| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard bookwork. Part (i) requires recalling a condition, parts (ii)-(iii) are routine probability calculations with given rates, and part (iv) involves a standard normal approximation—all typical S2 exercises requiring minimal problem-solving beyond following learned procedures. |
| 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 probabilities5.02l Poisson conditions: for modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Accidents occur independently or randomly | B1 | In context. Allow 'singly' |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-2.5} \times \frac{2.5^4}{4!}\) | M1 | Poisson P(4), allow any \(\lambda\) |
| \(= 0.134\) (3 sfs) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = \frac{25}{12}\) or \(2.08(333)\) | B1 | |
| \(1 - e^{-\frac{25}{12}}\left(1 + \frac{25}{12} + \frac{(25/12)^2}{2!} + \frac{(25/12)^3}{3!}\right)\) | M1 | \(1 -\) Poisson P(0,1,2,3), allow any \(\lambda\), allow one end error |
| \(= 0.158\) (3 sfs) | A1 | As final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(N\!\left(\frac{1825}{84}, \frac{1825}{84}\right)\) or \(N(21.7(26), 21.7(26))\) | B1 | Stated or implied |
| \(\frac{29.5 - \frac{1825}{84}}{\sqrt{\frac{1825}{84}}}\) | M1 | Allow with wrong or no cc with their mean/sd |
| \(\Phi(``1.668")\) | M1 | Correct area consistent with their working |
| \(= 0.952\) (3 sfs) | A1 |
## Question 6(i):
| Accidents occur independently or randomly | B1 | In context. Allow 'singly' |
---
## Question 6(ii):
| $e^{-2.5} \times \frac{2.5^4}{4!}$ | M1 | Poisson P(4), allow any $\lambda$ |
| $= 0.134$ (3 sfs) | A1 | |
---
## Question 6(iii):
| $\lambda = \frac{25}{12}$ or $2.08(333)$ | B1 | |
| $1 - e^{-\frac{25}{12}}\left(1 + \frac{25}{12} + \frac{(25/12)^2}{2!} + \frac{(25/12)^3}{3!}\right)$ | M1 | $1 -$ Poisson P(0,1,2,3), allow any $\lambda$, allow one end error |
| $= 0.158$ (3 sfs) | A1 | As final answer |
---
## Question 6(iv):
| $N\!\left(\frac{1825}{84}, \frac{1825}{84}\right)$ or $N(21.7(26), 21.7(26))$ | B1 | Stated or implied |
| $\frac{29.5 - \frac{1825}{84}}{\sqrt{\frac{1825}{84}}}$ | M1 | Allow with wrong or no cc with their mean/sd |
| $\Phi(``1.668")$ | M1 | Correct area consistent with their working |
| $= 0.952$ (3 sfs) | A1 | |
---6 Accidents on a particular road occur at a constant average rate of 1 every 4.8 weeks.\\
(i) State, in context, one condition for the number of accidents in a given period to be modelled by a Poisson distribution.\\
Assume now that a Poisson distribution is a suitable model.\\
(ii) Find the probability that exactly 4 accidents will occur during a randomly chosen 12-week period.\\
(iii) Find the probability that more than 3 accidents will occur during a randomly chosen 10 -week period.\\
(iv) Use a suitable approximating distribution to find the probability that fewer than 30 accidents will occur during a randomly chosen 2 -year period ( $104 \frac { 2 } { 7 }$ weeks).\\
\hfill \mbox{\textit{CAIE S2 2018 Q6 [10]}}