| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring standard formulas. Part (a) involves calculating a sample mean and applying the normal distribution CI formula with known σ. Part (b) requires working backwards from CI width to find σ, which is slightly less routine but still a direct application of the formula. Both parts are computational with no conceptual challenges beyond knowing the standard formulas. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{4590}{85} \pm z \times \frac{8.3}{\sqrt{85}}\) | M1 | For expression of correct form. Any \(z\) (but not \(\phi(z)\)). |
| \(z = 1.96\) | B1 | |
| \(52.2\) to \(55.8\) (3 sf) | A1 | Must be an interval. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1.96 \times \frac{s}{\sqrt{100}} = 1.4\) or \(2 \times 1.96 \times \frac{s}{\sqrt{100}} = 2.8\) | M1 | Equation of correct form (any \(z\)). Allow factor of 2 error (i.e. first equation = 2.8). |
| \(s = 7.14\) (3 sf) or \(\frac{50}{7}\) | A1 |
## Question 2:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4590}{85} \pm z \times \frac{8.3}{\sqrt{85}}$ | **M1** | For expression of correct form. Any $z$ (but not $\phi(z)$). |
| $z = 1.96$ | **B1** | |
| $52.2$ to $55.8$ (3 sf) | **A1** | Must be an interval. |
| | **3** | |
## Question 2(b):
$1.96 \times \frac{s}{\sqrt{100}} = 1.4$ or $2 \times 1.96 \times \frac{s}{\sqrt{100}} = 2.8$ | M1 | Equation of correct form (any $z$). Allow factor of 2 error (i.e. first equation = 2.8).
$s = 7.14$ (3 sf) or $\frac{50}{7}$ | A1 |
---2 The length, in minutes, of mathematics lectures at a certain college has mean $\mu$ and standard deviation 8.3.
\begin{enumerate}[label=(\alph*)]
\item The total length of a random sample of 85 lectures was 4590 minutes.
Calculate a 95\% confidence interval for $\mu$.\\
The length, in minutes, of history lectures at the college has mean $m$ and standard deviation $s$.
\item Using a random sample of 100 history lectures, a 95\% confidence interval for $m$ was found to have width 2.8 minutes.
Find the value of $s$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q2 [5]}}