| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparison involving sums or multiples |
| Difficulty | Standard +0.3 This question tests standard operations with normal distributions (sums and linear transformations) using straightforward formulas. Part (i) requires finding the distribution of 2C - R and computing a probability; part (ii) applies E(aX) and Var(aX) rules; part (iii) is a routine normal probability calculation. All steps are textbook applications with no novel insight required, making it slightly easier than average for Further Maths Statistics. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(R - (C_1 + C_2) \sim N(27.4, 228.5)\) or vv | B1 | Correct mean (accept un-simplified form) |
| B1 | Correct variance (accept un-simplified form) | |
| \(P\{(R - (C_1 + C_2)) < 0\} = \Phi\left(\frac{0 - 27.4}{\sqrt{228.5}}\right)\) | M1 | Considering \(P\{(R - (C_1 + C_2)) < 0\}\) o.e. |
| \(= 1 - \Phi(1.813) = 0.0349\) | M1, A1 | Standardising and finding correct area ie \(< 0.5\), Correct answer |
| (ii) Mean \(= 99.5(99.45)\) | B1 | Correct mean |
| Variance \(= 1.5^2 \times 7.1^2 = 113.4 (= 113)\) | M1 | Variance involving \(1.5^2\) |
| A1 | Correct variance (SR var=\(7.1^2+(1/2)^2(7.1)^2\) scores M1) | 3 marks total |
| (iii) \(P(1.5C' > 87) = 1 - \Phi\left(\frac{87 - 99.45}{\sqrt{113.4}}\right)\) | M1 | Standardising and finding correct area ie \(> 0.5\) |
| \(= 1 - \Phi(-1.169) = \Phi(1.169) = 0.879\) | A1 | Correct answer |
**(i)** $R - (C_1 + C_2) \sim N(27.4, 228.5)$ or vv | B1 | Correct mean (accept un-simplified form)
| B1 | Correct variance (accept un-simplified form)
$P\{(R - (C_1 + C_2)) < 0\} = \Phi\left(\frac{0 - 27.4}{\sqrt{228.5}}\right)$ | M1 | Considering $P\{(R - (C_1 + C_2)) < 0\}$ o.e.
$= 1 - \Phi(1.813) = 0.0349$ | M1, A1 | Standardising and finding correct area ie $< 0.5$, Correct answer | 5 marks total
**(ii)** Mean $= 99.5(99.45)$ | B1 | Correct mean
Variance $= 1.5^2 \times 7.1^2 = 113.4 (= 113)$ | M1 | Variance involving $1.5^2$
| A1 | Correct variance (SR var=$7.1^2+(1/2)^2(7.1)^2$ scores M1) | 3 marks total
**(iii)** $P(1.5C' > 87) = 1 - \Phi\left(\frac{87 - 99.45}{\sqrt{113.4}}\right)$ | M1 | Standardising and finding correct area ie $> 0.5$
$= 1 - \Phi(-1.169) = \Phi(1.169) = 0.879$ | A1 | Correct answer | 2 marks total
---5 Climbing ropes produced by a manufacturer have breaking strengths which are normally distributed with mean 160 kg and standard deviation 11.3 kg . A group of climbers have weights which are normally distributed with mean 66.3 kg and standard deviation 7.1 kg .\\
(i) Find the probability that a rope chosen randomly will break under the combined weight of 2 climbers chosen randomly.
Each climber carries, in a rucksack, equipment amounting to half his own weight.\\
(ii) Find the mean and variance of the combined weight of a climber and his rucksack.\\
(iii) Find the probability that the combined weight of a climber and his rucksack is greater than 87 kg .
\hfill \mbox{\textit{CAIE S2 2006 Q5 [10]}}