| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Same variable, two observations |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal distributions with standard formulas. Part (a) requires finding P(|X-Y|>2) using X-Y ~ N(1,5), and part (b)(i) uses T+1.5P ~ N(125, 401). Both are routine calculations once the distribution parameters are identified, slightly easier than average due to the mechanical nature of the work. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X-Y) = 1\), \(\text{Var}(X-Y) = 5\) | B1 | Seen or implied, OE e.g. \(X - Y - 2\) |
| \(\frac{2-1}{\sqrt{5}}\) \([= 0.447]\) and \(\frac{-2-1}{\sqrt{5}}\) \([= -1.342]\) | M1 | Standardising with their values must come from a combination |
| \(1 - \Phi(\text{'0.447'})\) and \(\Phi(\text{'}-1.342\text{'}) = 1 - \Phi(\text{'1.342'})\) | M1 | Correct probability area consistent with their values |
| \(= 0.327\) or \(0.328\) and \(= 0.0898\) or \(0.0899\) | A1 | Seen or implied |
| Probability that difference is more than \(2 = 0.417\) (3 sf) or \(0.418\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = 62 + 1.5(42)\) \([= 125]\) | B1 | OE |
| \(\text{Var}(X) = 158 + 1.5^2 \times 108\) \([= 401]\) | B1 | Correct expression OE |
| \(\frac{90 - \text{"125"}}{\sqrt{\text{"401"}}}\) \([= -1.748]\) | M1 | Correct standardisation using their \(E(X)\) and \(\text{Var}(X)\). Must both be from a combination attempt. Ignore any attempted continuity correction |
| \(\Phi(\text{'1.748'})\) | M1 | Correct probability area consistent with their stated values |
| \(= 0.960\) or \(96.0\%\) (3 sf) | A1 | Allow 0.96 or 96% |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Unlikely. A candidate who does well in Theory is likely to do well in Practical | B1 | Need both. Accept 'unlikely', 'not independent', 'dependent', 'not realistic', or similar; and accept 'both testing knowledge from the same syllabus', 'theory and practical share same content' or similar statement |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X-Y) = 1$, $\text{Var}(X-Y) = 5$ | B1 | Seen or implied, OE e.g. $X - Y - 2$ |
| $\frac{2-1}{\sqrt{5}}$ $[= 0.447]$ and $\frac{-2-1}{\sqrt{5}}$ $[= -1.342]$ | M1 | Standardising with their values must come from a combination |
| $1 - \Phi(\text{'0.447'})$ and $\Phi(\text{'}-1.342\text{'}) = 1 - \Phi(\text{'1.342'})$ | M1 | Correct probability area consistent with their values |
| $= 0.327$ or $0.328$ and $= 0.0898$ or $0.0899$ | A1 | Seen or implied |
| Probability that difference is more than $2 = 0.417$ (3 sf) or $0.418$ | A1 | |
---
## Question 5(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = 62 + 1.5(42)$ $[= 125]$ | B1 | OE |
| $\text{Var}(X) = 158 + 1.5^2 \times 108$ $[= 401]$ | B1 | Correct expression OE |
| $\frac{90 - \text{"125"}}{\sqrt{\text{"401"}}}$ $[= -1.748]$ | M1 | Correct standardisation using their $E(X)$ and $\text{Var}(X)$. Must both be from a combination attempt. Ignore any attempted continuity correction |
| $\Phi(\text{'1.748'})$ | M1 | Correct probability area consistent with their stated values |
| $= 0.960$ or $96.0\%$ (3 sf) | A1 | Allow 0.96 or 96% |
---
## Question 5(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Unlikely. A candidate who does well in Theory is likely to do well in Practical | B1 | Need both. Accept 'unlikely', 'not independent', 'dependent', 'not realistic', or similar; and accept 'both testing knowledge from the same syllabus', 'theory and practical share same content' or similar statement |
---5
\begin{enumerate}[label=(\alph*)]
\item Two random variables $X$ and $Y$ have the independent distributions $\mathrm { N } ( 7,3 )$ and $\mathrm { N } ( 6,2 )$ respectively. A random value of each variable is taken.
Find the probability that the two values differ by more than 2 .
\item Each candidate's overall score in a science test is calculated as follows. The mark for theory is denoted by $T$, the mark for practical is denoted by $P$, and the overall score is given by $T + 1.5 P$. The variables $T$ and $P$ are assumed to be independent with distributions $\mathrm { N } ( 62,158 )$ and $\mathrm { N } ( 42,108 )$ respectively. You should assume that no continuity corrections are needed when using these distributions.
\begin{enumerate}[label=(\roman*)]
\item A pass is awarded to candidates whose overall score is at least 90 .
Find the proportion of candidates who pass.
\item Comment on the assumption that the variables $T$ and $P$ are independent.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q5 [11]}}