IB Maths AA HL Topic 4 — Statistics & Probability Paper 1 & 2 HL only ~7 min read

Mean & Variance of a CRV

The mean (expected value) of a continuous random variable X is E(X) = ∫ x·f(x) dx — note the extra factor of x compared with the validity integral. Variance comes from Var(X) = E(X²) − [E(X)]², where E(X²) = ∫ x²·f(x) dx. For linear transformations, E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X) — the constant b shifts the mean but doesn’t affect spread.

📘 What you need to know

Mean: E(X) = ∫ x·f(x) dx over the pdf’s domain — given in the booklet.
Mean of squares: E(X²) = ∫ x²·f(x) dx — note the squared x, NOT the squared f.
Variance: Var(X) = E(X²) − [E(X)]² — booklet (alternative form ∫(x − μ)²f dx is rarely faster).
SD: σ = √Var(X) — square-root the variance.
Don’t confuse E(X²) with [E(X)]² — they differ by exactly the variance.
Symmetry shortcut: if f is symmetric about x = a, then E(X) = a automatically.
Linear mean: E(aX + b) = aE(X) + b.
Linear variance: Var(aX + b) = a²Var(X) — b drops out.

The mean E(X)

Mean of a continuous random variable μ = E(X) = ∫_−∞^∞ x·f(x) dx

The “−∞ to ∞” in the limits sounds scary but in practice means “over the values where f(x) ≠ 0″ — the rest contributes 0. So if f is nonzero only on [a, b], the integral is just ∫_a^b x·f(x) dx.

Symmetry shortcut: if the graph of f has axis of symmetry at x = a, then E(X) = a by inspection — no integration. The same applies for piecewise pdfs symmetric about a midpoint.

Variance and standard deviation

Variance — most practical form Var(X) = E(X²) − [E(X)]²

E(X²) — “mean of squares”

∫ x²·f(x) dx

integrate x² weighted by f

[E(X)]² — “square of mean”

(∫ x·f(x) dx)²

find the mean first, THEN square it

These two are different! Their difference is precisely the variance — confusing them is the single biggest exam slip.

Linear transformations

If a and b are constants, the random variable aX + b has:

Linear transformation rules E(aX + b) = a·E(X) + b and Var(aX + b) = a²·Var(X)

The constant b shifts every value by b, so the mean shifts too — but the spread stays the same. Multiplying by a stretches the distribution by a, so the SD is multiplied by |a| (and variance by a²).

🧭 Recipe — find mean, variance, SD

Identify f(x) and its domain; check for symmetry which can give E(X) free.
Mean: integrate x·f(x) over the domain.
E(X²): integrate x²·f(x) over the domain.
Variance: Var(X) = E(X²) − [E(X)]²; SD = √Var(X).
For aX + b: apply E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X).

Worked examples

WE 1

Mean of a simple polynomial pdf

The continuous random variable X has pdf f(x) = 3x² for 0 ≤ x ≤ 1 (and 0 otherwise). Find E(X).

Apply E(X) = ∫ x·f(x) dx E(X) = ∫₀¹ x · 3x² dx = ∫₀¹ 3x³ dx = [3x⁴/4]₀¹ = 3/4 E(X) = 3/4 most of the density sits near x = 1, so a mean > 1/2 (midpoint) makes sense

WE 2

Mean by symmetry — no integration needed

The continuous random variable X has pdf f(x) = 34(1 − x²) for −1 ≤ x ≤ 1 (and 0 otherwise). Find E(X).

Test for symmetry f(−x) = (3/4)(1 − (−x)²) = (3/4)(1 − x²) = f(x) → f is symmetric about x = 0 Symmetric pdf → E(X) = axis of symmetry E(X) = 0 verifying: x·f(x) is an odd function, so its integral on [−1, 1] is 0 ✓

WE 3

Full pipeline: E(X), E(X²), Var, SD

The continuous random variable X has pdf f(x) = 18x for 0 ≤ x ≤ 4 (and 0 otherwise). Find E(X), Var(X) and the standard deviation of X.

Step 1 — find E(X) E(X) = ∫₀⁴ x · (1/8)x dx = (1/8)∫₀⁴ x² dx = (1/8)[x³/3]₀⁴ = (1/8)(64/3) = 8/3 Step 2 — find E(X²) E(X²) = ∫₀⁴ x² · (1/8)x dx = (1/8)∫₀⁴ x³ dx = (1/8)[x⁴/4]₀⁴ = (1/8)(64) = 8 Step 3 — Var(X) = E(X²) − [E(X)]² Var(X) = 8 − (8/3)² = 8 − 64/9 = 72/9 − 64/9 = 8/9 Step 4 — SD = √Var σ = √(8/9) = (2√2)/3 ≈ 0.943 E(X) = 8/3; Var(X) = 8/9; σ = (2√2)/3 work in fractions throughout — 64/9 stays exact, 0.111… loses precision

WE 4

Linear transformation — E and Var rules

The continuous random variable X has E(X) = 5 and Var(X) = 4. Find: (a) E(3X − 7); (b) Var(3X − 7); (c) the standard deviation of 3X − 7.

(a) Apply E(aX + b) = aE(X) + b E(3X − 7) = 3(5) − 7 = 15 − 7 = 8 (b) Apply Var(aX + b) = a²Var(X) — the −7 vanishes Var(3X − 7) = 3² · Var(X) = 9(4) = 36 (c) SD = √Var σ(3X − 7) = √36 = 6 (a) 8; (b) 36; (c) 6 subtracting 7 shifts the mean but leaves the spread untouched — typical Paper 1 fact-check

WE 5

Mean of a piecewise pdf

The continuous random variable X has pdf:
f(x) = 23x for 0 ≤ x ≤ 1;
f(x) = 23 for 1 ≤ x ≤ 2;
f(x) = 0 otherwise.
Find E(X).

Quick validity check (it should integrate to 1) Triangle area + rectangle area = (1/2)(1)(2/3) + (1)(2/3) = 1/3 + 2/3 = 1 ✓ Split E(X) into two pieces E(X) = ∫₀¹ x · (2/3)x dx + ∫₁² x · (2/3) dx Piece 1 ∫₀¹ (2/3)x² dx = (2/3)[x³/3]₀¹ = (2/3)(1/3) = 2/9 Piece 2 (2/3)∫₁² x dx = (2/3)[x²/2]₁² = (2/3)(2 − 1/2) = (2/3)(3/2) = 1 Add them E(X) = 2/9 + 1 = 11/9 E(X) = 11/9 ≈ 1.222 always integrate piece-by-piece — there’s no shortcut around the junction

WE 6

Real-world: full pipeline + linear transformation

A delivery service’s wait time T (in hours, 0 ≤ T ≤ 2) has pdf f(t) = 38t² for 0 ≤ t ≤ 2 (and 0 otherwise). The courier’s pay (in dollars) is P = 5T − 2.

(a) Find E(T), Var(T) and σ(T). (b) Hence find E(P) and σ(P).

(a) Step 1 — E(T) E(T) = ∫₀² t · (3/8)t² dt = (3/8)∫₀² t³ dt = (3/8)[t⁴/4]₀² = (3/8)(4) = 3/2 Step 2 — E(T²) E(T²) = ∫₀² t² · (3/8)t² dt = (3/8)∫₀² t⁴ dt = (3/8)[t⁵/5]₀² = (3/8)(32/5) = 12/5 Step 3 — Variance and SD Var(T) = 12/5 − (3/2)² = 12/5 − 9/4 = 48/20 − 45/20 = 3/20 σ(T) = √(3/20) = √15 / 10 ≈ 0.387 (b) Apply linear transformation rules to P = 5T − 2 E(P) = 5·(3/2) − 2 = 15/2 − 2 = 11/2 = 5.5 Var(P) = 5²·Var(T) = 25·(3/20) = 75/20 = 15/4 σ(P) = √(15/4) = √15 / 2 ≈ 1.94 (a) E(T)=3/2, Var(T)=3/20, σ(T)≈0.387 (b) E(P)=$5.50, σ(P)≈$1.94 notice σ(P) = 5·σ(T) directly — multiplying by 5 multiplies the SD by |5|

💡 Top tips

Check for symmetry first — gives E(X) for free; saves an integral.
Work in fractions throughout — exact answers stay clean (e.g., 8/9 not 0.888…).
Keep constants outside the integral; multiply at the end.
For SD of aX + b, the answer is |a|·σ(X) — sign of a doesn’t matter (since variance squares it).
Sanity check: Variance is always ≥ 0; if you get a negative number, something’s wrong.

⚠ Common mistakes

Using E(X²) = [E(X)]² — they differ by exactly the variance.
Forgetting the extra x in E(X) = ∫ x·f(x) dx — without it you’re checking validity, not finding the mean.
Keeping the +b when computing Var(aX + b) — the constant doesn’t affect spread.
Reporting variance as SD — square-root the variance to get σ.
Skipping the piece-by-piece split for piecewise pdfs — you can’t integrate one expression across the junction.

That closes Topic 4 — Statistics & Probability. Up next is Topic 5 — Calculus, the longest and most heavily weighted HL topic. We’ll start with limits and continuity, then move through differentiation rules, applications (kinematics, optimisation, related rates), integration techniques, and finish with differential equations and Maclaurin series. Calculus quietly underpins half of what you’ve already used in Topics 3 and 4 (rates of change, areas, expected values for CRVs) — so it’ll feel like the theory catching up with the practice.

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