| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Multiple periods with binomial structure |
| Difficulty | Standard +0.3 This is a straightforward S2 Poisson question requiring standard techniques: scaling the parameter for different time periods, using tables/calculator for probabilities, applying binomial distribution with Poisson probability, and normal approximation to Poisson. All steps are routine applications of learned methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02d Binomial: mean np and variance np(1-p)5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(X\) be the random variable = number of houses sold; \(X \sim \text{Po}(8)\) | B1 | 1st B1 for writing or using Po(8) in either (i) or (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X \leq 3) - P(X \leq 2) = 0.0424 - 0.0138\) or \(\dfrac{e^{-8}8^3}{3!}\) | M1 | M1 writing or using \(P(X \leq 3) - P(X \leq 2)\) or \(\dfrac{e^{-8}8^3}{3!}\) |
| \(= 0.0286\) | A1 | awrt 0.0286 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 5) = 1 - P(X \leq 5) = 1 - 0.1912\) | M1 | M1 writing or using \(1 - P(X \leq 5)\) |
| \(= 0.8088\) | A1 | awrt 0.809 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(Y\) = number of periods where more than 5 houses sold; \(Y \sim B(12, 0.8088)\) | M1 | M1 writing or attempting to use B(12, their a(ii)); NB ft their a(ii) to at least 2sf |
| \(P(Y = 9) = (0.8088)^9(1-0.8088)^3 \dfrac{12!}{9!3!}\) | M1 | Allow \(^{12}C_3\) or \(^{12}C_9\) or 220; NB ft their a(ii) to at least 1sf but expression must be seen |
| \(= 0.228\) | A1 | awrt 0.228 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(N(20, 20)\) | M1A1 | 1st M1 for writing or using a normal approximation; 1st A1 for correct mean and sd |
| \(P(X > 25) = 1 - P\!\left(Z \leq \dfrac{25.5 - 20}{\sqrt{20}}\right)\) | M1, M1, A1 | 2nd M1 standardising using their mean and sd with \([24.5, 25, 25.5, 26, 26.5]\); 3rd M1 for continuity correction \((26 \pm 0.5)\); 2nd A1 for \(\pm\dfrac{25.5-20}{\sqrt{20}}\) or \(\pm\) awrt 1.2 or better |
| \(= 1 - P(Z \leq 1.23) = 1 - 0.8907\) | ||
| \(= 0.1093/0.1094\) | A1 | awrt 0.109 |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $X$ be the random variable = number of houses sold; $X \sim \text{Po}(8)$ | B1 | 1st B1 for writing or using Po(8) in either (i) or (ii) |
### Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X \leq 3) - P(X \leq 2) = 0.0424 - 0.0138$ or $\dfrac{e^{-8}8^3}{3!}$ | M1 | M1 writing or using $P(X \leq 3) - P(X \leq 2)$ or $\dfrac{e^{-8}8^3}{3!}$ |
| $= 0.0286$ | A1 | awrt 0.0286 |
### Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 5) = 1 - P(X \leq 5) = 1 - 0.1912$ | M1 | M1 writing or using $1 - P(X \leq 5)$ |
| $= 0.8088$ | A1 | awrt 0.809 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $Y$ = number of periods where more than 5 houses sold; $Y \sim B(12, 0.8088)$ | M1 | M1 writing or attempting to use B(12, their a(ii)); NB ft their a(ii) to at least 2sf |
| $P(Y = 9) = (0.8088)^9(1-0.8088)^3 \dfrac{12!}{9!3!}$ | M1 | Allow $^{12}C_3$ or $^{12}C_9$ or 220; NB ft their a(ii) to at least 1sf but expression must be seen |
| $= 0.228$ | A1 | awrt 0.228 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $N(20, 20)$ | M1A1 | 1st M1 for writing or using a normal approximation; 1st A1 for correct mean and sd |
| $P(X > 25) = 1 - P\!\left(Z \leq \dfrac{25.5 - 20}{\sqrt{20}}\right)$ | M1, M1, A1 | 2nd M1 standardising using their mean and sd with $[24.5, 25, 25.5, 26, 26.5]$; 3rd M1 for continuity correction $(26 \pm 0.5)$; 2nd A1 for $\pm\dfrac{25.5-20}{\sqrt{20}}$ or $\pm$ awrt 1.2 or better |
| $= 1 - P(Z \leq 1.23) = 1 - 0.8907$ | | |
| $= 0.1093/0.1094$ | A1 | awrt 0.109 |
---4. The number of houses sold by an estate agent follows a Poisson distribution, with a mean of 2 per week.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that in the next 4 weeks the estate agent sells,
\begin{enumerate}[label=(\roman*)]
\item exactly 3 houses,
\item more than 5 houses.
The estate agent monitors sales in periods of 4 weeks.
\end{enumerate}\item Find the probability that in the next twelve of these 4 week periods there are exactly nine periods in which more than 5 houses are sold.
The estate agent will receive a bonus if he sells more than 25 houses in the next 10 weeks.
\item Use a suitable approximation to estimate the probability that the estate agent receives a bonus.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2012 Q4 [14]}}