| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | March |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Hypothesis test on Poisson rate |
| Difficulty | Standard +0.8 This question requires understanding of Poisson distribution properties including scaling parameters over multiple days, summing independent Poisson variables, conditional probability with truncation, and hypothesis testing. Part (c) requires systematic enumeration of cases, and part (d) involves a one-tailed test with proper critical region determination. While the individual techniques are A-level standard, the combination and the need to work with sums of independent Poisson distributions makes this moderately challenging. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c 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 probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| \([e^{-0.2}] = 0.819\) (3 sf) | B1 | Accept \(e^{-0.2}\) as final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = 1.25\) | B1 | |
| \(e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2}\right)\) or \(e^{-1.25}(1 + 1.25 + 0.78125)\) or \(0.2865 + 0.3581 + 0.2238\) | M1 | Any \(\lambda\). Allow one end error. Must see expression (in any form). Accept correct \(\Sigma\) notation. |
| \(= 0.868\) (3 sf) | A1 | SC Answer with no working seen scores B1 (could be implied) |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-0.15} \times e^{-0.1}(0.1) = 0.077879\) | M1 | \(P(B=0)\times P(G=1) = 0.8607 \times 0.09048\) |
| \(e^{-0.15} \times e^{-0.1}\left(\frac{0.1^2}{2}\right) = 0.003894\) | \(P(B=0)\times P(G=2) = 0.8607 \times 0.004524\) | |
| \(e^{-0.15} \times 0.15 \times e^{-0.1} \times \frac{0.1^2}{2} = 0.0005841\) | \(P(B=1)\times P(G=2) = 0.1291 \times 0.004524\) | |
| \(e^{-0.15} \times e^{-0.1}(0.1) + e^{-0.15} \times e^{-0.1}\left(\frac{0.1^2}{2}\right) + e^{-0.15} \times 0.15 \times e^{-0.1} \times \frac{0.1^2}{2}\) \(= 0.077879 + 0.00389036 + 0.0005841\) | M1 | \(P(B=0)\times P(G=1) + P(B=0)\times P(G=2) + P(B=1)\times P(G=2)\). Three Poisson terms added (must be from complete attempt at all 3 terms). |
| \(= 0.0824\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B=0)\times P(G>0) = e^{-0.15} \times (1 - e^{-0.1})\) | M1 | For one expression seen |
| \(P(B=1)\times P(G>1) = e^{-0.15} \times 0.15 \times (1 - e^{-0.1}(1+0.1))\) | ||
| \(e^{-0.15}\times(1 - e^{-0.1}) + e^{-0.15}\times 0.15 \times (1 - e^{-0.1}(1+0.1))\) | M1 | For adding their expressions |
| \(= 0.0824\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \lambda = 1.25\) or \(0.25\) [per day]; \(H_1: \lambda > 1.25\) or \(0.25\) [per day] | B1 | Or \(\mu\) or 'population mean' |
| \(P(\geq 4\ \text{late}) = 1 - e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2} + \frac{1.25^3}{3!}\right)\) or \(1 - e^{-1.25}(1 + 1.25 + 0.7813 + 0.3255)\) or \(1 - (0.2865 + 0.3581 + 0.2238 + 0.09326)\) | M1 | Any \(\lambda\). No end errors. Expression must be seen (in any form). Accept correct \(\Sigma\) notation. |
| \(= 0.0383\) | A1 | SC 0.0383 with no working scores B1 |
| \(0.0383 < 0.05\) | M1 | For a valid comparison |
| [Reject \(H_0\)] 'Hence there is sufficient evidence to suggest that the teacher's claim is true' or 'There is sufficient evidence to suggest that more students are late on average'. | A1 FT | No contradictions. In context and not definite, e.g. not 'More students are late' or 'Claim is correct'. Ft their 0.0383. |
## Question 5(a):
| $[e^{-0.2}] = 0.819$ (3 sf) | B1 | Accept $e^{-0.2}$ as final answer |
---
## Question 5(b):
| $\lambda = 1.25$ | B1 | |
|---|---|---|
| $e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2}\right)$ or $e^{-1.25}(1 + 1.25 + 0.78125)$ or $0.2865 + 0.3581 + 0.2238$ | M1 | Any $\lambda$. Allow one end error. Must see expression (in any form). Accept correct $\Sigma$ notation. |
| $= 0.868$ (3 sf) | A1 | SC Answer with no working seen scores B1 (could be implied) |
---
## Question 5(c):
| $e^{-0.15} \times e^{-0.1}(0.1) = 0.077879$ | M1 | $P(B=0)\times P(G=1) = 0.8607 \times 0.09048$ |
|---|---|---|
| $e^{-0.15} \times e^{-0.1}\left(\frac{0.1^2}{2}\right) = 0.003894$ | | $P(B=0)\times P(G=2) = 0.8607 \times 0.004524$ |
| $e^{-0.15} \times 0.15 \times e^{-0.1} \times \frac{0.1^2}{2} = 0.0005841$ | | $P(B=1)\times P(G=2) = 0.1291 \times 0.004524$ |
| $e^{-0.15} \times e^{-0.1}(0.1) + e^{-0.15} \times e^{-0.1}\left(\frac{0.1^2}{2}\right) + e^{-0.15} \times 0.15 \times e^{-0.1} \times \frac{0.1^2}{2}$ $= 0.077879 + 0.00389036 + 0.0005841$ | M1 | $P(B=0)\times P(G=1) + P(B=0)\times P(G=2) + P(B=1)\times P(G=2)$. Three Poisson terms added (must be from complete attempt at all 3 terms). |
| $= 0.0824$ (3 sf) | A1 | |
**Alternative method:**
| $P(B=0)\times P(G>0) = e^{-0.15} \times (1 - e^{-0.1})$ | M1 | For one expression seen |
|---|---|---|
| $P(B=1)\times P(G>1) = e^{-0.15} \times 0.15 \times (1 - e^{-0.1}(1+0.1))$ | | |
| $e^{-0.15}\times(1 - e^{-0.1}) + e^{-0.15}\times 0.15 \times (1 - e^{-0.1}(1+0.1))$ | M1 | For adding their expressions |
| $= 0.0824$ (3 sf) | A1 | |
---
## Question 5(d):
| $H_0: \lambda = 1.25$ or $0.25$ [per day]; $H_1: \lambda > 1.25$ or $0.25$ [per day] | B1 | Or $\mu$ or 'population mean' |
|---|---|---|
| $P(\geq 4\ \text{late}) = 1 - e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2} + \frac{1.25^3}{3!}\right)$ or $1 - e^{-1.25}(1 + 1.25 + 0.7813 + 0.3255)$ or $1 - (0.2865 + 0.3581 + 0.2238 + 0.09326)$ | M1 | Any $\lambda$. No end errors. Expression must be seen (in any form). Accept correct $\Sigma$ notation. |
| $= 0.0383$ | A1 | SC 0.0383 with no working scores B1 |
| $0.0383 < 0.05$ | M1 | For a valid comparison |
| [Reject $H_0$] 'Hence there is sufficient evidence to suggest that the teacher's claim is true' or 'There is sufficient evidence to suggest that more students are late on average'. | A1 FT | No contradictions. In context and not definite, e.g. not 'More students are late' or 'Claim is correct'. Ft their 0.0383. |
---5 A teacher models the numbers of girls and boys who arrive late for her class on any day by the independent random variables $G \sim \operatorname { Po } ( 0.10 )$ and $B \sim \operatorname { Po } ( 0.15 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that during a randomly chosen 2-day period no girls arrive late.
\item Find the probability that during a randomly chosen 5-day period the total number of students who arrive late is less than 3 .
\item It is given that the values of $\mathrm { P } ( G = r )$ and $\mathrm { P } ( B = r )$ for $r \geqslant 3$ are very small and can be ignored. Find the probability that on a randomly chosen day more girls arrive late than boys.\\
Following a timetable change the teacher claims that on average more students arrive late than before the change. During a randomly chosen 5-day period a total of 4 students are late.
\item Test the teacher's claim at the $5 \%$ significance level.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q5 [12]}}