| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown distribution, CLT applied |
| Difficulty | Standard +0.3 This is a straightforward application of CLT to construct a confidence interval for a Poisson mean, requiring knowledge that for large samples the sample mean is approximately normal. The 'hence comment' part requires simple comparison of the interval to 7 calls/week. Slightly above average difficulty due to being Further Maths content and requiring CLT understanding, but the calculation itself is routine once the method is identified. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Estimate of weekly mean: \(\hat{\lambda} = \frac{108}{13} = \frac{108}{13} \approx 8.308\) | B1 | Correct estimate |
| 98% CI uses \(z = 2.326\) | B1 | Correct \(z\)-value |
| CI: \(\hat{\lambda} \pm z\sqrt{\hat{\lambda}}\) i.e. \(\frac{108}{13} \pm 2.326\sqrt{\frac{108}{13}}\) | M1 A1 | Correct structure using Poisson variance |
| \(= (2.61, 14.0)\) approximately \((2.61 \leq \lambda \leq 14.01)\) | A1 | Correct interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 1 call per day = 7 calls per week | B1 | Correct conversion |
| 7 lies within the confidence interval \((2.61, 14.0)\) | M1 | Comparison with CI |
| Therefore there is no significant evidence to contradict the officer's claim | A1 | Correct contextual conclusion |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Estimate of weekly mean: $\hat{\lambda} = \frac{108}{13} = \frac{108}{13} \approx 8.308$ | B1 | Correct estimate |
| 98% CI uses $z = 2.326$ | B1 | Correct $z$-value |
| CI: $\hat{\lambda} \pm z\sqrt{\hat{\lambda}}$ i.e. $\frac{108}{13} \pm 2.326\sqrt{\frac{108}{13}}$ | M1 A1 | Correct structure using Poisson variance |
| $= (2.61, 14.0)$ approximately $(2.61 \leq \lambda \leq 14.01)$ | A1 | Correct interval |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| 1 call per day = 7 calls per week | B1 | Correct conversion |
| 7 lies within the confidence interval $(2.61, 14.0)$ | M1 | Comparison with CI |
| Therefore there is no significant evidence to contradict the officer's claim | A1 | Correct contextual conclusion |
---2 The number of emergency calls received by a fire station may be modelled by a Poisson distribution.
During a given period of 13 weeks, the station received a total of 108 emergency calls.
\begin{enumerate}[label=(\alph*)]
\item Construct an approximate $98 \%$ confidence interval for the average weekly number of emergency calls received by the station.
\item Hence comment on the station officer's claim that the station receives an average of one emergency call per day.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2011 Q2 [7]}}