| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Joint probability of independent events |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard techniques: part (i)(a) uses independence of two Poisson processes with complementary probability, part (i)(b) combines processes into a single Poisson, and part (ii) applies normal approximation. All are routine S2 procedures requiring no novel insight, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) East \(P(\geq1) = 1 - e^{-2} = 0.8647\) | B1 | Correct mean of 2 or 1.25 used in Poisson expression |
| West \(P(\geq1) = 1 - e^{-1.25} = 0.7135\) | M1 | One Poisson expression \(P(\geq1) = 1 - P(0)\) or 1 - \(P(0) - P(1)\) any mean. |
| \(P(\text{Both}) = 0.8647 \times 0.7135 = 0.617\) | M1 | For multiplying their 2 probs together |
| A1 (4) | For correct answer | |
| (b) \(P(\text{total} \geq 2) = 1 - e^{-13/4}(1+13/4)\) | M1 | For attempt at summing their means and for 1 - their \(P(0, 1)\) or 1 - their \(P(0,1,2)\) or 1 - \(P(0E,0W) - P(1E,0W) - P(0E,1W)\) or equiv. expression incl.2 |
| \(= 0.835\) | A1 (2) | For correct answer |
| or \(P(\text{total} \geq 2) = P(2) + P(3) +...P(13)\) etc. | M1 | |
| (ii) \(T \sim N(156, 156)\) | B1 | For correct mean and variance |
| \(P(>175) = 1-\Phi\left(\frac{175.5-156}{\sqrt{156}}\right)\) | M1 | For standardising, with or without cc or sq rt |
| \(= 1 - \Phi(1.5612)\) | ||
| \(= 1 - 0.9407\) | ||
| \(= 0.0593/0.0592\) | A1 (3) | For correct answer |
**(i) (a)** East $P(\geq1) = 1 - e^{-2} = 0.8647$ | B1 | Correct mean of 2 or 1.25 used in Poisson expression
West $P(\geq1) = 1 - e^{-1.25} = 0.7135$ | M1 | One Poisson expression $P(\geq1) = 1 - P(0)$ or 1 - $P(0) - P(1)$ any mean.
$P(\text{Both}) = 0.8647 \times 0.7135 = 0.617$ | M1 | For multiplying their 2 probs together
| A1 (4) | For correct answer
**(b)** $P(\text{total} \geq 2) = 1 - e^{-13/4}(1+13/4)$ | M1 | For attempt at summing their means and for 1 - their $P(0, 1)$ or 1 - their $P(0,1,2)$ or 1 - $P(0E,0W) - P(1E,0W) - P(0E,1W)$ or equiv. expression incl.2
$= 0.835$ | A1 (2) | For correct answer
or $P(\text{total} \geq 2) = P(2) + P(3) +...P(13)$ etc. | M1 |
**(ii)** $T \sim N(156, 156)$ | B1 | For correct mean and variance
$P(>175) = 1-\Phi\left(\frac{175.5-156}{\sqrt{156}}\right)$ | M1 | For standardising, with or without cc or sq rt
$= 1 - \Phi(1.5612)$ | |
$= 1 - 0.9407$ | |
$= 0.0593/0.0592$ | A1 (3) | For correct answer6 At a petrol station cars arrive independently and at random times at constant average rates of 8 cars per hour travelling east and 5 cars per hour travelling west.\\
(i) Find the probability that, in a quarter-hour period,
\begin{enumerate}[label=(\alph*)]
\item one or more cars travelling east and one or more cars travelling west will arrive,
\item a total of 2 or more cars will arrive.\\
(ii) Find the approximate probability that, in a 12 -hour period, a total of more than 175 cars will arrive.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2005 Q6 [9]}}