| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Moderate -0.3 This is a straightforward application of Poisson distribution with standard calculations (PMF, cumulative probability) and a routine normal approximation for larger λ. Part (b) requires recognizing when to use normal approximation (λ=30 is large enough) and applying continuity correction, but this is a textbook technique covered in S2 with no novel problem-solving required. |
| 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 probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim \text{Po}(3)\) | B1 | Writing or using Po(3) in either (i) or (ii) |
| \(P(X=7) = P(X \leq 7) - P(X \leq 6)\) or \(\frac{e^{-3}3^7}{7!}\) | M1 | Writing or using \(P(X \leq 7) - P(X \leq 6)\) or \(\frac{e^{-\lambda}\lambda^7}{7!}\) |
| \(= 0.9881 - 0.9665\) | ||
| \(= 0.0216\) | A1 | awrt 0.0216 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X \geq 4) = 1 - P(X \leq 3)\) | M1 | Writing or using \(1 - P(X \leq 3)\). Do not accept \(1 - P(X < 4)\) unless they have used \(1 - P(X \leq 3)\) |
| \(= 1 - 0.6472\) | ||
| \(= 0.3528\) | A1 | awrt 0.353 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim \text{Po}(30)\) approximated by \(N(30, 30)\) | M1A1 | M1 for writing or using a normal approximation; A1 for correct mean and sd |
| \(P(X < 20) = P\!\left(Z < \frac{19.5-30}{\sqrt{30}}\right)\) | M1M1A1 | 2nd M1 standardising using their mean and sd and using [18.5, 19, 19.5, 20 or 20.5] and finding correct area by doing \(1 - P(Z \leq \text{"their 1.92"})\); 3rd M1 for attempting continuity correction (\(19 \pm 0.5\)) i.e. 18.5 or 19.5 only; A1 for \(\pm\frac{19.5-30}{\sqrt{30}}\) or \(\pm\) awrt 1.9 or better |
| \(= P(Z < -1.92)\) | ||
| \(= 1 - 0.9726\) | ||
| \(= 0.0274 - 0.0276\) | A1 | awrt 0.0274, 0.0275 or 0.0276 |
| SC using \(P(X < 20.5/19.5) - P(X < 19.5/18.5)\) can get M1A1 M0M1A0A0 |
## Question 2:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim \text{Po}(3)$ | B1 | Writing or using Po(3) in either (i) or (ii) |
| $P(X=7) = P(X \leq 7) - P(X \leq 6)$ or $\frac{e^{-3}3^7}{7!}$ | M1 | Writing or using $P(X \leq 7) - P(X \leq 6)$ or $\frac{e^{-\lambda}\lambda^7}{7!}$ |
| $= 0.9881 - 0.9665$ | | |
| $= 0.0216$ | A1 | awrt 0.0216 |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X \geq 4) = 1 - P(X \leq 3)$ | M1 | Writing or using $1 - P(X \leq 3)$. Do not accept $1 - P(X < 4)$ unless they have used $1 - P(X \leq 3)$ |
| $= 1 - 0.6472$ | | |
| $= 0.3528$ | A1 | awrt 0.353 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim \text{Po}(30)$ approximated by $N(30, 30)$ | M1A1 | M1 for writing or using a normal approximation; A1 for correct mean and sd |
| $P(X < 20) = P\!\left(Z < \frac{19.5-30}{\sqrt{30}}\right)$ | M1M1A1 | 2nd M1 standardising using their mean and sd **and** using [18.5, 19, 19.5, 20 or 20.5] **and** finding correct area by doing $1 - P(Z \leq \text{"their 1.92"})$; 3rd M1 for attempting continuity correction ($19 \pm 0.5$) i.e. 18.5 or 19.5 **only**; A1 for $\pm\frac{19.5-30}{\sqrt{30}}$ or $\pm$ awrt 1.9 or better |
| $= P(Z < -1.92)$ | | |
| $= 1 - 0.9726$ | | |
| $= 0.0274 - 0.0276$ | A1 | awrt 0.0274, 0.0275 or 0.0276 |
| SC using $P(X < 20.5/19.5) - P(X < 19.5/18.5)$ can get M1A1 M0M1A0A0 | | |
---2. In a village, power cuts occur randomly at a rate of 3 per year.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that in any given year there will be
\begin{enumerate}[label=(\roman*)]
\item exactly 7 power cuts,
\item at least 4 power cuts.
\end{enumerate}\item Use a suitable approximation to find the probability that in the next 10 years the number of power cuts will be less than 20
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2013 Q2 [11]}}