| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| 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 standard S3 question testing routine application of Poisson-to-Normal approximation and confidence intervals. Part (a) requires scaling the parameter and applying continuity correction—straightforward bookwork. Part (b) involves calculating a confidence interval using the standard formula √(λ/n) and making a simple comparison. No novel insight or complex reasoning required, just methodical application of learned techniques, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Letters/4-week \(\sim N(49, 49)\) | B1 | CAO; mean = variance = 49 |
| \(P(42 \leq X_P \leq 54) = P(41.5 < X_N < 54.5)\) | M1 | Use of \(\pm 0.5\) |
| \(= P\left(\frac{41.5-49}{7} < Z < \frac{54.5-49}{7}\right)\) | M1 | Standardising (41.5, 42 or 42.5) or (53.5, 54 or 54.5) with C's \(\mu\) and \(\sqrt{\mu}\) |
| \(= P(-1.07 < Z < 0.79)\) | ||
| \(= \Phi(0.79) - (1 - \Phi(1.07))\) | m1 | Area change |
| \(= 0.641\) to \(0.644\) | A1 | AWFW |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| 98% CI \(\Rightarrow z = 2.3263\) | B1 | AWFW 2.32 to 2.33 |
| CI for \(\lambda\)/16-week: \(\hat{\lambda} \pm z\sqrt{\hat{\lambda}}\) | M1 | Use of expression |
| \(248 \pm 2.3263 \times \sqrt{248}\) | A1\(\checkmark\) | \(\checkmark\) on \(z\) |
| or \(15.5 \pm 2.3263 \times \sqrt{\frac{15.5}{16}}\) | ||
| \(248 \pm 36.6\) or \(15.5 \pm 2.3\) | M1 | Division by 16 somewhere |
| \((13.2, 17.8)\) | A1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Value of \(12.25\ (196)\) is below CI | B1\(\checkmark\) \(\uparrow\) dep | \(\checkmark\) on CI; must use 12.25 (196) |
| Rosa's belief is supported | B1\(\checkmark\) | \(\checkmark\) on CI |
# Question 5:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Letters/4-week $\sim N(49, 49)$ | B1 | CAO; mean = variance = 49 |
| $P(42 \leq X_P \leq 54) = P(41.5 < X_N < 54.5)$ | M1 | Use of $\pm 0.5$ |
| $= P\left(\frac{41.5-49}{7} < Z < \frac{54.5-49}{7}\right)$ | M1 | Standardising (41.5, 42 or 42.5) or (53.5, 54 or 54.5) with C's $\mu$ and $\sqrt{\mu}$ |
| $= P(-1.07 < Z < 0.79)$ | | |
| $= \Phi(0.79) - (1 - \Phi(1.07))$ | m1 | Area change |
| $= 0.641$ to $0.644$ | A1 | AWFW |
## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| 98% CI $\Rightarrow z = 2.3263$ | B1 | AWFW 2.32 to 2.33 |
| CI for $\lambda$/16-week: $\hat{\lambda} \pm z\sqrt{\hat{\lambda}}$ | M1 | Use of expression |
| $248 \pm 2.3263 \times \sqrt{248}$ | A1$\checkmark$ | $\checkmark$ on $z$ |
| or $15.5 \pm 2.3263 \times \sqrt{\frac{15.5}{16}}$ | | |
| $248 \pm 36.6$ or $15.5 \pm 2.3$ | M1 | Division by 16 somewhere |
| $(13.2, 17.8)$ | A1 | AWRT |
## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Value of $12.25\ (196)$ **is below** CI | B1$\checkmark$ $\uparrow$ dep | $\checkmark$ on CI; must use 12.25 (196) |
| Rosa's belief **is supported** | B1$\checkmark$ | $\checkmark$ on CI |
---5 The number of letters per week received at home by Rosa may be modelled by a Poisson distribution with parameter 12.25.
\begin{enumerate}[label=(\alph*)]
\item Using a normal approximation, estimate the probability that, during a 4 -week period, Rosa receives at home at least 42 letters but at most 54 letters.
\item Rosa also receives letters at work. During a 16-week period, she receives at work a total of 248 letters.
\begin{enumerate}[label=(\roman*)]
\item Assuming that the number of letters received at work by Rosa may also be modelled by a Poisson distribution, calculate a $98 \%$ confidence interval for the average number of letters per week received at work by Rosa.
\item Hence comment on Rosa's belief that she receives, on average, fewer letters at home than at work.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2006 Q5 [12]}}