| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | State Poisson approximation with justification |
| Difficulty | Moderate -0.8 This is a straightforward application of the Poisson approximation to the binomial with standard conditions (n large, p small, np moderate). Part (a) requires calculating p, finding λ=np, and evaluating P(X<4) using tables. Part (b) is routine recall of the justification criteria (n≥50, p≤0.1, np<5). No problem-solving or novel insight required, just direct application of a standard technique. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(20540/6012300 = 0.0034163\) | B1 | |
| \(1000 \times 0.0034163 = 3.4163\) | ||
| \(\text{Po}(3.4163)\) | B1 | Could be implied by expression seen. |
| \(e^{-\lambda \cdot 3.4163}(1 + 3.4163 + \frac{3.4163^2}{2!} + \frac{3.4163^3}{3!})\) OR \(e^{-\lambda \cdot 3.4163}(1 + 3.4163 + 5.8356 + 6.6453)\) or \(0.03283 + 0.1122 + 0.1916 + 0.21819\) | M1 | Allow any \(\lambda\). Allow with one end error. Must see expression. |
| \(= 0.555\) (3sf) | A1 | CAO. SC No working: B1 B1 (Po must be stated) B1 correct answer (max 3/4). SC Binomial: B1 B0 B1 correct answer (max 2/4). |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(n = 1000 > 50\) | B1 | Must show comparison with 50. |
| \(np = 3.4163 < 5\) | B1 | Must show comparison with 5. |
| 2 | SC B1: \(n > 50\) (or n large), \(np < 5\). SC B1: n large, p small. |
## Question 1(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $20540/6012300 = 0.0034163$ | B1 | |
| $1000 \times 0.0034163 = 3.4163$ | | |
| $\text{Po}(3.4163)$ | B1 | Could be implied by expression seen. |
| $e^{-\lambda \cdot 3.4163}(1 + 3.4163 + \frac{3.4163^2}{2!} + \frac{3.4163^3}{3!})$ OR $e^{-\lambda \cdot 3.4163}(1 + 3.4163 + 5.8356 + 6.6453)$ or $0.03283 + 0.1122 + 0.1916 + 0.21819$ | M1 | Allow any $\lambda$. Allow with one end error. Must see expression. |
| $= 0.555$ (3sf) | A1 | CAO. **SC** No working: B1 B1 (Po must be stated) B1 correct answer (max 3/4). **SC** Binomial: B1 B0 B1 correct answer (max 2/4). |
| | **4** | |
## Question 1(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n = 1000 > 50$ | B1 | Must show comparison with 50. |
| $np = 3.4163 < 5$ | B1 | Must show comparison with 5. |
| | **2** | **SC B1**: $n > 50$ (or n large), $np < 5$. **SC B1**: n large, p small. |1 In a certain country, 20540 adults out of a population of 6012300 have a degree in medicine.\\[0pt]
\begin{enumerate}[label=(\alph*)]
\item Use an approximating distribution to calculate the probability that, in a random sample of 1000 adults in this country, there will be fewer than 4 adults who have a degree in medicine. [4]
\item Justify the approximating distribution used in part (a).
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q1 [6]}}