| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding minimum n for P(X≥k) threshold |
| Difficulty | Standard +0.8 Part (i) is routine Poisson probability calculation. Part (ii) requires recognizing that sum of independent Poissons is Poisson with adjusted rate for half-hour period. Part (iii) is more challenging, requiring students to set up inequality P(X≥1)≥0.99, convert to 1-P(0)≥0.99, then solve e^(-1.8t)≤0.01 for t using logarithms—this inverse problem with inequality manipulation goes beyond standard textbook exercises. |
| Spec | 5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - e^{-1.8}(1 + 1.8)\) | M1 | Accept any \(\lambda\). Accept \(1 - P(0,1,2)\) |
| \(= 0.537\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = 2.2\) | B1 | |
| \(e^{-2.2}\left(1 + 2.2 + \frac{2.2^2}{2!} + \frac{2.2^3}{3!} + \frac{2.2^4}{4!}\right)\) | M1 | Attempt expr'n for \(P(X \leqslant 4)\), allow one end error, allow any \(\lambda\) |
| \(= 0.928\) (3 sf) or \(0.927\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - e^{-1.8t} \geqslant 0.99\) or \(1 - e^{-\lambda} \geqslant 0.99\) | M1 | Condone = signs/incorrect inequality signs |
| \(e^{-1.8t} \leqslant 0.01\), \(-1.8t \leqslant \ln 0.01\) or \(e^{-\lambda} \leqslant 0.01\) | M1 | Valid attempt take logs (must have single term on each side) |
| \(t \geqslant 2.56\), She must watch for at least 2.56 (hours) | A1 | or 2 hours, 34 mins or better. No errors seen |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - e^{-1.8}(1 + 1.8)$ | M1 | Accept any $\lambda$. Accept $1 - P(0,1,2)$ |
| $= 0.537$ (3 sf) | A1 | |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 2.2$ | B1 | |
| $e^{-2.2}\left(1 + 2.2 + \frac{2.2^2}{2!} + \frac{2.2^3}{3!} + \frac{2.2^4}{4!}\right)$ | M1 | Attempt expr'n for $P(X \leqslant 4)$, allow one end error, allow any $\lambda$ |
| $= 0.928$ (3 sf) or $0.927$ | A1 | |
---
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - e^{-1.8t} \geqslant 0.99$ or $1 - e^{-\lambda} \geqslant 0.99$ | M1 | Condone = signs/incorrect inequality signs |
| $e^{-1.8t} \leqslant 0.01$, $-1.8t \leqslant \ln 0.01$ or $e^{-\lambda} \leqslant 0.01$ | M1 | Valid attempt take logs (must have single term on each side) |
| $t \geqslant 2.56$, She must watch for at least 2.56 (hours) | A1 | or 2 hours, 34 mins or better. No errors seen |
---5 The number of eagles seen per hour in a certain location has the distribution $\operatorname { Po } ( 1.8 )$. The number of vultures seen per hour in the same location has the independent distribution $\operatorname { Po } ( 2.6 )$.\\
(i) Find the probability that, in a randomly chosen hour, at least 2 eagles are seen.\\
(ii) Find the probability that, in a randomly chosen half-hour period, the total number of eagles and vultures seen is less than 5 .\\
Alex wants to be at least $99 \%$ certain of seeing at least 1 eagle.\\
(iii) Find the minimum time for which she should watch for eagles.\\
\hfill \mbox{\textit{CAIE S2 2019 Q5 [8]}}