| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for sampling distribution |
| Difficulty | Moderate -0.3 This is a straightforward application of standard formulas: calculating sample mean and variance (routine), finding probability for a sample mean using normal distribution (standard S2 content), and explaining why CLT isn't needed since R is already normal. All parts are textbook exercises requiring recall and direct application rather than problem-solving or insight. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(33.6\) | B1 | 33.6 clearly stated [not recoverable later] |
| \(\frac{115782.84}{100} - 33.6^2 = 28.8684\) | M1 | Correct formula used for biased estimate |
| \(\times \frac{100}{99} = \mathbf{29.16}\) | M1, A1 4 | \(\times \frac{100}{99}\), M's independent. Eg \(\frac{\Sigma r^2}{99}[-33.6^2]\). SR B1 variance in range \([29.1, 29.2]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{R} \sim N(33.6,\ 29.16/9) = N(33.6,\ 1.8^2)\) | M1, A1 | Normal, their \(\mu\) stated or implied. Variance \([\text{their (i)}] \div 9\) [*not* \(\div 100\)] |
| \(1 - \Phi\!\left(\dfrac{32-33.6}{\sqrt{3.24}}\right) = \Phi(0.8889)\) | M1 | Standardise & use \(\Phi\), 9 used, answer \(> 0.5\), allow \(\sqrt{}\) errors, allow cc 0.05 but *not* 0.5 |
| \(= \mathbf{0.8130}\) | A1 4 | Answer, art 0.813 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| No, distribution of \(R\) is normal so that of \(\bar{R}\) is normal | B2 2 | Must be saying this. Eg "9 is not large enough": B0. Both: B1 max, unless saying that \(n\) is irrelevant. |
# Question 6:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $33.6$ | B1 | 33.6 clearly stated [not recoverable later] |
| $\frac{115782.84}{100} - 33.6^2 = 28.8684$ | M1 | Correct formula used for biased estimate |
| $\times \frac{100}{99} = \mathbf{29.16}$ | M1, A1 **4** | $\times \frac{100}{99}$, M's independent. Eg $\frac{\Sigma r^2}{99}[-33.6^2]$. **SR** B1 variance in range $[29.1, 29.2]$ |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{R} \sim N(33.6,\ 29.16/9) = N(33.6,\ 1.8^2)$ | M1, A1 | Normal, their $\mu$ stated or implied. Variance $[\text{their (i)}] \div 9$ [*not* $\div 100$] |
| $1 - \Phi\!\left(\dfrac{32-33.6}{\sqrt{3.24}}\right) = \Phi(0.8889)$ | M1 | Standardise & use $\Phi$, 9 used, answer $> 0.5$, allow $\sqrt{}$ errors, allow cc 0.05 but *not* 0.5 |
| $= \mathbf{0.8130}$ | A1 **4** | Answer, art 0.813 |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| No, distribution of $R$ is normal so that of $\bar{R}$ is normal | B2 **2** | Must be saying this. Eg "9 is not large enough": B0. Both: B1 max, unless saying that $n$ is irrelevant. |
---6 The continuous random variable $R$ has the distribution $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$. The results of 100 observations of $R$ are summarised by
$$\Sigma r = 3360.0 , \quad \Sigma r ^ { 2 } = 115782.84 .$$
(i) Calculate an unbiased estimate of $\mu$ and an unbiased estimate of $\sigma ^ { 2 }$.\\
(ii) The mean of 9 observations of $R$ is denoted by $\bar { R }$. Calculate an estimate of $\mathrm { P } ( \bar { R } > 32.0 )$.\\
(iii) Explain whether you need to use the Central Limit Theorem in your answer to part (ii).
\hfill \mbox{\textit{OCR S2 2009 Q6 [10]}}