| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Recover sample stats from CI |
| Difficulty | Moderate -0.8 This question tests basic understanding of confidence interval structure with straightforward arithmetic. Part (a) requires finding the midpoint of an interval, part (b) involves rearranging the standard CI formula with given values, and part (c) tests conceptual knowledge that sample SD doesn't depend on population distribution. All parts are routine recall and simple calculation with no problem-solving or novel insight required. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(254\) [m] | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(263 = \text{'254'} + 1.96 \times \frac{\sigma}{\sqrt{50}}\) oe or \(2 \times 1.96 \times \frac{\sigma}{\sqrt{50}} = 18\) | M1 | ft their '254' accept 1.96 or 1.645 for M1. |
| \(\left[\sigma = \frac{9\sqrt{50}}{1.96} =\right]\) s.d. \(= 32.5\) [m] (3 sf) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No. Because the sample mean is approximately normally distributed [for large \(n\)] | B1 | Both needed. Or because of the Central Limit theorem. Or because \(n\) is large [accept \(\geqslant 30\) condone \(\geqslant 50\)]. |
| Total: 1 |
**Question 2(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $254$ [m] | B1 | |
| **Total: 1** | | |
---
**Question 2(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $263 = \text{'254'} + 1.96 \times \frac{\sigma}{\sqrt{50}}$ oe or $2 \times 1.96 \times \frac{\sigma}{\sqrt{50}} = 18$ | M1 | ft their '254' accept 1.96 or 1.645 for **M1**. |
| $\left[\sigma = \frac{9\sqrt{50}}{1.96} =\right]$ s.d. $= 32.5$ [m] (3 sf) | A1 | |
| **Total: 2** | | |
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**Question 2(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| No. Because the sample mean is approximately normally distributed [for large $n$] | B1 | Both needed. Or because of the Central Limit theorem. Or because $n$ is large [accept $\geqslant 30$ condone $\geqslant 50$]. |
| **Total: 1** | | |2 The lengths of a random sample of 50 roads in a certain region were measured.Using the results,a $95 \%$ confidence interval for the mean length,in metres,of all roads in this region was found to be[245,263].
\begin{enumerate}[label=(\alph*)]
\item Find the mean length of the 50 roads in the sample.
\item Calculate an estimate of the standard deviation of the lengths of roads in this region.
\item It is now given that the lengths of roads in this region are normally distributed.\\
State,with a reason,whether this fact would make any difference to your calculation in part(b).
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q2 [4]}}