| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Multiple independent time periods |
| Difficulty | Standard +0.3 This is a straightforward S2 Poisson question with standard parts: identifying the distribution (rate adjustment from year to 6 months), using tables for a tail probability, calculating P(X≥1) for one month, then applying binomial distribution. Part (d) combines Poisson and binomial but in a routine way. Slightly above average due to the two-distribution combination, but all steps are textbook standard with no novel insight required. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02i 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 | Marks | Guidance |
| \(X \sim Po(9)\) | M1A1 (2) | M1 for Poisson (accept Po). Condone P(9). A1 for mean of 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X>7) = 1 - P(X \leq 7) = [1-0.3239] = 0.6761\) | M1, A1 (2) | M1 for writing \(1-P(X \leq 7)\), may be implied by \(1-0.3239\) or correct answer. A1 for awrt 0.676 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([Y = \text{no. of accidents in a month}]\ Y \sim Po(1.5)\) | B1 | B1 for \(Po(1.5)\) written or used |
| \(P(Y \geq 1) = 1 - P(Y=0) = [1-0.2231] = 0.7769\) | M1, A1cso (3) | M1 for writing/using \(1-P(Y=0)\) or \(1-P(Y \leq 0)\) or \(1-e^{-\lambda}\). A1 for at least \((1-0.223)\) or better. Answer given so 0.777 does not imply all three marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([A = \text{no. of months with at least one accident}]\ A \sim B(6, 0.777)\) | M1 | 1st M1 for identifying binomial with \(n=6\) and \(p=0.777\) or better. Condone use of \(p=0.223\) |
| \(P(A=4) = \binom{6}{4}(0.777)^4(0.223)^2\) | M1 | 2nd M1 must have \(^6C_4\ (0.777)^4(1-0.777)^2\) |
| \(= 0.2719\ldots\) awrt 0.272 | A1 (3) | A1 for awrt 0.272 |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim Po(9)$ | M1A1 (2) | M1 for Poisson (accept Po). Condone P(9). A1 for mean of 9 |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X>7) = 1 - P(X \leq 7) = [1-0.3239] = 0.6761$ | M1, A1 (2) | M1 for writing $1-P(X \leq 7)$, may be implied by $1-0.3239$ or correct answer. A1 for awrt 0.676 |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[Y = \text{no. of accidents in a month}]\ Y \sim Po(1.5)$ | B1 | B1 for $Po(1.5)$ written or used |
| $P(Y \geq 1) = 1 - P(Y=0) = [1-0.2231] = 0.7769$ | M1, A1cso (3) | M1 for writing/using $1-P(Y=0)$ or $1-P(Y \leq 0)$ or $1-e^{-\lambda}$. A1 for at least $(1-0.223)$ or better. Answer given so 0.777 does not imply all three marks |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[A = \text{no. of months with at least one accident}]\ A \sim B(6, 0.777)$ | M1 | 1st M1 for identifying binomial with $n=6$ and $p=0.777$ or better. Condone use of $p=0.223$ |
| $P(A=4) = \binom{6}{4}(0.777)^4(0.223)^2$ | M1 | 2nd M1 must have $^6C_4\ (0.777)^4(1-0.777)^2$ |
| $= 0.2719\ldots$ awrt **0.272** | A1 (3) | A1 for awrt 0.272 |
---3. Accidents occur randomly at a road junction at a rate of 18 every year.
The random variable $X$ represents the number of accidents at this road junction in the next 6 months.
\begin{enumerate}[label=(\alph*)]
\item Write down the distribution of $X$.
\item Find $\mathrm { P } ( X > 7 )$.
\item Show that the probability of at least one accident in a randomly selected month is 0.777 (correct to 3 decimal places).
\item Find the probability that there is at least one accident in exactly 4 of the next 6 months.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q3 [10]}}