| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single variable sum probability |
| Difficulty | Standard +0.3 Part (a) requires recognizing that the sum of two independent normal variables is normal with mean 2μ and variance 2σ², then standardizing - a direct application of a key result. Part (b) involves finding the distribution of a linear combination C - ½S, requiring variance calculation for the scaled variable. Both are standard textbook exercises testing knowledge of normal distribution properties with straightforward computation, slightly above average due to the linear combination concept but no novel problem-solving required. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X_1 + X_2) = 2 \times 4.2 = 8.4\), \(\text{Var}(X_1 + X_2) = 2 \times 1.1^2 = 2.42\) | B1 | Both seen or implied (or sd \(= 1.56\)) |
| \(\frac{10 - 8.4}{\sqrt{2.42}} = 1.029\) | M1 | Standardising with their mean and var (no sd/var mix) |
| \(1 - \phi(\text{``}1.029\text{''})\) | M1 | For area consistent with their working |
| \(= 0.152\) (3 sf) | A1 | |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = 20.5\) | B1 | |
| \(\text{Var}(X) = 105 + 0.5^2 \times 15 = 108.75\) | M1 | Correct expression oe |
| \(\frac{0 - \text{``}20.5\text{''}}{\sqrt{\text{``}108.75\text{''}}} = -1.966\) | M1 | Correct standardisation using their \(E\) & \(V\) (no sd/var mix), ignore any attempted cc |
| \(\phi(\text{``}-1.966\text{''}) = 1 - \phi(\text{``}1.966\text{''}) = (1 - 0.9754)\) | M1 | For area consistent with their working |
| \(= 0.0246\) or \(2.46\%\) (3 sf) | A1 | Accept \(0.0247\) |
| Total: | 5 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X_1 + X_2) = 2 \times 4.2 = 8.4$, $\text{Var}(X_1 + X_2) = 2 \times 1.1^2 = 2.42$ | B1 | Both seen or implied (or sd $= 1.56$) |
| $\frac{10 - 8.4}{\sqrt{2.42}} = 1.029$ | M1 | Standardising with their mean and var (no sd/var mix) |
| $1 - \phi(\text{``}1.029\text{''})$ | M1 | For area consistent with their working |
| $= 0.152$ (3 sf) | A1 | |
| **Total:** | **4** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = 20.5$ | B1 | |
| $\text{Var}(X) = 105 + 0.5^2 \times 15 = 108.75$ | M1 | Correct expression oe |
| $\frac{0 - \text{``}20.5\text{''}}{\sqrt{\text{``}108.75\text{''}}} = -1.966$ | M1 | Correct standardisation using their $E$ & $V$ (no sd/var mix), ignore any attempted cc |
| $\phi(\text{``}-1.966\text{''}) = 1 - \phi(\text{``}1.966\text{''}) = (1 - 0.9754)$ | M1 | For area consistent with their working |
| $= 0.0246$ or $2.46\%$ (3 sf) | A1 | Accept $0.0247$ |
| **Total:** | **5** | |7
\begin{enumerate}[label=(\alph*)]
\item A random variable $X$ is normally distributed with mean 4.2 and standard deviation 1.1. Find the probability that the sum of two randomly chosen values of $X$ is greater than 10 .
\item Each candidate's overall score for an essay is calculated as follows. The mark for creativity is denoted by $C$, the penalty mark for spelling errors is denoted by $S$ and the overall score is defined by $C - \frac { 1 } { 2 } S$. The variables $C$ and $S$ are independent and have distributions $\mathrm { N } ( 29,105 )$ and $\mathrm { N } ( 17,15 )$ respectively. Find the proportion of candidates receiving a negative overall score.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2017 Q7 [9]}}