Key Idea: This topic finds the mean and variance of a random variable. A discrete one uses a table; a continuous one uses a pdf f(x), where probability is area. It is on Paper 1 and Paper 2.
Discrete random variables: E(X) & Var(X)
- Sum to 1 — ΣP = 1 finds any missing probability.
- Mean — E(X) = Σ x·P.
- Variance — Var(X) = E(X²) − [E(X)]², where E(X²) = Σ x²·P.
Continuous random variables: the pdf
- A pdf has f(x) ≥ 0 and total area 1.
- P(a < X < b) = area under f from a to b.
- Find k first: integrate over the support and set it equal to 1.
- P(X = a) = 0, so < and ≤ give the same answer.
| Measure | How |
|---|---|
| Mean | E(X) = ∫ x·f dx |
| Median m | ∫ up to m of f = 0.5 |
| Mode | Highest point of f (f′(x) = 0, or an endpoint) |
| Variance | ∫ x²·f dx − [E(X)]² |
Linear change
- E(aX + b) = a·E(X) + b.
- Var(aX + b) = a²·Var(X).
For a continuous variable, every probability is an area.
P(X=0) = 0.5, P(X=1) = p, P(X=2) = 0.2. Find p. p = 0.3.
For that table, E(X) and Var(X)? E(X) = 0.7, Var(X) = 1.1 − 0.49 = 0.61.
f(x) = k(4 − x) on [0, 4]. Find k. 8k = 1, so k = 1/8.
f(x) = ½x on [0, 2]. Median? m²/4 = 0.5, so m = √2.
Var(X) = 4. What is Var(3X + 1)? 9 × 4 = 36.