| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Combined Poisson approximation and exact calculation |
| Difficulty | Standard +0.3 This is a straightforward S3 question requiring standard Poisson calculations and normal approximation. Part (a) uses tables/calculator for exact Poisson probabilities with λ=15, while part (b) applies the routine normal approximation (with continuity correction) for λ=30. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim \text{Po}(2.5 \times 6) = \text{Po}(15)\) | M1 | |
| \(P(X \leq 18) = 0.8195\) (awrt \(0.819\) or \(0.820\)) | A1 | Use of tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Require \(P(X > n) \leq 0.05\), i.e. \(P(X \leq n) \geq 0.95\) | M1 | |
| \(P(X \leq 19) = 0.8752\), \(P(X \leq 20) = 0.9170\)... \(P(X \leq 22) = 0.9780 \geq 0.95\) | ||
| Minimum number \(= 22\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 2 weeks: \(\lambda = 15 \times 2 = 30\); use Normal approximation \(N(30, 30)\) | M1 | |
| Continuity correction: \(P(X > 35) \approx P\!\left(Z > \dfrac{35.5 - 30}{\sqrt{30}}\right)\) | M1 | Continuity correction |
| \(= P(Z > 1.004)\) | A1 | |
| \(= 1 - \Phi(1.004) = 1 - 0.8424 = 0.1576\) (awrt \(0.158\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 2 weeks: \(\lambda = 30\); use \(N(30,30)\); require \(P(X > n) \leq 0.05\) | M1 | |
| \(\dfrac{n + 0.5 - 30}{\sqrt{30}} \geq 1.6449\) | M1 | |
| \(n \geq 30 + 1.6449\sqrt{30} - 0.5 = 38.51\) | A1ft | |
| Minimum \(n = 39\) | A1 |
## Question 6:
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim \text{Po}(2.5 \times 6) = \text{Po}(15)$ | M1 | |
| $P(X \leq 18) = 0.8195$ (awrt $0.819$ or $0.820$) | A1 | Use of tables |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Require $P(X > n) \leq 0.05$, i.e. $P(X \leq n) \geq 0.95$ | M1 | |
| $P(X \leq 19) = 0.8752$, $P(X \leq 20) = 0.9170$... $P(X \leq 22) = 0.9780 \geq 0.95$ | | |
| Minimum number $= 22$ | A1 | |
### Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| 2 weeks: $\lambda = 15 \times 2 = 30$; use Normal approximation $N(30, 30)$ | M1 | |
| Continuity correction: $P(X > 35) \approx P\!\left(Z > \dfrac{35.5 - 30}{\sqrt{30}}\right)$ | M1 | Continuity correction |
| $= P(Z > 1.004)$ | A1 | |
| $= 1 - \Phi(1.004) = 1 - 0.8424 = 0.1576$ (awrt $0.158$) | A1 | |
### Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| 2 weeks: $\lambda = 30$; use $N(30,30)$; require $P(X > n) \leq 0.05$ | M1 | |
| $\dfrac{n + 0.5 - 30}{\sqrt{30}} \geq 1.6449$ | M1 | |
| $n \geq 30 + 1.6449\sqrt{30} - 0.5 = 38.51$ | A1ft | |
| Minimum $n = 39$ | A1 | |6 The demand for a WWSatNav at a superstore may be modelled by a Poisson distribution with a mean of 2.5 per day. The superstore is open 6 days each week, from Monday morning to Saturday evening.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine the probability that the demand for WWSatNavs during a particular week is at most 18 .
\item The superstore receives a delivery of WWSatNavs on each Sunday evening. The manager, Meena, requires that the probability of WWSatNavs being out of stock during a week should be at most $5 \%$.
Determine the minimum number of WWSatNavs that Meena requires to be in stock after a delivery.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use a distributional approximation to estimate the probability that the demand for WWSatNavs during a period of $\mathbf { 2 }$ weeks is more than 35.
\item Changes to the superstore's delivery schedule result in it receiving a delivery of WWSatNavs on alternate Sunday evenings. Meena now requires that the probability of WWSatNavs being out of stock during the 2 weeks following a delivery should be at most $5 \%$.
Use a distributional approximation to determine the minimum number of WWSatNavs that Meena now requires to be in stock after a delivery.\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2013 Q6 [11]}}