| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Different variables, one observation each |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal variables. Students need to form G - M, find its mean (36.42 - 42.65 = -6.23) and variance (6.87² + 10.25²), then calculate P(G - M ≥ 10) using standardization. It's slightly easier than average because it's a direct textbook application with clear setup and no conceptual traps, though it does require understanding independence and variance addition. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Mean | Standard deviation | |
| \(G\) | 36.42 | 6.87 |
| \(M\) | 42.65 | 10.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(G - M \sim N(-6.23, \sigma^2)\) | M1 | Or \(G\text{-}M\text{-}10 \sim N(-16.23, \sigma^2)\) |
| \(\sigma^2 = 6.87^2 + 10.25^2\) | A1 | |
| \(z = (16.23)/\sigma\) | M1 | |
| \(= 1.315\) | A1 | |
| Probability \(= 0.0942\) or \(0.0943\) | A1 [5] | Accept 0.094 |
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $G - M \sim N(-6.23, \sigma^2)$ | M1 | Or $G\text{-}M\text{-}10 \sim N(-16.23, \sigma^2)$ |
| $\sigma^2 = 6.87^2 + 10.25^2$ | A1 | |
| $z = (16.23)/\sigma$ | M1 | |
| $= 1.315$ | A1 | |
| Probability $= 0.0942$ or $0.0943$ | A1 **[5]** | Accept 0.094 |
---2 In a Year 8 internal examination in a large school the Geography marks, $G$, and Mathematics marks, $M$, had means and standard deviations as follows.
\begin{center}
\begin{tabular}{ l c c }
& Mean & Standard deviation \\
$G$ & 36.42 & 6.87 \\
$M$ & 42.65 & 10.25 \\
\end{tabular}
\end{center}
Assuming that $G$ and $M$ have independent normal distributions, find the probability that a randomly chosen Geography candidate scores at least 10 marks more than a randomly chosen Mathematics candidate. Do not use a continuity correction.
\hfill \mbox{\textit{OCR S3 2011 Q2 [5]}}