| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparing two journey times |
| Difficulty | Standard +0.8 Part (i) is standard application of summing normal distributions (routine for S2). Part (ii) requires the insight to form the linear combination X - 2Y > 0, find its distribution parameters, then calculate the probability - this non-obvious setup and multi-step execution elevates it above average difficulty but remains within typical Further Maths Statistics scope. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(T) = 4.5 + 2.3 = 6.8\), \(\text{Var}(T) = 1.1^2 + 0.7^2 = 1.7\) | M1 | Both methods seen or implied |
| \(\frac{8.5 - "6.8"}{\sqrt{"1.7"}} = 1.304\) | M1 | Correct standardisation using their \(\mu\) and \(\sigma^2\), must be a combination of the two variables |
| \(\phi("1.304")\) | M1 | Area consistent with their working |
| \(= 0.904\) (3 sf) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(D) = 4.5 - 2 \times 2.3 = -0.1\) | M1 | |
| \(\text{Var}(D) = 1.1^2 + 2^2 \times 0.7^2 = 3.17\) | M1 | Both can be seen or implied |
| \(\frac{0 - ('-0.1')}{\sqrt{'3.17'}} = 0.056\) | M1 | Correct standardisation using their \(\mu\) and \(\sigma^2\), combination of the two variables |
| \(1 - \phi("0.056")\) | M1 | Area consistent with their working |
| \(= 0.478\) (3 sf) | A1 | |
| Total: 5 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(T) = 4.5 + 2.3 = 6.8$, $\text{Var}(T) = 1.1^2 + 0.7^2 = 1.7$ | M1 | Both methods seen or implied |
| $\frac{8.5 - "6.8"}{\sqrt{"1.7"}} = 1.304$ | M1 | Correct standardisation using their $\mu$ and $\sigma^2$, must be a combination of the two variables |
| $\phi("1.304")$ | M1 | Area consistent with their working |
| $= 0.904$ (3 sf) | A1 | |
| **Total: 4** | | |
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(D) = 4.5 - 2 \times 2.3 = -0.1$ | M1 | |
| $\text{Var}(D) = 1.1^2 + 2^2 \times 0.7^2 = 3.17$ | M1 | Both can be seen or implied |
| $\frac{0 - ('-0.1')}{\sqrt{'3.17'}} = 0.056$ | M1 | Correct standardisation using their $\mu$ and $\sigma^2$, combination of the two variables |
| $1 - \phi("0.056")$ | M1 | Area consistent with their working |
| $= 0.478$ (3 sf) | A1 | |
| **Total: 5** | | |6 The times, in minutes, taken to complete the two parts of a task are normally distributed with means 4.5 and 2.3 respectively and standard deviations 1.1 and 0.7 respectively.\\
(i) Find the probability that the total time taken for the task is less than 8.5 minutes.\\
(ii) Find the probability that the time taken for the first part of the task is more than twice the time taken for the second part.\\
\hfill \mbox{\textit{CAIE S2 2018 Q6 [9]}}