| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unbiased estimator from summary statistics |
| Difficulty | Standard +0.3 This is a straightforward application of standard formulas for unbiased estimators and the Central Limit Theorem. Part (i) requires direct substitution into memorized formulas (sample mean and s²), while part (ii) applies CLT with a normal approximation—both are routine procedures for S2 level with no conceptual challenges or novel problem-solving required. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = \frac{7520}{150} = 50.1\) (3 sf) | B1 | |
| \(s^2 = \frac{150}{149}\left(\frac{413540}{150} - \left(\frac{7520}{150}\right)^2\right)\) | M1 | Attempt at unbiased variance (either formula) |
| \(= 245\) or \(246\) (3 sf) | A1 (3) | Allow \(s^2 = 15.7^2\) (3 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{53 - \frac{7520}{150}}{\sqrt{\frac{245.217}{80}}} = 1.637\) to \(1.638\) | M1 | For standardising (\(\pm\)) with their mean and their variance must have \(\sqrt{80}\) (ignore cc) |
| \(1 - \Phi(1.637)\) | M1 | Correct area consistent with their working |
| \(= 0.0488\) to \(0.0509\) | A1 (3) | Correct working only |
## Question 3:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = \frac{7520}{150} = 50.1$ (3 sf) | B1 | |
| $s^2 = \frac{150}{149}\left(\frac{413540}{150} - \left(\frac{7520}{150}\right)^2\right)$ | M1 | Attempt at unbiased variance (either formula) |
| $= 245$ or $246$ (3 sf) | A1 (3) | Allow $s^2 = 15.7^2$ (3 sf) |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{53 - \frac{7520}{150}}{\sqrt{\frac{245.217}{80}}} = 1.637$ to $1.638$ | M1 | For standardising ($\pm$) with their mean and their variance must have $\sqrt{80}$ (ignore cc) |
| $1 - \Phi(1.637)$ | M1 | Correct area consistent with their working |
| $= 0.0488$ to $0.0509$ | A1 (3) | Correct working only |
---3 The lengths, $x \mathrm {~mm}$, of a random sample of 150 insects of a certain kind were found. The results are summarised by $\Sigma x = 7520$ and $\Sigma x ^ { 2 } = 413540$.\\
(i) Calculate unbiased estimates of the population mean and variance of the lengths of insects of this kind.\\
(ii) Using the values found in part (i), calculate an estimate of the probability that the mean length of a further random sample of 80 insects of this kind is greater than 53 mm .
\hfill \mbox{\textit{CAIE S2 2012 Q3 [6]}}