| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Difficulty | Challenging +1.2 This question requires understanding of linear combinations of normal variables and the law of total probability. Part (a) involves forming L - 2S ~ N(μ, σ²) and finding P(L - 2S > 0), which is a standard technique. Part (b) requires finding distributions for 10L and 20S separately, then using conditional probability with given proportions (60%/40%), which adds moderate complexity but follows established methods for this topic. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(D = L - 2S \quad E(D) = 410 - 2(206) = -2\) | B1 | SOI. OE using \(2S - L\). |
| \(\text{Var}(D) = 3.6^2 + 4 \times 3.7^2 = 67.72\) | B1 | SOI |
| \(\frac{0-(-2)}{\sqrt{67.72}} = 0.243\) | M1 | For standardising using their values. |
| \(1 - \phi(\text{their } 0.243)\) | M1 | For probability area consistent with their values. |
| \(= 0.404\) (3 sf) | A1 | As final answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T_L \sim N(4100,\ 10 \times 3.6^2)\) and \(T_S \sim N(4120,\ 20 \times 3.7^2)\) | B1 | One of \(N(4100, 129.6)\) or \(N(4120, 273.8)\) USED (unchanged) in a standardising equation. |
| \(\frac{4080-4100}{\sqrt{129.6}} = -1.757\) and \(\frac{4080-4120}{\sqrt{273.8}} = -2.417\) | M1 | Standardising with either their \(N(4100, 129.6)\) or \(N(4120, 273.8)\) or their \(N(\ldots,\ldots)\) (could be from a combination). |
| \(1-\phi(-1.757) = \phi(1.757)\) and \(1-\phi(-2.417) = \phi(2.417)\) | M1 | One area consistent with their working (could be from a combination). Do not ISW. |
| \(= 0.9605\) or \(0.961\) and \(= 0.9921\) or \(0.9922\) or \(0.992\) | A1 | Both of these correct. Do not ISW. |
| \(0.6 \times \text{their } 0.9605 + 0.4 \times \text{their } 0.9921\) | M1 | Must be using probabilities. |
| \(= 0.973\) (3 sf) | A1 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $D = L - 2S \quad E(D) = 410 - 2(206) = -2$ | B1 | SOI. OE using $2S - L$. |
| $\text{Var}(D) = 3.6^2 + 4 \times 3.7^2 = 67.72$ | B1 | SOI |
| $\frac{0-(-2)}{\sqrt{67.72}} = 0.243$ | M1 | For standardising using their values. |
| $1 - \phi(\text{their } 0.243)$ | M1 | For probability area consistent with their values. |
| $= 0.404$ (3 sf) | A1 | As final answer. |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T_L \sim N(4100,\ 10 \times 3.6^2)$ and $T_S \sim N(4120,\ 20 \times 3.7^2)$ | B1 | One of $N(4100, 129.6)$ or $N(4120, 273.8)$ USED (unchanged) in a standardising equation. |
| $\frac{4080-4100}{\sqrt{129.6}} = -1.757$ and $\frac{4080-4120}{\sqrt{273.8}} = -2.417$ | M1 | Standardising with either their $N(4100, 129.6)$ or $N(4120, 273.8)$ or their $N(\ldots,\ldots)$ (could be from a combination). |
| $1-\phi(-1.757) = \phi(1.757)$ and $1-\phi(-2.417) = \phi(2.417)$ | M1 | One area consistent with their working (could be from a combination). Do not ISW. |
| $= 0.9605$ or $0.961$ and $= 0.9921$ or $0.9922$ or $0.992$ | A1 | Both of these correct. Do not ISW. |
| $0.6 \times \text{their } 0.9605 + 0.4 \times \text{their } 0.9921$ | M1 | Must be using probabilities. |
| $= 0.973$ (3 sf) | A1 | |
---5 The masses, in grams, of large and small packets of Maxwheat cereal have the independent distributions $\mathrm { N } \left( 410.0,3.6 ^ { 2 } \right)$ and $\mathrm { N } \left( 206.0,3.7 ^ { 2 } \right)$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen large packet has a mass that is more than double the mass of a randomly chosen small packet.\\
The packets are placed in boxes. The boxes are identical in appearance. $60 \%$ of the boxes contain exactly 10 randomly chosen large packets. 40\% of the boxes contain exactly 20 randomly chosen small packets.
\item Find the probability that a randomly chosen box contains packets with a total mass of more than 4080 grams.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q5 [11]}}