Continuous random variables (HL only)

Exactly 1 (ΣP(X=x) = 1). Use this to find any unknown probability.

E(X) = Σ x·P(X=x) — each value times its probability, all added.

E(X²) = Σ x²·P(X=x) — square each value, then weight by its probability.

Var(X) = E(X²) − [E(X)]² — mean of the squares minus the square of the mean.

Set the sum of all probabilities equal to 1 and solve for k.

E(X) = 1(0.2)+2(0.3)+3(0.4)+4(0.1) = 2.4.

Var = 6.6 − 2.4² = 6.6 − 5.76 = 0.84.

E(aX + b) = a·E(X) + b. (And Var(aX + b) = a²·Var(X).)

f(x) ≥ 0 everywhere, and the total area ∫ over all x of f(x) dx = 1.

Integrate the pdf: P(a<X<b) = ∫ from a to b of f(x) dx (the area under the curve).

A single point has zero width, so zero area; probability comes from intervals (areas).

Integrate f over its support, set the result equal to 1, and solve for k.

∫₀⁴ kx dx = 8k = 1, so k = 1/8.

∫₀³ kx² dx = 9k = 1, so k = 1/9.

Yes — it's a density, not a probability. Only the total AREA must equal 1.

No — endpoints have probability 0, so < and ≤ give the same area.

E(X) = ∫ x·f(x) dx over the support (the continuous version of Σ x·P).

Solve ∫ from the bottom of the support up to m of f(x) dx = 0.5 (half the area to the left).

It's where f is tallest: solve f'(x) = 0 (a maximum), or read the peak off the curve.

Var(X) = ∫ x²·f(x) dx − [E(X)]² — mean of the squares minus the square of the mean.

E(X) = ∫₀² x·(3/8)x² dx = ∫₀² (3/8)x³ dx = 3/2.

∫₀ᵐ (3/8)x² dx = m³/8 = 0.5, so m³ = 4 and m = ∛4 ≈ 1.59.

At the endpoint where f is largest (e.g. x = b if f is increasing) — there's no interior peak.

E(Y) = a·E(X) + b and Var(Y) = a²·Var(X).