| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | March |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for confidence intervals |
| Difficulty | Moderate -0.8 Part (i) is a routine confidence interval calculation with given σ and large n, requiring only formula substitution. Part (ii) tests basic understanding that CLT applies due to large sample size (n=200), but this is straightforward recall of when CLT is needed rather than any conceptual challenge. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = 2.326\) | B1 | |
| \(62.3 \pm z \dfrac{13.2}{\sqrt{200}}\) | M1 | Any \(z\). Expression of correct form. Must be a '\(z\)' |
| \(60.1\) to \(64.5\) (3 sfs) | A1 | Must be an interval |
| Answer | Marks | Guidance |
|---|---|---|
| Yes, because pop not (given to be) normal, or pop distribution unknown | B1 | No contradictions |
**Question 1(i):**
$z = 2.326$ | B1 |
$62.3 \pm z \dfrac{13.2}{\sqrt{200}}$ | M1 | Any $z$. Expression of correct form. Must be a '$z$'
$60.1$ to $64.5$ (3 sfs) | A1 | Must be an interval
**Total: 3 marks**
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**Question 1(ii):**
Yes, because pop not (given to be) normal, or pop distribution unknown | B1 | No contradictions
**Total: 1 mark**
---1 The masses of a certain variety of plums are known to have standard deviation 13.2 g . A random sample of 200 of these plums is taken and the mean mass of the plums in the sample is found to be 62.3 g .\\
(i) Calculate a $98 \%$ confidence interval for the population mean mass.\\
(ii) State with a reason whether it was necessary to use the Central Limit theorem in the calculation in part (i).\\
\hfill \mbox{\textit{CAIE S2 2019 Q1 [4]}}