| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Joint probability of separate processes |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with clearly stated rates and standard probability calculations. Parts (a)-(c) are direct single-distribution lookups, (d) requires summing independent Poisson events, and (e) uses complement rule with independence. All techniques are standard S2 material with no novel problem-solving required, making it slightly easier than average. |
| 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 |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | 0.703 | B1 |
| (b) | \(e^{-2.3} + e^{-2.3} \times 2.3 = 0.100... + 0.231...) = 0.331\) | M1 A1 |
| \(1 - (e^{-2.3} + e^{-2.3} \times 2.3) = 0.669\) | A1 | |
| Total: 3 | ||
| (c) | Use of Po(6) | M1 |
| 0.9799 for top value; -0.8472 for bottom value | A1 | A1 for either of these values or this expression OE |
| Or \(e^{-6}\left(\frac{6^0}{9!} + \frac{6^{10}}{10!} + \frac{6^{11}}{11!}\right) = 0.1327 = 0.133\) | A1 | AWRT 0.133 |
| Total: 3 | ||
| (d) | Use of Po(0.8) | M1 |
| \(e^{-0.8} \times 0.8^2/2\) or 0.9526 – 0.8088 = 0.1438 = 0.144 | m1 A1 | AWRT 0.144 |
| Total: 3 | ||
| (e) | \((1 - 0.0111) \times (1 - e^{-2.3}) \times (1 - 0.3012)\) | B1 |
| M1 | ||
| \(0.9889 \times 0.8997 \times 0.6988 = 0.622\) | A1 | AWRT |
| Total: 3 | ||
| TOTAL FOR Q1: 13 |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | 0.703 | B1 | AWRT |
| (b) | $e^{-2.3} + e^{-2.3} \times 2.3 = 0.100... + 0.231...) = 0.331$ | M1 A1 | PI for either P(X=0) or P(X=1); For both P(X=0) and P(X=1) or sum AWRT 0.669 |
| | $1 - (e^{-2.3} + e^{-2.3} \times 2.3) = 0.669$ | A1 | |
| | **Total: 3** | | |
| (c) | Use of Po(6) | M1 | 0.7440, 0.8472, 0.9161, 0.9799, 0.9912 or 0.9964 seen; AWFW 0.9798 to 0.9800; AWFW 0.8470 to 0.8473 |
| | 0.9799 for top value; -0.8472 for bottom value | A1 | A1 for either of these values or this expression OE |
| | Or $e^{-6}\left(\frac{6^0}{9!} + \frac{6^{10}}{10!} + \frac{6^{11}}{11!}\right) = 0.1327 = 0.133$ | A1 | AWRT 0.133 |
| | **Total: 3** | | |
| (d) | Use of Po(0.8) | M1 | Correct formula or 0.9526 or 0.8088 seen |
| | $e^{-0.8} \times 0.8^2/2$ or 0.9526 – 0.8088 = 0.1438 = 0.144 | m1 A1 | AWRT 0.144 |
| | **Total: 3** | | |
| (e) | $(1 - 0.0111) \times (1 - e^{-2.3}) \times (1 - 0.3012)$ | B1 | For any one of these seen or PI by the correct value seen (3 sf or better); For all three correct and multiplied PI |
| | | M1 | |
| | $0.9889 \times 0.8997 \times 0.6988 = 0.622$ | A1 | AWRT |
| | **Total: 3** | | |
| | **TOTAL FOR Q1: 13** | | |
---The water in a pond contains three different species of a spherical green algae:
Volvox globator, at an average rate of 4.5 spheres per 1 cm³;
Volvox aureus, at an average rate of 2.3 spheres per 1 cm³;
Volvox tertius, at an average rate of 1.2 spheres per 1 cm³.
Individual Volvox spheres may be considered to occur randomly and independently of all other Volvox spheres.
Random samples of water are collected from this pond.
Find the probability that:
\begin{enumerate}[label=(\alph*)]
\item a 1 cm³ sample contains no more than 5 Volvox globator spheres; [1 mark]
\item a 1 cm³ sample contains at least 2 Volvox aureus spheres; [3 marks]
\item a 5 cm³ sample contains more than 8 but fewer than 12 Volvox tertius spheres; [3 marks]
\item a 0.1 cm³ sample contains a total of exactly 2 Volvox spheres; [3 marks]
\item a 1 cm³ sample contains at least 1 sphere of each of the three different species of algae. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2016 Q1 [13]}}