| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Fixed container with random contents |
| Difficulty | Standard +0.8 This question requires understanding of variance properties for linear combinations of independent normal variables (part a: variance of sum with coefficient), then applying this to find the distribution of a more complex linear combination (3Y - X) and calculating a probability (part b). While the concepts are standard S2 material, the multi-step reasoning, careful handling of independence assumptions, and the non-trivial '3 times' comparison in part (b) elevate this above routine exercises. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2.0^2 + 20 \times 3.4^2\) | M1 | |
| \(= 235.2\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(C - 3B) = 50 + 20 \times 1010 - 3 \times 6730\) or \(60\) | B1 | |
| \(\text{Var}(C - 3B) = '235.2' + 9 \times 15^2\) or \(2260.2\) | M1 | FT *their* values from (a). |
| \([C - 3B \sim N('60',\ '2260.2')]\), \(= \frac{0 - 60}{\sqrt{2260.2}}\) \([= -1.262]\) | M1 | Standardising with their values (could be implied). |
| \(1 - \Phi('-1.262') = \Phi('1.262')\) | M1 | Probability area consistent with their values. |
| \(= 0.897\) (3 sf) | A1 | |
| Total: 5 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2.0^2 + 20 \times 3.4^2$ | M1 | |
| $= 235.2$ | A1 | |
| **Total: 2** | | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(C - 3B) = 50 + 20 \times 1010 - 3 \times 6730$ or $60$ | B1 | |
| $\text{Var}(C - 3B) = '235.2' + 9 \times 15^2$ or $2260.2$ | M1 | FT *their* values from **(a)**. |
| $[C - 3B \sim N('60',\ '2260.2')]$, $= \frac{0 - 60}{\sqrt{2260.2}}$ $[= -1.262]$ | M1 | Standardising with their values (could be implied). |
| $1 - \Phi('-1.262') = \Phi('1.262')$ | M1 | Probability area consistent with their values. |
| $= 0.897$ (3 sf) | A1 | |
| **Total: 5** | | |5 Large packets of rice are packed in cartons, each containing 20 randomly chosen packets. The masses of these packets are normally distributed with mean 1010 g and standard deviation 3.4 g . The masses of the cartons, when empty, are independently normally distributed with mean 50 g and standard deviation 2.0 g .
\begin{enumerate}[label=(\alph*)]
\item Find the variance of the masses of full cartons.\\
Small packets of rice are packed in boxes. The total masses of full boxes are normally distributed with mean 6730 g and standard deviation 15.0 g . The masses of the boxes and cartons are distributed independently of each other.
\item Find the probability that the mass of a randomly chosen full carton is more than three times the mass of a randomly chosen full box.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q5 [7]}}