| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Conditional probability with Poisson |
| Difficulty | Standard +0.8 This question requires understanding of Poisson distribution properties (part i), addition of independent Poisson variables (part ii), and conditional probability with Poisson distributions (part iii). Part (iii) is particularly challenging as it requires P(F=0|M+F>3) = P(F=0 and M+F>3)/P(M+F>3), involving complementary probability calculations. While the individual techniques are A-level standard, the multi-step conditional probability reasoning and careful probability manipulation elevate this above average difficulty. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| No of males leaving (to do eng) each year has constant mean, or Males leave (to do eng) independently of other males leaving (to do eng), or Males leave (to do eng) at random | B1 | One of these or any equivalent statement in context |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 3.9\) | B1 | |
| \(1 - e^{-3.9}\left(1 + 3.9 + \frac{3.9^2}{2!} + \frac{3.9^3}{3!}\right)\) | M1 | Any \(\lambda\). Allow one end error or extra term |
| \(0.546753\) or \(0.547\) (3 sf) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(F=0 \text{ and } M>3) = e^{-0.8} \times \left[1 - e^{-3.1}\left(1 + 3.1 + \frac{3.1^2}{2!} + \frac{3.1^3}{3!}\right)\right]\) \((= 0.16857)\) | M1 | Attempt \(P(F=0) \times P(M>3)\), allow one end error for \(P(M>3)\) provided \(\lambda = 3.1\) |
| \(\frac{P(F=0 \text{ and } M>3)}{P(M+F>3)} = \frac{"0.16857"}{" 0.54675"}\) | M1 | Attempted, allow any probability/their (ii) provided answer is \(<1\) |
| \(= 0.308\) (3 sf) | A1 | |
| Total: 3 |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| No of males leaving (to do eng) each year has constant mean, or Males leave (to do eng) independently of other males leaving (to do eng), or Males leave (to do eng) at random | B1 | One of these or any equivalent statement in context |
| **Total: 1** | | |
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 3.9$ | B1 | |
| $1 - e^{-3.9}\left(1 + 3.9 + \frac{3.9^2}{2!} + \frac{3.9^3}{3!}\right)$ | M1 | Any $\lambda$. Allow one end error or extra term |
| $0.546753$ or $0.547$ (3 sf) | A1 | |
| **Total: 3** | | |
## Question 4(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(F=0 \text{ and } M>3) = e^{-0.8} \times \left[1 - e^{-3.1}\left(1 + 3.1 + \frac{3.1^2}{2!} + \frac{3.1^3}{3!}\right)\right]$ $(= 0.16857)$ | M1 | Attempt $P(F=0) \times P(M>3)$, allow one end error for $P(M>3)$ provided $\lambda = 3.1$ |
| $\frac{P(F=0 \text{ and } M>3)}{P(M+F>3)} = \frac{"0.16857"}{" 0.54675"}$ | M1 | Attempted, allow any probability/their (ii) provided answer is $<1$ |
| $= 0.308$ (3 sf) | A1 | |
| **Total: 3** | | |4 The numbers, $M$ and $F$, of male and female students who leave a particular school each year to study engineering have means 3.1 and 0.8 respectively.\\
(i) State, in context, one condition required for $M$ to have a Poisson distribution.\\
Assume that $M$ and $F$ can be modelled by independent Poisson distributions.\\
(ii) Find the probability that the total number of students who leave to study engineering in a particular year is more than 3 .\\
(iii) Given that the total number of students who leave to study engineering in a particular year is more than 3 , find the probability that no female students leave to study engineering in that year.\\
\hfill \mbox{\textit{CAIE S2 2018 Q4 [7]}}