| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Find parameter from given probability |
| Difficulty | Standard +0.8 This question requires understanding Poisson distributions, normal approximation with continuity correction, and working backwards from a given probability to find an unknown parameter using inverse normal tables. Part (a) is routine Poisson calculation, but part (b) involves multiple steps: setting up the rate for n minutes, applying continuity correction correctly, using z-tables in reverse, and solving for n. This reverse-engineering with approximations is more demanding than standard forward calculations. |
| 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 probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([X \sim Po(6)]\); \(P(X=7) = P(X \leqslant 7) - P(X \leqslant 6)\) | M1 | For writing or using \(P(X \leqslant 7) - P(X \leqslant 6)\) or \(\frac{e^{-\lambda}\lambda^7}{7!}\) |
| \(= 0.7440 - 0.6063 = 0.13767697...\) awrt \(\underline{0.138}\) | A1 | Correct answer only 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 7) = 1 - P(X \leqslant 7)\) | M1 | For writing or using \(1 - P(X \leqslant 7)\) |
| \(= 1 - 0.744 = 0.2560202...\) awrt \(\underline{0.256}\) | A1 | Correct answer only 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Y \sim N(\lambda, \lambda)\) | M1 | For stating Normal with mean = variance |
| \(\lambda = 0.6n\) | B1 | For using \(\lambda = 0.6n\) somewhere |
| \(P(Y > 40) = 0.2266\); \(\frac{40.5 - \lambda}{\sqrt{\lambda}} = 0.75\) | M1M1 B1 A1 | 2nd M1: continuity correction on 40; 3rd M1: standardising with 40.5 and \(\sqrt{\lambda}\); B1 for \(z = \pm 0.75\) or better; A1 for fully correct equation |
| \(\lambda + 0.75\sqrt{\lambda} - 40.5 = 0\) | ||
| \(\sqrt{\lambda} = 6\) | M1A1 | 4th M1 for solving 3-term quadratic; A1 for \(\sqrt{\lambda} = 6\) or \(\lambda = 36\) |
| \(n = 60\) | A1 |
## Question 5:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[X \sim Po(6)]$; $P(X=7) = P(X \leqslant 7) - P(X \leqslant 6)$ | M1 | For writing or using $P(X \leqslant 7) - P(X \leqslant 6)$ or $\frac{e^{-\lambda}\lambda^7}{7!}$ |
| $= 0.7440 - 0.6063 = 0.13767697...$ awrt $\underline{0.138}$ | A1 | Correct answer only 2/2 |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 7) = 1 - P(X \leqslant 7)$ | M1 | For writing or using $1 - P(X \leqslant 7)$ |
| $= 1 - 0.744 = 0.2560202...$ awrt $\underline{0.256}$ | A1 | Correct answer only 2/2 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y \sim N(\lambda, \lambda)$ | M1 | For stating Normal with mean = variance |
| $\lambda = 0.6n$ | B1 | For using $\lambda = 0.6n$ somewhere |
| $P(Y > 40) = 0.2266$; $\frac{40.5 - \lambda}{\sqrt{\lambda}} = 0.75$ | M1M1 B1 A1 | 2nd M1: continuity correction on 40; 3rd M1: standardising with 40.5 and $\sqrt{\lambda}$; B1 for $z = \pm 0.75$ or better; A1 for fully correct equation |
| $\lambda + 0.75\sqrt{\lambda} - 40.5 = 0$ | | |
| $\sqrt{\lambda} = 6$ | M1A1 | 4th M1 for solving 3-term quadratic; A1 for $\sqrt{\lambda} = 6$ or $\lambda = 36$ |
| $n = 60$ | A1 | |
---5. Cars stop at a service station randomly at a rate of 3 every 5 minutes.
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability that in a randomly selected 10 minute period,
\begin{enumerate}[label=(\roman*)]
\item exactly 7 cars will stop at the service station,
\item more than 7 cars will stop at the service station.
Using a normal approximation, the probability that more than 40 cars will stop at the service station during a randomly selected $n$ minute period is 0.2266 correct to 4 significant figures.
\end{enumerate}\item Find the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q5 [13]}}