1?
Yes, for a complete distribution. It is a good check before you calculate.