| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Direct comparison with scalar multiple (different variables) |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for linear combinations of normal random variables. Part (a) requires simple scaling properties (mean and variance under multiplication by a constant), while part (b) involves forming a difference of normals and finding a probability—both are routine S2 techniques with no novel insight required. The 6-mark allocation for part (b) reflects multiple computational steps rather than conceptual difficulty. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | E(A income) = [10.3 × 2.50 ] = 25.75 [$] | B1 |
| Var(A income) = [5.76 × 2.502 ] = 36 [$2] | B1 |
| Answer | Marks |
|---|---|
| 4(b) | B income ~ N(37.05, 101.506) |
| Answer | Marks | Guidance |
|---|---|---|
| E(B income) = 37.05 and Var(B income) = 101.51 | B1 | Or N(37.1, 102) soi. |
| A income – B income ~ N(‘25.75’ – ‘37.05’, ‘36’ + ‘101.506’) | M1 | Ft their values for A and B. |
| = N(–11.3, 137.506) | A1 | Accept 3sf. |
| Answer | Marks | Guidance |
|---|---|---|
| '137.506' | M1 | Standardising with their values from attempt at A |
| Answer | Marks | Guidance |
|---|---|---|
| 1 – ɸ('0.964') | M1 | For area consistent with their values. |
| = 0.168 or 0.167 (3 sf) | A1 | cwo |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
--- 4(a) ---
4(a) | E(A income) = [10.3 × 2.50 ] = 25.75 [$] | B1 | Accept 3sf.
Var(A income) = [5.76 × 2.502 ] = 36 [$2] | B1
2
--- 4(b) ---
4(b) | B income ~ N(37.05, 101.506)
or
E(B income) = 37.05 and Var(B income) = 101.51 | B1 | Or N(37.1, 102) soi.
A income – B income ~ N(‘25.75’ – ‘37.05’, ‘36’ + ‘101.506’) | M1 | Ft their values for A and B.
= N(–11.3, 137.506) | A1 | Accept 3sf.
0−(−'11.3')
[= 0.964]
'137.506' | M1 | Standardising with their values from attempt at A
income – B income.
1 – ɸ('0.964') | M1 | For area consistent with their values.
= 0.168 or 0.167 (3 sf) | A1 | cwo
6
Question | Answer | Marks | GuidanceThe masses, in kilograms, of chemicals $A$ and $B$ produced per day by a factory are modelled by the independent random variables $X$ and $Y$ respectively, where $X \sim$ N(10.3, 5.76) and $Y \sim$ N(11.4, 9.61). The income generated by the chemicals is \$2.50 per kilogram for $A$ and \$3.25 per kilogram for $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the mean and variance of the daily income generated by chemical $A$. [2]
\item Find the probability that, on a randomly chosen day, the income generated by chemical $A$ is greater than the income generated by chemical $B$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q4 [8]}}