Question 3 - A-Level Maths

CAIE S2 2020 November — Question 3 6 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2020
Session	November
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear combinations of normal random variables
Type	Direct comparison with scalar multiple (different variables)
Difficulty	Standard +0.8 This question requires students to construct a new random variable (F - 0.5M) from two independent normal distributions, find its mean and variance using linear combination properties, then calculate a probability using standardization. While the individual steps are standard S2 techniques, the setup requires careful algebraic manipulation and the insight to form the correct comparison inequality, making it moderately challenging but within typical A-level scope.
Spec	5.04b Linear combinations: of normal distributions

3 The masses, in kilograms, of female and male animals of a certain species have the distributions \(\mathrm { N } \left( 102,27 ^ { 2 } \right)\) and \(\mathrm { N } \left( 170,55 ^ { 2 } \right)\) respectively. Find the probability that a randomly chosen female has a mass that is less than half the mass of a randomly chosen male. \includegraphics[max width=\textwidth, alt={}, center]{65b50bfb-5fd8-4cf3-ae3b-cffc12e23cd8-06_76_1659_484_244}

Show mark scheme Show mark scheme source

Question 3:

Answer	Marks	Guidance
Answer	Mark	Guidance
\(F - 0.5M\)	M1	SOI
\(\sim N(17,\ 27^2 + 0.25 \times 55^2)\)	B1	for \(102 - 0.5(170)\) (= 17) or 34
B1	for \(27^2 + 0.25 \times 55^2\) (= 1485.25) or \(2^2 \times 27^2 + 55^2\) (= 5941)
\(\frac{0 - \text{'17'}}{\sqrt{\text{'1485.25'}}}\) (= −0.4411)	M1	Must have an attempt at combining \(F\) and \(M\). No standard deviation/variance errors
\(P(F - 0.5M < 0) = \phi(\text{'-0.4411'}) = 1 - \phi(\text{'0.4411'})\)	M1	Correct area consistent with their figures
\(= 0.330\) (3 sf)	A1	Allow 0.33 if no greater accuracy given

Alternative method for Question 3:

Answer	Marks	Guidance
Answer	Mark	Guidance
\(2F - M\)
\(\sim N(34,\ 2^2 \times 27^2 + 55^2)\)	B1	for \(102 - 0.5(170)\) (= 17) or 34
B1	for \(27^2 + 0.25 \times 55^2\) (= 1485.25) or \(2^2 \times 27^2 + 55^2\) (= 5941)
\(\frac{0 - \text{'34'}}{\sqrt{\text{'5941'}}}\) (= −0.4411)	M1	Must have an attempt at combining \(F\) and \(M\). No standard deviation/variance errors
\(P(2F - M < 0) = \phi(\text{'-0.4411'}) = 1 - \phi(\text{'0.4411'})\)	M1	Correct area consistent with their figures
\(= 0.330\) (3 sf)	A1	Allow 0.33 if no greater accuracy given

## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| $F - 0.5M$ | M1 | SOI |
| $\sim N(17,\ 27^2 + 0.25 \times 55^2)$ | B1 | for $102 - 0.5(170)$ (= 17) or 34 |
| | B1 | for $27^2 + 0.25 \times 55^2$ (= 1485.25) or $2^2 \times 27^2 + 55^2$ (= 5941) |
| $\frac{0 - \text{'17'}}{\sqrt{\text{'1485.25'}}}$ (= −0.4411) | M1 | Must have an attempt at combining $F$ and $M$. No standard deviation/variance errors |
| $P(F - 0.5M < 0) = \phi(\text{'-0.4411'}) = 1 - \phi(\text{'0.4411'})$ | M1 | Correct area consistent with *their* figures |
| $= 0.330$ (3 sf) | A1 | Allow 0.33 if no greater accuracy given |

**Alternative method for Question 3:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $2F - M$ | | |
| $\sim N(34,\ 2^2 \times 27^2 + 55^2)$ | B1 | for $102 - 0.5(170)$ (= 17) or 34 |
| | B1 | for $27^2 + 0.25 \times 55^2$ (= 1485.25) or $2^2 \times 27^2 + 55^2$ (= 5941) |
| $\frac{0 - \text{'34'}}{\sqrt{\text{'5941'}}}$ (= −0.4411) | M1 | Must have an attempt at combining $F$ and $M$. No standard deviation/variance errors |
| $P(2F - M < 0) = \phi(\text{'-0.4411'}) = 1 - \phi(\text{'0.4411'})$ | M1 | Correct area consistent with *their* figures |
| $= 0.330$ (3 sf) | A1 | Allow 0.33 if no greater accuracy given |

---

Show LaTeX source

3 The masses, in kilograms, of female and male animals of a certain species have the distributions $\mathrm { N } \left( 102,27 ^ { 2 } \right)$ and $\mathrm { N } \left( 170,55 ^ { 2 } \right)$ respectively.

Find the probability that a randomly chosen female has a mass that is less than half the mass of a randomly chosen male.\\
\includegraphics[max width=\textwidth, alt={}, center]{65b50bfb-5fd8-4cf3-ae3b-cffc12e23cd8-06_76_1659_484_244}\\

\hfill \mbox{\textit{CAIE S2 2020 Q3 [6]}}

This paper (6 questions)

View full paper

Q1 7 Q2 7 Q3 6 Q4 5 Q5 13 Q6 12