| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Rescale rate then sum Poissons |
| Difficulty | Standard +0.3 This is a straightforward application of standard Poisson distribution techniques: part (i) is direct calculation with λ=2.4, part (ii) is routine normal approximation with continuity correction, and part (iii) uses the sum of independent Poisson variables. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| 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 probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{-2.4} \times \frac{2.4^2}{2!}\) | M1 | Allow incorrect \(\lambda\) |
| \(= 0.261\) (3 sf) | A1 | |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(N(60, 60)\) | B1 | Seen or implied |
| \(\frac{54.5 - 60}{\sqrt{60}} = -0.710\) | M1 | Allow with wrong or missing cc |
| \(1 - \phi(\text{``}-0.710\text{''}) = \phi(\text{``}0.710\text{''})\) | M1 | For area consistent with their working |
| \(= 0.761\) (3 sf) | A1 | |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = 3.6 + 12 \div 7 = \frac{186}{35} = 5.314\) | M1 | |
| \(e^{-5.314}\left(1 + 5.314 + \frac{5.314^2}{2} + \frac{5.314^3}{3!}\right)\) | M1 | Allow incorrect \(\lambda\). Allow one end error. |
| \(= 0.224\) (3 sf) | A1 | |
| Total: | 3 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{-2.4} \times \frac{2.4^2}{2!}$ | M1 | Allow incorrect $\lambda$ |
| $= 0.261$ (3 sf) | A1 | |
| **Total:** | **2** | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $N(60, 60)$ | B1 | Seen or implied |
| $\frac{54.5 - 60}{\sqrt{60}} = -0.710$ | M1 | Allow with wrong or missing cc |
| $1 - \phi(\text{``}-0.710\text{''}) = \phi(\text{``}0.710\text{''})$ | M1 | For area consistent with their working |
| $= 0.761$ (3 sf) | A1 | |
| **Total:** | **4** | |
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 3.6 + 12 \div 7 = \frac{186}{35} = 5.314$ | M1 | |
| $e^{-5.314}\left(1 + 5.314 + \frac{5.314^2}{2} + \frac{5.314^3}{3!}\right)$ | M1 | Allow incorrect $\lambda$. Allow one end error. |
| $= 0.224$ (3 sf) | A1 | |
| **Total:** | **3** | |
---6 Old televisions arrive randomly and independently at a recycling centre at an average rate of 1.2 per day.\\
(i) Find the probability that exactly 2 televisions arrive in a 2-day period.\\
(ii) Use an appropriate approximating distribution to find the probability that at least 55 televisions arrive in a 50-day period.\\
Independently of televisions, old computers arrive randomly and independently at the same recycling centre at an average rate of 4 per 7-day week.\\
(iii) Find the probability that the total number of televisions and computers that arrive at the recycling centre in a 3-day period is less than 4.\\
\hfill \mbox{\textit{CAIE S2 2017 Q6 [9]}}