| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Fixed container with random contents |
| Difficulty | Standard +0.3 This question tests standard application of linear combinations of normal random variables with straightforward calculations. Part (i) requires finding the distribution of 12X + Y and computing a probability. Part (ii) involves finding the distribution of W - 2V and another probability calculation. Both parts are routine applications of the formula for sums/differences of independent normals with no conceptual challenges beyond remembering variance rules. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(W \sim N(6210,\ 171.88)\) | B2 | seen or implied. B1 each parameter |
| \(\frac{6200 - "6210"}{\sqrt{"171.88"}} \quad (= -0.763)\) | M1 | Standardising with their values. No sd/var mix |
| \(1 - \Phi("0.763")\) | M1 | For area consistent with their mean |
| \(= 0.223\) (3 sfs) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(C - 2B) = -50\) | M1 | "6210" \(- 2(3130)\) (or \(E(2B-C)=50\)) |
| \(\text{Var}(C - 2B) = \text{"171.88"} + 2^2 \times 12.1^2 = 757.52\) | M1 | |
| \(\dfrac{0-(-50)}{\sqrt{\text{"757.52"}}} = 1.817\) | M1 | Standardising with their values |
| \(\Phi(\text{"1.817"})\) | M1 | For area consistent with their mean |
| \(= 0.965\) (3 sfs) | A1 | |
| Total: | 5 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $W \sim N(6210,\ 171.88)$ | B2 | seen or implied. **B1** each parameter |
| $\frac{6200 - "6210"}{\sqrt{"171.88"}} \quad (= -0.763)$ | M1 | Standardising with their values. No sd/var mix |
| $1 - \Phi("0.763")$ | M1 | For area consistent with their mean |
| $= 0.223$ (3 sfs) | A1 | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(C - 2B) = -50$ | **M1** | "6210" $- 2(3130)$ (or $E(2B-C)=50$) |
| $\text{Var}(C - 2B) = \text{"171.88"} + 2^2 \times 12.1^2 = 757.52$ | **M1** | |
| $\dfrac{0-(-50)}{\sqrt{\text{"757.52"}}} = 1.817$ | **M1** | Standardising with their values |
| $\Phi(\text{"1.817"})$ | **M1** | For area consistent with their mean |
| $= 0.965$ (3 sfs) | **A1** | |
| **Total:** | **5** | |
---5 Large packets of sugar are packed in cartons, each containing 12 packets. The weights of these packets are normally distributed with mean 505 g and standard deviation 3.2 g . The weights of the cartons, when empty, are independently normally distributed with mean 150 g and standard deviation 7 g .\\
(i) Find the probability that the total weight of a full carton is less than 6200 g .\\
Small packets of sugar are packed in boxes. The total weight of a full box has a normal distribution with mean 3130 g and standard deviation 12.1 g .\\
(ii) Find the probability that the weight of a randomly chosen full carton is less than double the weight of a randomly chosen full box.\\
\hfill \mbox{\textit{CAIE S2 2017 Q5 [10]}}