| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Consecutive non-overlapping periods |
| Difficulty | Moderate -0.8 This question tests basic recall of Poisson distribution conditions and straightforward application of the distribution with simple parameter scaling. Part (i) requires stating a standard condition (constant mean rate), while parts (ii)-(iv) involve routine calculations using the Poisson formula with adjusted parameters—no problem-solving insight or complex multi-step reasoning required. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Constant mean (or average) rate | B1 [1] | Constant mean per day (or week, etc.) o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{-\frac{4}{7}} \times \frac{(\frac{4}{7})^2}{2!}\) or \(e^{-0.571} \times \frac{0.571^2}{2!}\) | M1 | Expression for P(2); allow any \(\lambda\) |
| \(= 0.0922\) or 0.0921 (3 s.f.) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda = \frac{40}{7}\) or 5.71… | B1 | |
| \(1 - e^{-\frac{40}{7}}\left(1 + \frac{40}{7} + \frac{(\frac{40}{7})^2}{2!} + \frac{(\frac{40}{7})^3}{3!}\right)\) | M1 | Allow any \(\lambda\); allow one end error |
| \(= 0.821\) (3 s.f.) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{24}{7}\) o.e. 3 s.f. or better seen | B1 | |
| \(e^{-\frac{4}{7}} \times e^{-\frac{24}{7}} \times \frac{(\frac{24}{7})^5}{5!}\) | M1 | M1 for \(P(0) \times P(5)\) any consistent \(\lambda\) |
| \(= 0.0723\) (3 s.f.) | A1 [3] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Constant mean (or average) rate | B1 [1] | Constant mean per day (or week, etc.) o.e. |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{-\frac{4}{7}} \times \frac{(\frac{4}{7})^2}{2!}$ or $e^{-0.571} \times \frac{0.571^2}{2!}$ | M1 | Expression for P(2); allow any $\lambda$ |
| $= 0.0922$ or 0.0921 (3 s.f.) | A1 [2] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = \frac{40}{7}$ or 5.71… | B1 | |
| $1 - e^{-\frac{40}{7}}\left(1 + \frac{40}{7} + \frac{(\frac{40}{7})^2}{2!} + \frac{(\frac{40}{7})^3}{3!}\right)$ | M1 | Allow any $\lambda$; allow one end error |
| $= 0.821$ (3 s.f.) | A1 [3] | |
### Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{24}{7}$ o.e. 3 s.f. or better seen | B1 | |
| $e^{-\frac{4}{7}} \times e^{-\frac{24}{7}} \times \frac{(\frac{24}{7})^5}{5!}$ | M1 | M1 for $P(0) \times P(5)$ any consistent $\lambda$ |
| $= 0.0723$ (3 s.f.) | A1 [3] | |
---7 A Lost Property office is open 7 days a week. It may be assumed that items are handed in to the office randomly, singly and independently.\\
(i) State another condition for the number of items handed in to have a Poisson distribution.
It is now given that the number of items handed in per week has the distribution $\operatorname { Po } ( 4.0 )$.\\
(ii) Find the probability that exactly 2 items are handed in on a particular day.\\
(iii) Find the probability that at least 4 items are handed in during a 10-day period.\\
(iv) Find the probability that, during a certain week, 5 items are handed in altogether, but no items are handed in on the first day of the week.
\hfill \mbox{\textit{CAIE S2 2014 Q7 [9]}}