| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Three or more independent Poisson sums |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson approximation to binomial with standard parameters (n large, p small, np moderate). All three parts follow directly from calculating λ = np and applying basic Poisson probability formulas or normal approximation in part (iii). No conceptual challenges beyond recognizing when to use which approximation. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(\text{Po}(2.8)\) | M1 | May be implied |
| \(1 - e^{-2.8}\left(1 + 2.8 + \frac{2.8^2}{2}\right)\) | M1 | Any \(\lambda\) allowing one end error |
| \(= 0.531\) or \(0.53(0)\) (3 sf) | A1 | SC Binomial 0.534 B1 |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(\text{Po}(5.8)\) | M1 | May be implied |
| \(e^{-5.8} \times \frac{5.8^6}{6!}\) | M1 | Any \(\lambda\) |
| \(= 0.16(0)\) (3 sf) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(N(58, 58)\) | M1 | May be implied or \(N(58, 55.38)\) |
| \(\frac{50.5 - \text{"58"}}{\sqrt{\text{"58"}}} \ (= -0.985)\) | M1 | Standardised with their values, allow wrong or incorrect cc |
| \(\Phi(\text{"0.985"})\) | M1 | Correct area consistent with their working; or \(\Phi(\text{"1.008"})\) |
| \(= 0.838\) (3 sf) | A1 | or 0.843 |
| Total: 4 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $\text{Po}(2.8)$ | M1 | May be implied |
| $1 - e^{-2.8}\left(1 + 2.8 + \frac{2.8^2}{2}\right)$ | M1 | Any $\lambda$ allowing one end error |
| $= 0.531$ or $0.53(0)$ (3 sf) | A1 | SC Binomial 0.534 B1 |
| **Total: 3** | | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $\text{Po}(5.8)$ | M1 | May be implied |
| $e^{-5.8} \times \frac{5.8^6}{6!}$ | M1 | Any $\lambda$ |
| $= 0.16(0)$ (3 sf) | A1 | |
| **Total: 3** | | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $N(58, 58)$ | M1 | May be implied or $N(58, 55.38)$ |
| $\frac{50.5 - \text{"58"}}{\sqrt{\text{"58"}}} \ (= -0.985)$ | M1 | Standardised with their values, allow wrong or incorrect cc |
| $\Phi(\text{"0.985"})$ | M1 | Correct area consistent with their working; or $\Phi(\text{"1.008"})$ |
| $= 0.838$ (3 sf) | A1 | or 0.843 |
| **Total: 4** | | |7 Each day at a certain doctor's surgery there are 70 appointments available in the morning and 60 in the afternoon. All the appointments are filled every day. The probability that any patient misses a particular morning appointment is 0.04 , and the probability that any patient misses a particular afternoon appointment is 0.05 . All missed appointments are independent of each other.
Use suitable approximating distributions to answer the following.\\
(i) Find the probability that on a randomly chosen morning there are at least 3 missed appointments.\\
(ii) Find the probability that on a randomly chosen day there are a total of exactly 6 missed appointments.\\
(iii) Find the probability that in a randomly chosen 10-day period there are more than 50 missed appointments.\\
\hfill \mbox{\textit{CAIE S2 2019 Q7 [10]}}