| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Conditional probability given total |
| Difficulty | Standard +0.3 This is a straightforward multi-part Poisson distribution question requiring standard techniques: calculating probabilities from tables, working backwards with inequalities, using the additive property for non-overlapping periods, and applying normal approximation. All parts follow textbook methods with no novel insight required, making it slightly easier than average for S2. |
| Spec | 2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([X\sim\text{Po}(10)]\), \(P(X>12) = 1-P(X\leq 12) = 1-0.7916 = 0.2084\) awrt 0.208 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X\leq k) < 1-2\times 0.2084\), \(P(X\leq k)<0.5832...\), \(k=\mathbf{10}\) | M1 A1cao | M1 for \(P(X\leq k)<1-2p\) or \(P(X |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(W\sim\text{Po}(5)\) | B1 | B1 for writing/using Po(5) |
| \([P(W=4)]^2 = \left(\frac{e^{-5}5^4}{4!}\right)^2 = (0.4405-0.2650)^2 = 0.030788...\) awrt 0.0308 | M1 A1 | M1 for \([P(W=4)]^2\) or either correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P((X_1\geq 10\cap X_2\geq 10)\mid Y=21) = \dfrac{\frac{e^{-10}10^{10}}{10!}\times\frac{e^{-10}10^{11}}{11!}+\frac{e^{-10}10^{11}}{11!}\times\frac{e^{-10}10^{10}}{10!}}{\frac{e^{-20}20^{21}}{21!}}\) | M1 M1 | 1st M1 use of Po(10) with \(X=10\) or \(X=11\); 2nd M1 correct expression for \(2\times P(X=10)\times P(X=11)\) from Po(10) |
| \(= 0.336376...\) using tables: \(\frac{2\times(0.5830-0.4579)(0.6968-0.5830)}{\frac{e^{-20}20^{21}}{21!}} = 0.336537...\) awrt 0.336/7 | M1 A1 | 3rd M1 ratio with correct denominator (awrt 0.0846) and num<denom |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(L\sim\text{Po}(40)\approx\text{N}(40,40)\) | B1 B1 | 1st B1 for Po(40); 2nd B1 for normal with mean and variance 40 |
| \(P(L>27) = P\!\left(Z>\frac{27.5-40}{\sqrt{40}}\right)\) | M1 M1 | 1st M1 continuity correction \((27\pm 0.5)\); 2nd M1 standardising with their mean and sd |
| \(P(Z>-1.98) = 0.9761\) awrt 0.976 | A1 |
## Question 5:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[X\sim\text{Po}(10)]$, $P(X>12) = 1-P(X\leq 12) = 1-0.7916 = 0.2084$ **awrt 0.208** | B1 | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X\leq k) < 1-2\times 0.2084$, $P(X\leq k)<0.5832...$, $k=\mathbf{10}$ | M1 A1cao | M1 for $P(X\leq k)<1-2p$ or $P(X<k+1)<1-2p$; A1 $k=10$ |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $W\sim\text{Po}(5)$ | B1 | B1 for writing/using Po(5) |
| $[P(W=4)]^2 = \left(\frac{e^{-5}5^4}{4!}\right)^2 = (0.4405-0.2650)^2 = 0.030788...$ **awrt 0.0308** | M1 A1 | M1 for $[P(W=4)]^2$ or either correct expression |
### Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P((X_1\geq 10\cap X_2\geq 10)\mid Y=21) = \dfrac{\frac{e^{-10}10^{10}}{10!}\times\frac{e^{-10}10^{11}}{11!}+\frac{e^{-10}10^{11}}{11!}\times\frac{e^{-10}10^{10}}{10!}}{\frac{e^{-20}20^{21}}{21!}}$ | M1 M1 | 1st M1 use of Po(10) with $X=10$ or $X=11$; 2nd M1 correct expression for $2\times P(X=10)\times P(X=11)$ from Po(10) |
| $= 0.336376...$ using tables: $\frac{2\times(0.5830-0.4579)(0.6968-0.5830)}{\frac{e^{-20}20^{21}}{21!}} = 0.336537...$ **awrt 0.336/7** | M1 A1 | 3rd M1 ratio with correct denominator (awrt 0.0846) and num<denom |
### Part (e)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $L\sim\text{Po}(40)\approx\text{N}(40,40)$ | B1 B1 | 1st B1 for Po(40); 2nd B1 for normal with mean and variance 40 |
| $P(L>27) = P\!\left(Z>\frac{27.5-40}{\sqrt{40}}\right)$ | M1 M1 | 1st M1 continuity correction $(27\pm 0.5)$; 2nd M1 standardising with their mean and sd |
| $P(Z>-1.98) = 0.9761$ **awrt 0.976** | A1 | |
---5. A delivery company loses packages randomly at a mean rate of 10 per month.
The probability that the delivery company loses more than 12 packages in a randomly selected month is $p$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $p$
The probability that the delivery company loses more than $k$ packages in a randomly selected month is at least $2 p$
\item Find the largest possible value of $k$
In a randomly selected month,
\item find the probability that exactly 4 packages were lost in each half of the month.
In a randomly selected two-month period, 21 packages were lost.
\item Find the probability that at least 10 packages were lost in each of these two months.
\item Using a suitable approximation, find the probability that more than 27 packages are lost during a randomly selected 4-month period.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q5 [15]}}