| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Moderate -0.3 This is a straightforward application of standard results for linear combinations of independent normal variables (sum means, sum variances) followed by a routine sampling distribution calculation. While it requires understanding of these key concepts, the execution is mechanical with no problem-solving insight needed—slightly easier than a typical A-level question due to its direct application of formulas. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Mean \(= 3.5 + 2.9 + 3.1 = 9.5\); Variance \(= 0.3^2 + 0.25^2 + 0.35^2 = 0.275\); Std dev \(= 0.524\) | B1 M1 A1 | For 9.5 as final answer; For summing three squared deviations; For correct answer |
| 3 | ||
| (ii) \(z = \frac{9 - 9.5}{\sqrt{\text{their var}}} = -1.907\) OR \(z = \frac{36-38}{\sqrt{4 \times \text{their var}}} = -1.907\) | M1 M1 | For standardising, no cc; For \(\sqrt{\frac{\text{their var}}{4}}\) or \(\sqrt{4 \times \text{their var}}\) in denom - no 'mixed' methods |
| \(\Phi(1.907) = 0.9717 = 0.972\) | A1 | For correct answer |
| 3 |
**(i)** Mean $= 3.5 + 2.9 + 3.1 = 9.5$; Variance $= 0.3^2 + 0.25^2 + 0.35^2 = 0.275$; Std dev $= 0.524$ | B1 M1 A1 | For 9.5 as final answer; For summing three squared deviations; For correct answer
| | **3** | |
**(ii)** $z = \frac{9 - 9.5}{\sqrt{\text{their var}}} = -1.907$ OR $z = \frac{36-38}{\sqrt{4 \times \text{their var}}} = -1.907$ | M1 M1 | For standardising, no cc; For $\sqrt{\frac{\text{their var}}{4}}$ or $\sqrt{4 \times \text{their var}}$ in denom - no 'mixed' methods
$\Phi(1.907) = 0.9717 = 0.972$ | A1 | For correct answer
| | **3** | |2 In athletics matches the triple jump event consists of a hop, followed by a step, followed by a jump. The lengths covered by Albert in each part are independent normal variables with means $3.5 \mathrm {~m} , 2.9 \mathrm {~m}$, 3.1 m and standard deviations $0.3 \mathrm {~m} , 0.25 \mathrm {~m} , 0.35 \mathrm {~m}$ respectively. The length of the triple jump is the sum of the three parts.\\
(i) Find the mean and standard deviation of the length of Albert's triple jumps.\\
(ii) Find the probability that the mean of Albert's next four triple jumps is greater than 9 m .
\hfill \mbox{\textit{CAIE S2 2004 Q2 [6]}}