| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Expectation and variance with context application |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for linear combinations of independent normal variables. Part (a) requires scaling by 3 and combining means/variances using basic formulas. Part (b) involves forming a linear combination (1.5X - 0.2Y) and calculating a single probability. All techniques are routine for S2 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Mean} = [3(2500 + 3700)] = 18600\) (kg) | B1 | |
| \(\text{Var(Total profit)} = 3(120^2 + 130^2)\) or \(93900\) | M1 | or \(\sqrt{}\) of this stated. |
| \(\text{sd} = 306\) (kg) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(1.5X - 0.2Y) = 1.5 \times 2500 - 0.20 \times 3700 = [3010]\) | B1 | Give at early stage. |
| \(\text{Var}(1.5X - 0.2Y) = 1.5^2 \times 120^2 + 0.2^2 \times 130^2 \quad [= 33076]\) | B1 | Correct expression or result or sd \(= 182\) (3 sf) seen. |
| \(\dfrac{3000 - 3010}{\sqrt{\textit{their } `33076`}} \quad [= -0.055]\) | M1 | Ignore continuity correction attempts. \(E(X)\) and Var must come from a combination attempt. Can be implied. |
| \(\Phi(\textit{their } `-0.055`) = 1 - \Phi(\textit{their } `0.055`)\) | M1 | For area consistent with their values. Can be implied. |
| \(= 0.478\) (3 sf) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Mean} = [3(2500 + 3700)] = 18600$ (kg) | B1 | |
| $\text{Var(Total profit)} = 3(120^2 + 130^2)$ or $93900$ | M1 | or $\sqrt{}$ of this stated. |
| $\text{sd} = 306$ (kg) (3 sf) | A1 | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(1.5X - 0.2Y) = 1.5 \times 2500 - 0.20 \times 3700 = [3010]$ | B1 | Give at early stage. |
| $\text{Var}(1.5X - 0.2Y) = 1.5^2 \times 120^2 + 0.2^2 \times 130^2 \quad [= 33076]$ | B1 | Correct expression or result or sd $= 182$ (3 sf) seen. |
| $\dfrac{3000 - 3010}{\sqrt{\textit{their } `33076`}} \quad [= -0.055]$ | M1 | Ignore continuity correction attempts. $E(X)$ and Var must come from a combination attempt. Can be implied. |
| $\Phi(\textit{their } `-0.055`) = 1 - \Phi(\textit{their } `0.055`)$ | M1 | For area consistent with their values. Can be implied. |
| $= 0.478$ (3 sf) | A1 | |
---4 Each month a company sells $X \mathrm {~kg}$ of brown sugar and $Y \mathrm {~kg}$ of white sugar, where $X$ and $Y$ have the independent distributions $\mathrm { N } \left( 2500,120 ^ { 2 } \right)$ and $\mathrm { N } \left( 3700,130 ^ { 2 } \right)$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the mean and standard deviation of the total amount of sugar that the company sells in 3 randomly chosen months.\\
The company makes a profit of $\$ 1.50$ per kilogram of brown sugar sold and makes a loss of $\$ 0.20$ per kilogram of white sugar sold.
\item Find the probability that, in a randomly chosen month, the total profit is less than $\$ 3000$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q4 [8]}}