| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Normal approximation to summed Poisson |
| Difficulty | Challenging +1.2 Part (a) is routine Poisson probability calculation. Part (b) requires knowing that sum of independent Poissons is Poisson—a standard S2 result. Part (c) requires recognizing that sums of Poissons become normal by CLT, then comparing two normal distributions (difference of normals), which is more sophisticated but still a textbook application of standard results with clear signposting. |
| 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 | Mark | Guidance |
| \(e^{-3.7}\left(\frac{3.7^3}{3!} + \frac{3.7^4}{4!} + \frac{3.7^5}{5!}\right) = e^{-3.7}(8.44217 + 7.80900 + 5.77866) = 0.20872 + 0.19307 + 0.14287\) | M1 | Expression must be seen or implied by correct figures. Any \(\lambda\). Allow one end error. Accept fully correct \(\Sigma\) notation. |
| \(= 0.545\) (3 sf) | A1 | SC 0.545 unsupported scores B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\lambda] = 6.3\) | B1 | |
| \(e^{-6.3}\left(1 + 6.3 + \frac{6.3^2}{2!} + \frac{6.3^3}{3!}\right) = e^{-6.3}(1 + 6.3 + 19.845 + 41.6745) = 0.0018363 + 0.011569 + 0.0364415 + 0.076527\) | M1 | Expression must be seen or implied by correct figures. Any \(\lambda\). Allow one end error. Accept fully correct \(\Sigma\) notation. |
| \(= 0.126\) (3 sf) | A1 | SC 0.126 unsupported scores B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(L \sim N(37, 37)\), \(O \sim N(26, 26)\) | B1 | SOI |
| \((O - L) \sim N(-11, 63)\) | B1 | For \(N(\pm 11, \ldots)\) SOI |
| M1 | For \(\text{var} = 37 + 26\) SOI | |
| \(\frac{0 - {-}11}{\sqrt{63}}\) \([=1.386]\) or \(\frac{0 + 0.5 - {-}11}{\sqrt{63}}\) \([=1.449]\) | M1 | Standardising with their values (wrong cc scores M1) |
| \(1 - \Phi(1.386)\) or \(1 - \Phi(1.449)\) | M1 | For area consistent with their working |
| \(= 0.0828\) or \(0.0829\) (3 sf) or \(= 0.0737\) or \(0.0736\) (3 sf) | A1 | \(SC_1\) 10 used twice \(N(37,37)\), \(N(26,26)\) and use of \(10O - 10L > 0\) Apply MR rules max B1 B1 M1 M1 M1 A0; \(SC_2\) \(P(11)\) giving \(N(11,11)\) scores B0 B1 M0 M1 M1 A0 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-3.7}\left(\frac{3.7^3}{3!} + \frac{3.7^4}{4!} + \frac{3.7^5}{5!}\right) = e^{-3.7}(8.44217 + 7.80900 + 5.77866) = 0.20872 + 0.19307 + 0.14287$ | M1 | Expression must be seen or implied by correct figures. Any $\lambda$. Allow one end error. Accept fully correct $\Sigma$ notation. |
| $= 0.545$ (3 sf) | A1 | SC 0.545 unsupported scores **B1** |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\lambda] = 6.3$ | B1 | |
| $e^{-6.3}\left(1 + 6.3 + \frac{6.3^2}{2!} + \frac{6.3^3}{3!}\right) = e^{-6.3}(1 + 6.3 + 19.845 + 41.6745) = 0.0018363 + 0.011569 + 0.0364415 + 0.076527$ | M1 | Expression must be seen or implied by correct figures. Any $\lambda$. Allow one end error. Accept fully correct $\Sigma$ notation. |
| $= 0.126$ (3 sf) | A1 | SC 0.126 unsupported scores **B1 B1** |
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $L \sim N(37, 37)$, $O \sim N(26, 26)$ | B1 | SOI |
| $(O - L) \sim N(-11, 63)$ | B1 | For $N(\pm 11, \ldots)$ SOI |
| | M1 | For $\text{var} = 37 + 26$ SOI |
| $\frac{0 - {-}11}{\sqrt{63}}$ $[=1.386]$ or $\frac{0 + 0.5 - {-}11}{\sqrt{63}}$ $[=1.449]$ | M1 | Standardising with their values (wrong cc scores M1) |
| $1 - \Phi(1.386)$ or $1 - \Phi(1.449)$ | M1 | For area consistent with their working |
| $= 0.0828$ or $0.0829$ (3 sf) or $= 0.0737$ or $0.0736$ (3 sf) | A1 | $SC_1$ 10 used twice $N(37,37)$, $N(26,26)$ and use of $10O - 10L > 0$ Apply MR rules max **B1 B1 M1 M1 M1 A0**; $SC_2$ $P(11)$ giving $N(11,11)$ scores **B0 B1 M0 M1 M1 A0** |
---5 A machine puts sweets into bags at random. The numbers of lemon and orange sweets in a bag have the independent distributions $\operatorname { Po } ( 3.7 )$ and $\operatorname { Po } ( 2.6 )$ respectively.
A bag of sweets is chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the number of lemon sweets in the bag is more than 2 but not more than 5 .
\item Find the probability that the total number of lemon and orange sweets in the bag is less than 4 .\\
\includegraphics[max width=\textwidth, alt={}, center]{9ac74d4c-f5e0-4c5d-ab25-5692dfb06f0b-06_2725_47_107_2002}\\
\includegraphics[max width=\textwidth, alt={}, center]{9ac74d4c-f5e0-4c5d-ab25-5692dfb06f0b-07_2716_29_107_22}
10 bags of sweets are chosen at random.
\item Use approximating distributions to find the probability that the total number of lemon sweets in the 10 bags is less than the total number of orange sweets in the 10 bags.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q5 [11]}}