| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | March |
| Marks | 7 |
| 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 standard confidence interval calculation with given summary statistics requiring only formula application (z-value lookup, standard error calculation). Part (b) tests understanding that CLT justifies using normal distribution when population distribution is unknown, but this is routine bookwork recall. The question involves no problem-solving or novel insight—just mechanical computation and stating a standard justification. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Est}(\mu) = \dfrac{4820}{60}\) or \(\dfrac{241}{3}\) or \(80.3\) (3 sf) | B1 | |
| \(\text{Est}(\sigma^2) = \dfrac{60}{59}\left(\dfrac{392050}{60} - \left(\dfrac{4820}{60}\right)^2\right)\) | M1 | Use of biased (80.72) score M0 A0 |
| \(82.0904\left(\dfrac{14530}{177}\right)\) to \(82.635\) or \(SD = 9.0604\) to \(9.0904\) (3sf) | A1 | |
| \(z = 2.326\) | B1 | |
| \(\dfrac{4820}{60} \pm z \times \sqrt{\dfrac{82.0904}{60}}\) | M1 | Expression of the correct form – must be \(z\) value |
| \(77.6\) to \(83.1\) (3 sf) | A1 | CWO; Use of biased \(77.6\) to \(83.0(3)\) can score B1M1A1 (max 4/6) |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Population distribution of times unknown | B1 | Accept 'not normal' |
| 1 |
## Question 1:
### Part 1(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Est}(\mu) = \dfrac{4820}{60}$ or $\dfrac{241}{3}$ or $80.3$ (3 sf) | **B1** | |
| $\text{Est}(\sigma^2) = \dfrac{60}{59}\left(\dfrac{392050}{60} - \left(\dfrac{4820}{60}\right)^2\right)$ | **M1** | Use of biased (80.72) score M0 A0 |
| $82.0904\left(\dfrac{14530}{177}\right)$ to $82.635$ or $SD = 9.0604$ to $9.0904$ (3sf) | **A1** | |
| $z = 2.326$ | **B1** | |
| $\dfrac{4820}{60} \pm z \times \sqrt{\dfrac{82.0904}{60}}$ | **M1** | Expression of the correct form – must be $z$ value |
| $77.6$ to $83.1$ (3 sf) | **A1** | CWO; Use of biased $77.6$ to $83.0(3)$ can score B1M1A1 (max 4/6) |
| | **6** | |
### Part 1(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Population distribution of times unknown | **B1** | Accept 'not normal' |
| | **1** | |1 A construction company notes the time, $t$ days, that it takes to build each house of a certain design. The results for a random sample of 60 such houses are summarised as follows.
$$\Sigma t = 4820 \quad \Sigma t ^ { 2 } = 392050$$
\begin{enumerate}[label=(\alph*)]
\item Calculate a 98\% confidence interval for the population mean time.
\item Explain why it was necessary to use the Central Limit theorem in part (a).
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q1 [7]}}