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. While the individual steps are standard S2 techniques, combining them in this context (comparing one variable to half of another) requires problem-solving insight beyond routine textbook exercises.
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]{6346fd4b-7bc9-4205-94db-67368b9415fe-06_76_1659_484_244}

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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:

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:**

| 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]{6346fd4b-7bc9-4205-94db-67368b9415fe-06_76_1659_484_244}\\

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

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