Cambridge International AS & A Level Mathematics 9709: Probability & Statistics 2 – Revision Notes

These notes condense topic 6, Probability & Statistics 2, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4): sections 6.1 to 6.5. All of it is examined on Paper 6 (1 hour 15 minutes, 50 marks, 20% of the A Level; A Level only, and not combinable with Paper 4). Probability & Statistics 1 and the Pure Mathematics 3 calculus are assumed. For full explanations and worked examples, use the Probability & Statistics 2 study guide.

Other links: Probability & Statistics 2 practice questions, the Probability & Statistics 1 revision notes, the A Level Mathematics hub, the printable 9709 checklist and the free A Level diagnostic.

Formulas

Result	Formula	In MF19?
Poisson Po(λ)	P(X = r) = e^(−λ) λʳ / r!; mean λ; variance λ	Yes
Linear function	E(aX + b) = aE(X) + b; Var(aX + b) = a²Var(X)	No
Sum of variables	E(aX + bY) = aE(X) + bE(Y)	No
Variance of sum (independent)	Var(aX + bY) = a²Var(X) + b²Var(Y)	No
Continuous mean	E(X) = ∫ x f(x) dx	Yes
Continuous variance	Var(X) = ∫ x² f(x) dx − {E(X)}²	Yes
Unbiased mean	x̄ = Σx / n	Yes
Unbiased variance	s² = (1/(n − 1))(Σx² − (Σx)²/n)	Yes
Central Limit Theorem	X̄ ~ N(μ, σ²/n) approximately	Yes
Sample proportion	approximately N(p, p(1 − p)/n)	Yes
CI for a mean	x̄ ± z σ/√n (or s/√n, large sample)	No
CI for a proportion	p̂ ± z √(p̂(1 − p̂)/n)	No

6.1 The Poisson distribution

Conditions for Po(λ). Events occur singly, independently, at random, at a constant mean rate. Scale λ with the interval: rate 0.8 per metre → 2.4 in 3 metres.

Approximations.

From	To	When (syllabus guide)	Continuity correction?
B(n, p)	Po(np)	n > 50 and np < 5, approximately	No
Po(λ)	N(λ, λ)	λ > 15, approximately	Yes
B(n, p)	N(np, npq)	np > 5 and nq > 5	Yes

Worked reminder (normal approximation). X ~ Po(18). Find P(X ≤ 12).

λ = 18 > 15, so use N(18, 18)
P(X ≤ 12) ≈ P(Y < 12.5)
z = (12.5 − 18)/√18 = −1.296
P = 1 − Φ(1.296) = 0.0974 (3 s.f.)

Sums. Independent X ~ Po(λ₁) and Y ~ Po(λ₂) give X + Y ~ Po(λ₁ + λ₂). Only sums: 2X is not Poisson.

6.2 Linear combinations

Method in steps.

1. Write the combination in symbols first (T = X₁ + X₂ + Y, or D = 2B − A).
2. Mean: combine the means with the same coefficients.
3. Variance: square each coefficient, then add (independent variables).
4. State the distribution (normal if the parts are normal).
5. Standardise and use Φ.

Must-know distinction.

X₁ + X₂ + X₃ (three separate items)	3X (one item, tripled)
mean 3μ	mean 3μ
variance 3σ²	variance 9σ²

Small reminder: if Var(X) = 3, then Var(5 − 2X) = (−2)² × 3 = 12. The constant 5 does not change the variance.

6.3 Continuous random variables

Checklist for a PDF on one interval.

1. f(x) ≥ 0 everywhere on the interval.
2. Total area = 1 (use it to find k).
3. P(a < X < b) = area between a and b.
4. E(X) and Var(X) from the MF19 integrals.
5. Median m: area from the lower limit up to m = 1/2. Quartiles: 1/4 and 3/4.

Worked reminder. f(x) = (3/8)x² for 0 ≤ x ≤ 2.

area:   (3/8) × [x³/3]₀² = (3/8)(8/3) = 1   ✓
E(X)  = (3/8) ∫₀² x³ dx = (3/8)(4) = 1.5
E(X²) = (3/8) ∫₀² x⁴ dx = (3/8)(32/5) = 2.4
Var(X) = 2.4 − 1.5² = 0.15
median: (3/8)(m³/3) = 1/2  →  m³ = 4  →  m = 1.59 (3 s.f.)

For an infinite domain, e.g. x ≥ 2, evaluate the improper integral as a limit: terms such as 1/x³ → 0 as x → ∞.

The cumulative distribution function is not required explicitly; integrate f(x) directly.

6.4 Sampling and estimation

Unbiased, in simple terms. Each sample gives a different estimate, but the method is right on average. That is why s² divides by n − 1: dividing by n would, on average, underestimate the population variance.

Why randomness matters. A random sample gives every member an equal chance of selection, so results are not biased. A method is unsatisfactory when part of the population cannot be chosen (one time, one place, volunteers only).

Distribution of X̄.

Population	n	X̄
Normal	any	exactly N(μ, σ²/n)
Not normal or unknown	large	approximately N(μ, σ²/n) by the CLT
Not normal or unknown	small	no result available

Confidence interval, method in steps.

1. Find x̄ (or p̂) and, if needed, s² with divisor n − 1.
2. Pick z from the critical values table: 90% → 1.645, 95% → 1.960, 98% → 2.326, 99% → 2.576.
3. Standard error: σ/√n, s/√n, or √(p̂(1 − p̂)/n).
4. Interval = estimate ± z × standard error, to 3 s.f.
5. Interpret: about 95% of intervals built this way contain the true value.

Worked reminder (sample size). σ = 5. How large must n be for a 95% interval for μ to have width less than 2?

width = 2 × 1.96 × 5/√n < 2  →  √n > 9.8  →  n > 96.04
smallest n = 97

6.5 Hypothesis tests

Method in steps (every test).

1. State H₀ and H₁ in terms of the parameter (p, λ or μ), not the sample value.
2. State the distribution under H₀.
3. Calculate the tail probability (or the z-value).
4. Compare with the significance level (or critical value). For a two-tailed test, compare the tail with half the level.
5. Conclude in context, without overstating: “evidence that…” or “insufficient evidence that…”.

Which test?

Situation	Test statistic
Single observation, binomial	Direct P(X ≤ x) or P(X ≥ x); normal approximation if n large
Single observation, Poisson	Direct tail probability; normal approximation if λ large
Population mean, σ known, normal population	z = (x̄ − μ)/(σ/√n)
Population mean, large sample	z = (x̄ − μ)/(s/√n)

Finding a discrete rejection region. H₀: p = 0.4, H₁: p > 0.4, n = 15, 5% level. Under H₀, P(X ≥ 9) = 0.0950 and P(X ≥ 10) = 0.0338. The first tail below 0.05 starts at 10, so the rejection region is X ≥ 10 and P(Type I error) = 0.0338.

Errors.

	H₀ true	H₀ false
Reject H₀	Type I error	Correct
Accept H₀	Correct	Type II error

• P(Type I) = P(in rejection region | H₀ true). For a normal test it equals the significance level; for binomial or Poisson it is the actual tail probability of the rejection region.
• P(Type II) = P(in acceptance region | a stated alternative value is true).

Quick self-test

1. X ~ Po(4). Find P(X = 4).
2. X ~ Po(1.5) and Y ~ Po(2.3) are independent. Find P(X + Y = 0).
3. Var(X) = 3. Find Var(5 − 2X).
4. X ~ N(20, 9) and Y ~ N(15, 16) are independent. State the distribution of X − Y.
5. f(x) = kx for 0 ≤ x ≤ 2. Find k and E(X).
6. Find unbiased estimates of μ and σ² from the sample 4, 7, 9, 10, 5.
7. A random sample of 25 is taken from N(60, 10²). State the distribution of X̄.
8. What value of z is used for a 99% confidence interval?
9. Define a Type I error.
10. Which distribution approximates B(100, 0.03)? Justify it.
11. Which distribution approximates Po(20)?

Answers

1. e^(−4) × 4⁴/4! = 0.195 (3 s.f.).
2. X + Y ~ Po(3.8), so P = e^(−3.8) = 0.0224 (3 s.f.).
3. (−2)² × 3 = 12.
4. X − Y ~ N(5, 25): mean 20 − 15, variance 9 + 16.
5. 2k = 1, so k = 1/2; E(X) = ∫₀² x²/2 dx = 4/3.
6. x̄ = 35/5 = 7; s² = (9 + 0 + 4 + 9 + 4)/4 = 6.5.
7. X̄ ~ N(60, 4), since 100/25 = 4. It is exactly normal because the population is normal.
8. 2.576.
9. Rejecting H₀ when H₀ is true.
10. Po(3): n = 100 > 50 and np = 3 < 5.
11. N(20, 20), with a continuity correction, because λ > 15.

Where marks are usually lost

• λ not rescaled when the interval changes.
• Normal approximation to a Poisson used without a continuity correction.
• Variances subtracted for a difference of two variables.
• nX used instead of X₁ + … + Xₙ for separate items.
• k found, but E(X) then worked out with f(x) missing the k.
• A median equation set equal to 0.5 over the wrong limits.
• Biased variance (divisor n) given when the unbiased estimate was asked for.
• Hypotheses written in terms of x̄ or the observed count instead of μ, p or λ.
• Two-tailed test compared with the full significance level in one tail.
• Type II error probability found under H₀ instead of the stated alternative value.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 6, Probability & Statistics 2 (for Paper 6): sections 6.1 to 6.5.