Mathematics 9709 · syllabus 6.2
Cambridge A Level Mathematics 9709 — Linear combinations of random variables topical past papers
Expectation and variance of aX + b and of sums and differences of independent random variables, with the normal and Poisson results that follow.
- Questions
- 89
- Years
- 2012–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 30
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Linear combinations of random variables covers
Section 6.2 of Paper 6 is about combining random variables. Scaling and shifting one variable gives E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X): adding a constant moves a distribution without changing its spread. For two variables, E(aX + bY) = aE(X) + bE(Y) always, and Var(aX + bY) = a²Var(X) + b²Var(Y) when X and Y are independent.
The variance of a difference adds, because subtracting a variable still adds uncertainty: Var(X − Y) = Var(X) + Var(Y). You also need to tell X₁ + X₂, the total of two separate observations, from 2X, one observation doubled; they have the same mean but variances 2Var(X) and 4Var(X). A linear combination of independent normal variables is normal, and a sum of independent Poisson variables is Poisson, so you can find probabilities as well as means and variances.
Typical questions compare two quantities, such as whether one random mass is more than twice another, by forming a new variable like X − 2Y and finding P(X − 2Y > 0). The topic depends on the normal distribution of Paper 5 and the Poisson distribution of section 6.1, and leads into the distribution of a sample mean in section 6.4.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 6.2 Linear combinations of random variables
What you should be able to do
- Find the mean and variance of aX + b
- Find the mean and variance of sums and differences of independent random variables
- Distinguish between X₁ + X₂ + … + Xₙ and nX
- Use the fact that linear combinations of independent normal variables are normal to find probabilities
- Use the fact that a sum of independent Poisson variables is Poisson
Key ideas
- E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X)
- E(aX + bY) = aE(X) + bE(Y)
- Var(aX + bY) = a²Var(X) + b²Var(Y) for independent X and Y, so Var(X − Y) = Var(X) + Var(Y)
- Var(X₁ + X₂ + … + Xₙ) = nVar(X) for independent observations, but Var(nX) = n²Var(X)
- If X ~ Po(λ) and Y ~ Po(μ) are independent, then X + Y ~ Po(λ + μ)
How Linear combinations of random variables has been examined
From 2022 to 2026, Linear combinations of random variables was on Paper 6 · Probability & Statistics 2 in 12 of 14 exam series, worth about 13% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 6 · Probability & Statistics 2 | 30 | 13% | 12 of 14 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 8.
Filed in Momentum under: Linear Combination of Random Variables.
A worked example
Bags of rice have masses X ~ N(1010, 3²) grams and empty crates have masses Y ~ N(400, 8²) grams, all independent. A crate is packed with 4 bags. Find the probability that the total mass of a packed crate is more than 4460 g.
- Let T = X₁ + X₂ + X₃ + X₄ + Y, the total of four separate bags and one crate.
- E(T) = 4 × 1010 + 400 = 4440.
- Var(T) = 4 × 3² + 8² = 36 + 64 = 100, so the standard deviation is 10. Using 4X would wrongly give 16 × 9 = 144 for the bags.
- T is a sum of independent normal variables, so T ~ N(4440, 100).
- P(T > 4460) = P(Z > (4460 − 4440)/10) = P(Z > 2) = 1 − 0.9772.
Answer: 0.0228
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Subtracting variances for a difference: Var(X − Y) = Var(X) + Var(Y), because b² = 1 when b = −1.
- Writing Var(3X) = 3Var(X): the multiplier is squared, so Var(3X) = 9Var(X).
- Using 4X when a question means the total of four separate items, X₁ + X₂ + X₃ + X₄, whose variance is 4Var(X), not 16Var(X).
- Adding standard deviations instead of variances.
- Treating 2X as a Poisson variable: only a sum of independent Poisson variables is Poisson.
Exam technique
- Define the new variable in words and symbols first, such as 'let T = X₁ + X₂ + X₃ − Y', then find E(T) and Var(T) on separate lines.
- Turn a comparison such as 'X is more than twice Y' into P(X − 2Y > 0) before standardising.
- Keep variances, not standard deviations, until the last step, and take the square root only when you standardise.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.