| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Multi-period repeated application |
| Difficulty | Standard +0.3 This is a standard S2 Poisson question covering routine applications: direct probability calculation, independence over multiple trials, normal approximation, and a straightforward hypothesis test. Part (d) requires careful setup of hypotheses but follows a standard template. Slightly above average due to the multi-part nature and the hypothesis test component, but all techniques are textbook applications without novel insight. |
| 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 = lambda5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim \text{Po}(8)\); \(\{P(X \neq 8)\} = 1 - P(X=8)\) | M1 | Can be implied by \(1 - \frac{e^{-8}8^8}{8!}\) or \(1-(P(X \leq 8)-P(X \leq 7))\) or \(1-(0.5925-0.4530)\) or \(1-0.1395\) or \(P(X \leq 7)+1-P(X \leq 8)\) |
| \(= 0.860413\ldots\) or \(0.8605\) | A1 | \(0.86\) or awrt \(0.860\) or awrt \(0.861\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{P(X \geq 8)\} = 1 - 0.453\) | B1 | \(1-0.453\) or awrt \(0.547\) (calculator gives \(0.5470391905\ldots\)) |
| \(\left[\{P(X \geq 8)\}^4\right] = (1-0.453)^4\ \{=(0.547)^4\}\) | M1 | Applying \([\text{their } P(X \geq 8)]^4\) |
| \(= 0.089526\ldots\) | A1 | \(0.09\) or awrt \(0.090\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Y =\) number of chocolate chips in 9 biscuits; \(Y \sim \text{Po}(72) \approx Y \sim \text{N}(72, 72)\) | M1, A1 | Normal or N\((72, 72)\) |
| \(\{P(Y>75)\} \approx P(Y>75.5)\) | M1 | For either \(74.5\) or \(75.5\) |
| \(= P\!\left(Z > \frac{75.5-72}{\sqrt{72}}\right)\) | M1 | Standardising \((\pm)\) with their mean, their standard deviation and either \(75.5\) or \(75\) or \(74.5\) |
| \(= P(Z > 0.41\ldots) = 1 - 0.6591\) | — | — |
| \(= 0.3409\) (from calculator \(0.339994\ldots\)) | A1 | awrt \(0.341\) or awrt \(0.340\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 1.5,\ H_1: \lambda > 1.5\) or \(H_0: \lambda = 6,\ H_1: \lambda > 6\) | B1 | Both hypotheses stated correctly; allow \(\mu\) instead of \(\lambda\); B0 for \(H_0: \lambda=6,\ H_1>6\) or \(H_0:x=6,\ H_1:x>6\) etc. |
| Under \(H_0\), for 4 hours: \(X \sim \text{Po}(6)\) | — | — |
| Probability Method: \(P(X \geq 11) = 1 - P(X \leq 10) = 1 - 0.9574 = 0.0426\) | M1 | For use of \(X \sim \text{Po}(6)\); may be implied by \(0.9161, 0.9574, 0.9799, 0.0839, 0.0426\) or \(0.0201\); condone \(\frac{e^{-6}(6)^{11}}{11!}\) |
| Critical Region Method: \(P(X \leq 9) = 0.9161\) or \(P(X \geq 10)=0.0839\); \(P(X \leq 10)=0.9574\) or \(P(X \geq 11)=0.0426\) | M1 | — |
| \(P(X \geq 11) = 0.0426\) \ | \(\text{CR}: X \geq 11\) | A1 |
| Reject \(H_0\) or significant or 11 lies in CR | dM1 | Dependent on previous M; correct follow-through comparison based on probability/CR and significance level compatible with stated \(H_1\); do not allow non-contextual conflicting statements |
| Conclude: the rate of sales of packets of biscuits has increased OR the mean number of packets of biscuits sold has increased | A1 cso | Correct conclusion in context; award for correct solution only with all previous marks scored; using either "rate of sales and increased" or "mean sold and increased" |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim \text{Po}(8)$; $\{P(X \neq 8)\} = 1 - P(X=8)$ | M1 | Can be implied by $1 - \frac{e^{-8}8^8}{8!}$ or $1-(P(X \leq 8)-P(X \leq 7))$ or $1-(0.5925-0.4530)$ or $1-0.1395$ or $P(X \leq 7)+1-P(X \leq 8)$ |
| $= 0.860413\ldots$ or $0.8605$ | A1 | $0.86$ or awrt $0.860$ or awrt $0.861$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{P(X \geq 8)\} = 1 - 0.453$ | B1 | $1-0.453$ or awrt $0.547$ (calculator gives $0.5470391905\ldots$) |
| $\left[\{P(X \geq 8)\}^4\right] = (1-0.453)^4\ \{=(0.547)^4\}$ | M1 | Applying $[\text{their } P(X \geq 8)]^4$ |
| $= 0.089526\ldots$ | A1 | $0.09$ or awrt $0.090$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y =$ number of chocolate chips in 9 biscuits; $Y \sim \text{Po}(72) \approx Y \sim \text{N}(72, 72)$ | M1, A1 | Normal or N$(72, 72)$ |
| $\{P(Y>75)\} \approx P(Y>75.5)$ | M1 | For either $74.5$ or $75.5$ |
| $= P\!\left(Z > \frac{75.5-72}{\sqrt{72}}\right)$ | M1 | Standardising $(\pm)$ with their mean, their standard deviation and either $75.5$ or $75$ or $74.5$ |
| $= P(Z > 0.41\ldots) = 1 - 0.6591$ | — | — |
| $= 0.3409$ (from calculator $0.339994\ldots$) | A1 | awrt $0.341$ or awrt $0.340$ |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 1.5,\ H_1: \lambda > 1.5$ or $H_0: \lambda = 6,\ H_1: \lambda > 6$ | B1 | Both hypotheses stated correctly; allow $\mu$ instead of $\lambda$; B0 for $H_0: \lambda=6,\ H_1>6$ or $H_0:x=6,\ H_1:x>6$ etc. |
| Under $H_0$, for 4 hours: $X \sim \text{Po}(6)$ | — | — |
| **Probability Method:** $P(X \geq 11) = 1 - P(X \leq 10) = 1 - 0.9574 = 0.0426$ | M1 | For use of $X \sim \text{Po}(6)$; may be implied by $0.9161, 0.9574, 0.9799, 0.0839, 0.0426$ or $0.0201$; condone $\frac{e^{-6}(6)^{11}}{11!}$ |
| **Critical Region Method:** $P(X \leq 9) = 0.9161$ or $P(X \geq 10)=0.0839$; $P(X \leq 10)=0.9574$ or $P(X \geq 11)=0.0426$ | M1 | — |
| $P(X \geq 11) = 0.0426$ \| $\text{CR}: X \geq 11$ | A1 | Either $P(X \geq 11)=0.0426$ or $\text{CR}: X \geq 11$ or $\text{CR}: X > 10$; award $1^{\text{st}}$ M1 $1^{\text{st}}$ A1 for writing down $\text{CR}: X \geq 11$ or $\text{CR}: X > 10$ from no working |
| Reject $H_0$ or significant or 11 lies in CR | dM1 | Dependent on previous M; correct follow-through comparison based on probability/CR and significance level compatible with stated $H_1$; do not allow non-contextual conflicting statements |
| Conclude: the **rate of sales** of packets of biscuits has **increased** OR the **mean** number of packets of biscuits **sold** has **increased** | A1 cso | Correct conclusion in context; award for correct solution only with all previous marks scored; using either "rate of sales and increased" or "mean sold and increased" |
---2. A company produces chocolate chip biscuits. The number of chocolate chips per biscuit has a Poisson distribution with mean 8
\begin{enumerate}[label=(\alph*)]
\item Find the probability that one of these biscuits, selected at random, does not contain 8 chocolate chips.
A small packet contains 4 of these biscuits, selected at random.
\item Find the probability that each biscuit in the packet contains at least 8 chocolate chips.
A large packet contains 9 of these biscuits, selected at random.
\item Use a suitable approximation to find the probability that there are more than 75 chocolate chips in the packet.
A shop sells packets of biscuits, randomly, at a rate of 1.5 packets per hour. Following an advertising campaign, 11 packets are sold in 4 hours.
\item Test, at the $5 \%$ level of significance, whether or not there is evidence that the rate of sales of packets of biscuits has increased. State your hypotheses clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2015 Q2 [15]}}