| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Two independent Poisson sums |
| Difficulty | Moderate -0.3 This is a straightforward application of standard Poisson distribution properties with routine calculations. Part (a) requires recall of conditions, parts (b) and (d) use basic Poisson probability formulas, and part (c) applies the standard normal approximation technique. All steps are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| 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 probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Books received independently or singly or randomly | B1 | OE. Must be in context. If more than one condition given, ignore extras |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{-15.3} \times \frac{15.3^{10}}{10!}\) | M1 | Allow incorrect \(\lambda\) |
| \(= 0.0439\) (3sf) | A1 | SC No working shown but correct answer seen scores B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(N(153, 153)\) | B1 | Seen or implied |
| \(\frac{180.5 - 153}{\sqrt{153}}\) \([= 2.223]\) | M1 | For standardising with their values (can be implied). Allow with wrong or missing continuity correction |
| \(1 - \Phi(\text{'2.223'})\) | M1 | For correct probability area consistent with their values |
| \(= 0.0131\) (3sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\lambda =) 5.1 + 2.5\) \([= 7.6]\) | B1 | Give at early stage (seen or implied) |
| \(1 - e^{-7.6}(1 + 7.6 + \frac{7.6^2}{2} + \frac{7.6^3}{3!}) = 1 - e^{-7.6}(1+7.6+28.88+73.16)\) \(= 1-(0.0005005+0.003803+0.01445+0.03661)\) | M1 | Allow incorrect \(\lambda\). Allow one end error. Must see an expression (accept correct sigma notation) |
| \(= 0.945\) (3sf) | A1 | SC No working, 0.945 B1 (could be implied) SC B1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Books received independently or singly or randomly | B1 | OE. Must be in context. If more than one condition given, ignore extras |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{-15.3} \times \frac{15.3^{10}}{10!}$ | M1 | Allow incorrect $\lambda$ |
| $= 0.0439$ (3sf) | A1 | SC No working shown but correct answer seen scores B1 |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $N(153, 153)$ | B1 | Seen or implied |
| $\frac{180.5 - 153}{\sqrt{153}}$ $[= 2.223]$ | M1 | For standardising with their values (can be implied). Allow with wrong or missing continuity correction |
| $1 - \Phi(\text{'2.223'})$ | M1 | For correct probability area consistent with their values |
| $= 0.0131$ (3sf) | A1 | |
---
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\lambda =) 5.1 + 2.5$ $[= 7.6]$ | B1 | Give at early stage (seen or implied) |
| $1 - e^{-7.6}(1 + 7.6 + \frac{7.6^2}{2} + \frac{7.6^3}{3!}) = 1 - e^{-7.6}(1+7.6+28.88+73.16)$ $= 1-(0.0005005+0.003803+0.01445+0.03661)$ | M1 | Allow incorrect $\lambda$. Allow one end error. Must see an expression (accept correct sigma notation) |
| $= 0.945$ (3sf) | A1 | SC No working, 0.945 B1 (could be implied) SC B1 |
---4 The number, $X$, of books received at a charity shop has a constant mean of 5.1 per day.
\begin{enumerate}[label=(\alph*)]
\item State, in context, one condition for $X$ to be modelled by a Poisson distribution.\\
Assume now that $X$ can be modelled by a Poisson distribution.
\item Find the probability that exactly 10 books are received in a 3-day period.
\item Use a suitable approximating distribution to find the probability that more than 180 books are received in a 30-day period.\\
The number of DVDs received at the same shop is modelled by an independent Poisson distribution with mean 2.5 per day.
\item Find the probability that the total number of books and DVDs that are received at the shop in 1 day is more than 3 .
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q4 [10]}}