| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for confidence intervals |
| Difficulty | Moderate -0.5 Part (a) is a straightforward reverse-calculation from a confidence interval formula (using 1.96 × SE = half-width) requiring only algebraic manipulation. Part (b) tests conceptual understanding that CLT justifies using normal distribution for large samples regardless of population distribution, but this is a standard bookwork point. The question requires recall and basic calculation rather than problem-solving or insight. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3.12 + z \times \frac{\sigma}{\sqrt{150}} = 3.23\) | M1 | OE – correct expression. Any \(z\), but must be a \(z\) value. |
| \(z = 1.96\) | B1 | |
| \(\sigma = 0.687\) (3sf) [cm] | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Yes, because population [of widths] not given to be normally distributed | B1 | Or 'underlying distribution' instead of population. Allow 'yes, because population distribution not known'. Need both statements. |
| 1 |
## Question 2:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3.12 + z \times \frac{\sigma}{\sqrt{150}} = 3.23$ | M1 | OE – correct expression. Any $z$, but must be a $z$ value. |
| $z = 1.96$ | B1 | |
| $\sigma = 0.687$ (3sf) [cm] | A1 | |
| | **3** | |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Yes, because population [of widths] not given to be normally distributed | B1 | Or 'underlying distribution' instead of population. Allow 'yes, because population distribution not known'. Need both statements. |
| | **1** | |2 The widths, $w \mathrm {~cm}$, of a random sample of 150 leaves of a certain kind were measured. The sample mean of $w$ was found to be 3.12 cm .
Using this sample, an approximate $95 \%$ confidence interval for the population mean of the widths in centimetres was found to be [3.01, 3.23].
\begin{enumerate}[label=(\alph*)]
\item Calculate an estimate of the population standard deviation.
\item Explain whether it was necessary to use the Central Limit theorem in your answer to part (a). [1]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q2 [4]}}