Question 4 - A-Level Maths

CAIE S2 2019 June — Question 4 7 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2019
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear combinations of normal random variables
Type	Direct comparison with scalar multiple (different variables)
Difficulty	Challenging +1.8 This question requires students to construct a new random variable from a comparison of two independent normals (finding P(X₁ ≥ 1.5X₂ or X₂ ≥ 1.5X₁)), which involves forming linear combinations, calculating variance of the difference, and careful interpretation of the inequality. It goes beyond standard textbook exercises on sums/differences of normals and requires problem-solving insight to set up correctly.
Spec	5.04b Linear combinations: of normal distributions

4 The heights of a certain variety of plant are normally distributed with mean 110 cm and variance \(1050 \mathrm {~cm} ^ { 2 }\). Two plants of this variety are chosen at random. Find the probability that the height of one of these plants is at least 1.5 times the height of the other.

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Question 4:

Answer	Marks	Guidance
Answer	Marks	Guidance
Use of \(1.5X_1 - X_2\) or similar	B1
\(E(1.5X_1 - X_2) = 1.5(110) - 110\ (= 55)\)	B1	or \(E(X_1 - 1.5X_2) = 110 - 1.5(110)\ (= -55)\)
\(\text{Var}(1.5X_1 - X_2) = 1.5^2 \times 1050 + 1050\) (or 3412.5)	M1	Correct expression or result
\(\frac{0-55}{\sqrt{3412.5}}\) or \(\frac{0-(-55)}{\sqrt{3412.5}}\ (= \pm 0.942)\)	M1	Their '55'. Allow incorrect var (dep \(> 0\) and \(\neq 1050\))
\(1 - \Phi(0.942)\)	M1	Area consistent with their working
\(= 0.173\)	A1
Ans 0.346 (3 sf)	B1	FT double their prob (must be \(<1\))

## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $1.5X_1 - X_2$ or similar | B1 | |
| $E(1.5X_1 - X_2) = 1.5(110) - 110\ (= 55)$ | B1 | or $E(X_1 - 1.5X_2) = 110 - 1.5(110)\ (= -55)$ |
| $\text{Var}(1.5X_1 - X_2) = 1.5^2 \times 1050 + 1050$ (or 3412.5) | M1 | Correct expression or result |
| $\frac{0-55}{\sqrt{3412.5}}$ or $\frac{0-(-55)}{\sqrt{3412.5}}\ (= \pm 0.942)$ | M1 | Their '55'. Allow incorrect var (dep $> 0$ and $\neq 1050$) |
| $1 - \Phi(0.942)$ | M1 | Area consistent with their working |
| $= 0.173$ | A1 | |
| Ans 0.346 (3 sf) | B1 | FT double their prob (must be $<1$) |

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4 The heights of a certain variety of plant are normally distributed with mean 110 cm and variance $1050 \mathrm {~cm} ^ { 2 }$. Two plants of this variety are chosen at random. Find the probability that the height of one of these plants is at least 1.5 times the height of the other.\\

\hfill \mbox{\textit{CAIE S2 2019 Q4 [7]}}

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