| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Finding n from sample mean distribution |
| Difficulty | Moderate -0.8 Part (i) is a direct application of the formula σ_F̄ = σ/√n requiring only algebraic rearrangement (12/√n = 1.5 gives n=64). Part (ii) requires stating the CLT result that F̄ is approximately normal for large n, which is standard recall. This is simpler than average A-level questions as it involves minimal problem-solving and straightforward formula application. |
| Spec | 5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{12.0}{\sqrt{n}} = 1.50 \Rightarrow \sqrt{n} = \frac{12.0}{1.50} = 8 \Rightarrow n = 64\) | B1 | For any correct equation involving \(n\) |
| M1 | For correct solution method for \(n\) or \(\sqrt{n}\) | |
| A1 | For correct answer 64 | |
| (ii) \(n\) is large, the distribution of \(\bar{F}\) can be taken to be normal, according to the Central Limit Theorem | M1 | For relating the size of \(n\) to normality |
| A1 | For reference to the CLT |
(i) $\frac{12.0}{\sqrt{n}} = 1.50 \Rightarrow \sqrt{n} = \frac{12.0}{1.50} = 8 \Rightarrow n = 64$ | B1 | For any correct equation involving $n$
| M1 | For correct solution method for $n$ or $\sqrt{n}$
| A1 | For correct answer 64
(ii) $n$ is large, the distribution of $\bar{F}$ can be taken to be normal, according to the Central Limit Theorem | M1 | For relating the size of $n$ to normality
| A1 | For reference to the CLT
**Total: 5 marks**
---1 The standard deviation of a random variable $F$ is 12.0. The mean of $n$ independent observations of $F$ is denoted by $\bar { F }$.\\
(i) Given that the standard deviation of $\bar { F }$ is 1.50 , find the value of $n$.\\
(ii) For this value of $n$, state, with justification, what can be said about the distribution of $\bar { F }$.
\hfill \mbox{\textit{OCR S2 Q1 [5]}}