Cambridge International AS & A Level Mathematics 9709: Probability & Statistics 2 – Study Guide

This guide teaches topic 6, Probability & Statistics 2, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). It covers syllabus sections 6.1 to 6.5. This content is examined on Paper 6: 1 hour 15 minutes, 50 marks, 6 to 8 structured questions, worth 20% of the A Level. Paper 6 is offered only as part of the full A Level, and it cannot be combined with Paper 4 (Mechanics). The syllabus assumes you know all of Probability & Statistics 1 and the calculus in Pure Mathematics 3. A scientific calculator is allowed, and you get MF19 (formulae and statistical tables).

Related pages: the Probability & Statistics 2 revision notes, the Probability & Statistics 2 practice questions, the Probability & Statistics 1 study guide, the A Level Mathematics hub, the printable 9709 checklist and the free A Level diagnostic.

What this unit covers

Section	What you must be able to do	Paper
6.1 The Poisson distribution	Po(λ) probabilities; mean = variance = λ; model random events; approximate B(n, p) by Poisson; approximate Po(λ) by a normal	6
6.2 Linear combinations of random variables	E and Var of aX + b and aX + bY; normal and Poisson sums	6
6.3 Continuous random variables	Properties of a probability density function (single interval); probabilities, mean, variance, median and percentiles	6
6.4 Sampling and estimation	Samples and randomness; distribution of X̄; Central Limit Theorem; unbiased estimates; confidence intervals for a mean and a proportion	6
6.5 Hypothesis tests	One- and two-tailed tests for binomial, Poisson and normal means; Type I and Type II errors	6

6.1 The Poisson distribution

X ~ Po(λ) models the number of events in a fixed interval of time or space when events occur singly, independently, at random, and at a constant average rate. From MF19:

P(X = r) = e^(−λ) λʳ / r!        mean = λ        variance = λ

If the rate is λ per unit, the count in t units is Po(λt). A mean and variance that are roughly equal in data is evidence that a Poisson model may fit.

Worked example. Calls reach a help desk at random at a mean rate of 3.2 per hour.

X ~ Po(3.2)
P(X = 2) = e^(−3.2) × 3.2² / 2 = 0.209 (3 s.f.)
P(X ≥ 3) = 1 − e^(−3.2)(1 + 3.2 + 3.2²/2) = 0.620 (3 s.f.)
P(no calls in 30 minutes): Y ~ Po(1.6), P(Y = 0) = e^(−1.6) = 0.202 (3 s.f.)

Poisson approximation to the binomial

If X ~ B(n, p) with n large and p small (the syllabus gives n > 50 and np < 5, approximately), use X ≈ Po(np).

Worked example. 1.5% of items are faulty; a box holds 200. n = 200 > 50 and np = 3 < 5, so use Po(3).

P(at most 2 faulty) ≈ e^(−3)(1 + 3 + 9/2) = 8.5e^(−3) = 0.423 (3 s.f.)

Normal approximation to the Poisson

If λ is large (λ > 15, approximately), use X ≈ N(λ, λ) with a continuity correction.

Worked example. X ~ Po(25). Find P(X > 30).

N(25, 25):  P(X > 30) = P(X ≥ 31) ≈ P(Y > 30.5)
z = (30.5 − 25)/5 = 1.1
P = 1 − Φ(1.1) = 1 − 0.8643 = 0.136 (3 s.f.)

6.2 Linear combinations of random variables

These results are not in MF19. Learn them; proofs are not required.

E(aX + b) = aE(X) + b              Var(aX + b) = a² Var(X)
E(aX + bY) = aE(X) + bE(Y)
Var(aX + bY) = a² Var(X) + b² Var(Y)     (X, Y independent)

Variances always add, even for a difference: Var(X − Y) = Var(X) + Var(Y).

• If X is normal, aX + b is normal.
• If X and Y are independent normals, aX + bY is normal.
• If X and Y are independent Poissons, X + Y is Poisson with mean λ₁ + λ₂.

Two separate items or one item scaled? The total mass of 4 separate bags is X₁ + X₂ + X₃ + X₄, with variance 4σ². Four times the mass of one bag is 4X, with variance 16σ².

Worked example. Bags of flour have mass X ~ N(1010, 8²) grams. A crate has mass Y ~ N(150, 5²) grams. Four bags are packed into one crate. Find the probability that the total mass exceeds 4200 g.

T = X₁ + X₂ + X₃ + X₄ + Y
E(T) = 4 × 1010 + 150 = 4190
Var(T) = 4 × 64 + 25 = 281
P(T > 4200): z = (4200 − 4190)/√281 = 0.5965
P = 1 − Φ(0.5965) = 0.275 (3 s.f.)

6.3 Continuous random variables

A probability density function f(x) defined on one interval must satisfy f(x) ≥ 0 and ∫ f(x) dx = 1 over that interval. Probabilities are areas: P(a < X < b) = ∫ from a to b of f(x) dx. For a continuous variable, P(X = a) = 0, so < and ≤ give the same answer. From MF19:

E(X) = ∫ x f(x) dx        Var(X) = ∫ x² f(x) dx − {E(X)}²

The median m solves ∫ from the lower limit to m of f(x) dx = 1/2. Other percentiles work the same way. The syllabus excludes explicit use of the cumulative distribution function, so find these by direct integration of f(x).

Worked example. f(x) = k(4x − x²) for 0 ≤ x ≤ 3, and 0 otherwise.

∫₀³ (4x − x²) dx = [2x² − x³/3]₀³ = 18 − 9 = 9,  so k = 1/9
E(X)  = (1/9) ∫₀³ (4x² − x³) dx = (1/9)(36 − 81/4) = 7/4
E(X²) = (1/9) ∫₀³ (4x³ − x⁴) dx = (1/9)(81 − 243/5) = 18/5
Var(X) = 18/5 − (7/4)² = 0.5375 = 0.538 (3 s.f.)
median: (1/9)(2m² − m³/3) = 1/2  →  m³ − 6m² + 13.5 = 0
        root in [0, 3]: m = 1.79 (3 s.f.)

Infinite domain. The syllabus allows domains such as x ≥ 2. If f(x) = k/x⁴ for x ≥ 2, then ∫₂^∞ kx⁻⁴ dx = k/24, so k = 24. E(X) = ∫₂^∞ 24x⁻³ dx = 3. The median solves 1 − 8/m³ = 1/2, so m³ = 16 and m = 2.52 (3 s.f.).

6.4 Sampling and estimation

A population is the whole group; a sample is the part you observe. A sample should be random so that every member has an equal chance of selection and the results are not biased. A method is unsatisfactory if some members cannot be chosen (for example, only surveying people at one place or time). Random numbers can produce a random sample: number the population, generate random numbers, and select the matching members, ignoring repeats and numbers out of range. Named methods such as quota or stratified sampling are not required.

The sample mean

X̄ is a random variable with E(X̄) = μ and Var(X̄) = σ²/n.

• If X is normal, X̄ is exactly normal: X̄ ~ N(μ, σ²/n).
• If X is not normal but n is large, the Central Limit Theorem says X̄ is approximately N(μ, σ²/n).

Worked example. A population has mean 30 and standard deviation 6. A random sample of 50 is taken. n is large, so by the CLT X̄ ≈ N(30, 36/50).

P(X̄ > 31): z = (31 − 30)/(6/√50) = 1.179
P = 1 − Φ(1.179) = 0.119 (3 s.f.)

Unbiased estimates

From MF19:

x̄ = Σx / n        s² = (1/(n − 1)) ( Σx² − (Σx)²/n )

“Unbiased” means that although individual estimates vary, the method gives the right value on average.

Confidence intervals

For a mean with known σ (normal population), or a large sample (use s):

x̄ ± z × σ/√n

For a proportion from a large sample, with p̂ the sample proportion:

p̂ ± z √( p̂(1 − p̂)/n )

Common z values from the critical values table: 90% → 1.645, 95% → 1.960, 98% → 2.326, 99% → 2.576.

Worked example. For a sample of 40 values, Σx = 2260 and Σx² = 128 400. Find a 95% confidence interval for μ.

x̄ = 2260/40 = 56.5
s² = (128400 − 2260²/40)/39 = 710/39 = 18.21
interval: 56.5 ± 1.96 × √(18.21/40) = 56.5 ± 1.322
          (55.2, 57.8)

The sample is large, so the CLT means you do not need to assume the population is normal.

Proportion. In a sample of 200, 72 have a property. p̂ = 0.36. A 90% interval is 0.36 ± 1.645 × √(0.36 × 0.64/200) = 0.36 ± 0.0558, giving (0.304, 0.416).

6.5 Hypothesis tests

Vocabulary: null hypothesis H₀ (the parameter has its usual value), alternative hypothesis H₁ (it has changed: <, > or ≠), significance level, test statistic, rejection (critical) region, acceptance region. A one-tailed test looks for change in one direction; a two-tailed test splits the significance level between both tails. Conclusions must be in context and not overstated: “there is evidence at the 5% level that…” or “there is insufficient evidence that…”.

Binomial, direct evaluation. A claim says p = 0.3. In 20 trials there are 2 successes. Test H₀: p = 0.3 against H₁: p < 0.3 at the 5% level.

X ~ B(20, 0.3) under H₀
P(X ≤ 2) = 0.0355 < 0.05
Reject H₀: evidence at the 5% level that p is less than 0.3.

Poisson, direct evaluation. A shop sells a mean of 4.5 umbrellas per week. In one week after an advert it sells 9. Test at 5% whether the mean has increased.

H₀: λ = 4.5    H₁: λ > 4.5
P(X ≥ 9) = 1 − P(X ≤ 8) = 0.0403 < 0.05
Reject H₀: evidence at the 5% level that mean weekly sales have increased.

For large n or large λ, the syllabus also allows a normal approximation, with a continuity correction, in place of direct evaluation.

Normal mean. Lengths have σ = 12 cm. H₀: μ = 500, H₁: μ ≠ 500, 5% level, sample of 36 with x̄ = 496.

z = (496 − 500)/(12/√36) = −2.000
two-tailed critical values ±1.960; −2.000 < −1.960
Reject H₀: evidence at the 5% level that the mean length is not 500 cm.

Type I and Type II errors

• Type I error: rejecting H₀ when it is true. P(Type I) = probability of landing in the rejection region when H₀ is true. For a discrete test, find the actual value, which is usually below the stated significance level.
• Type II error: accepting H₀ when it is false. You need a specific alternative value of the parameter.

Worked example (Poisson). In the umbrella test, P(X ≥ 8) = 0.0866 and P(X ≥ 9) = 0.0403 under H₀, so the rejection region is X ≥ 9 and P(Type I) = 0.0403. If the true mean is now 7, P(Type II) = P(X ≤ 8 | λ = 7) = 0.729 (3 s.f.).

Worked example (normal). In the lengths test, H₀ is accepted when 500 − 1.96 × 2 < x̄ < 500 + 1.96 × 2, that is 496.08 < x̄ < 503.92. If in fact μ = 505:

P(Type II) = Φ((503.92 − 505)/2) − Φ((496.08 − 505)/2)
           = Φ(−0.54) − Φ(−4.46) = 0.295 (3 s.f.)

Common errors

• Rate not scaled to the interval (a rate per hour used for a 30-minute count).
• Subtracting variances for X − Y.
• Using 4X for four independent items.
• Forgetting to check that k makes the total area 1 before finding E(X).
• Dividing by n instead of n − 1 for an unbiased variance estimate.
• Using σ²/n for a proportion instead of p̂(1 − p̂)/n.
• Comparing a single probability, P(X = 9), with 5% instead of the tail P(X ≥ 9).
• A conclusion that “proves” H₁, or that has no context.

Next steps

Test yourself with the Probability & Statistics 2 practice questions, then keep the revision notes for the final weeks.

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.