| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Confidence interval for Poisson mean |
| Difficulty | Standard +0.3 This is a straightforward application of the normal approximation to Poisson for confidence intervals (standard S3 content). Part (a) requires knowing the formula and looking up a z-value; part (b) involves simple comparison of rates. The calculations are routine with no conceptual challenges beyond recognizing when means are comparable. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \frac{392}{12} = 32.\overline{6}\) (mean per shift) | B1 | Accept 392/12 |
| For Poisson, variance = mean, so use \(\hat{\lambda} = 32.\overline{6}\) | ||
| \(\bar{x} \pm z \sqrt{\frac{\bar{x}}{n}}\) | M1 | Correct structure of CI |
| \(z = 2.3263\) for 98% CI | B1 | Accept 2.326 or 2.33 |
| \(32.\overline{6} \pm 2.3263\sqrt{\frac{32.\overline{6}}{12}}\) | M1 | Correct substitution |
| \((29.6, 35.7)\) | A1 | Accept equivalent correct interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mean calls per hour for weekend shifts: \(\frac{136.8}{48} = 2.85\) | B1 | Correct rate per hour |
| Mean calls per hour for weekday night shifts: \(\frac{32.\overline{6}}{14} = 2.333...\) | B1 | Correct rate per hour |
| Since \(2.85 > 2.33\), the claim appears supported; however \(2.85\) lies within the 98% CI for weekday rate (scaled to per hour), so there is insufficient evidence to support the claim | B1 | Comment must compare rates per hour and relate to uncertainty/CI |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \frac{392}{12} = 32.\overline{6}$ (mean per shift) | B1 | Accept 392/12 |
| For Poisson, variance = mean, so use $\hat{\lambda} = 32.\overline{6}$ | | |
| $\bar{x} \pm z \sqrt{\frac{\bar{x}}{n}}$ | M1 | Correct structure of CI |
| $z = 2.3263$ for 98% CI | B1 | Accept 2.326 or 2.33 |
| $32.\overline{6} \pm 2.3263\sqrt{\frac{32.\overline{6}}{12}}$ | M1 | Correct substitution |
| $(29.6, 35.7)$ | A1 | Accept equivalent correct interval |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean calls per hour for weekend shifts: $\frac{136.8}{48} = 2.85$ | B1 | Correct rate per hour |
| Mean calls per hour for weekday night shifts: $\frac{32.\overline{6}}{14} = 2.333...$ | B1 | Correct rate per hour |
| Since $2.85 > 2.33$, the claim appears supported; however $2.85$ lies within the 98% CI for weekday rate (scaled to per hour), so there is insufficient evidence to support the claim | B1 | Comment must compare rates per hour and relate to uncertainty/CI |
---1 The number of telephone calls per hour to an out-of-hours doctors' service may be modelled by a Poisson distribution.
The total number of telephone calls received during a random sample of 12 weekday night shifts, all of the same duration, was 392.
\begin{enumerate}[label=(\alph*)]
\item Calculate an approximate $98 \%$ confidence interval for the mean number of calls received per weekday night shift.
\item The mean number of calls received during weekend shifts of 48 hours' total duration is 136.8 .
Comment on a claim that the mean number of calls per hour during weekend shifts is greater than that during weekday night shifts, which are each of $\mathbf { 1 4 }$ hours' duration.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2013 Q1 [8]}}