| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Sum of independent Poisson variables |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution properties (sum of independent Poissons) and normal approximation. Part (i) requires basic Poisson probability calculation with λ=6.1, while part (ii) applies the standard normal approximation with continuity correction for λ=61. Both parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02n Sum of Poisson variables: is Poisson5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = 10 \times 0.25 + 10 \times 0.36\ (= 6.1)\) | B1 | |
| \(1 - e^{-6.1}\left(1 + 6.1 + \frac{6.1^2}{2} + \frac{6.1^3}{3!}\right)\) | M1 | \(1 - P(X \leqslant 3)\), any \(\lambda\). Allow one end error |
| \(= 0.857\) | A1 | Allow 0.858 |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = 61\) | B1 ft | Ft from (i) |
| \(N(61', 61')\) | M1 | N with \(\mu = \lambda\), any \(\lambda\). May be implied |
| \(\frac{59.5 - 61}{\sqrt{61'}}\ (= -0.192)\) | M1 | Standardise with their mean and variance. Allow no or wrong cc. not 61/100 |
| \(\Phi(-0.192') = 1 - \Phi(0.192')\) | M1 | Correct area consistent with their working |
| \(= 0.424\) | A1 | |
| Total: 5 |
## Question 4:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 10 \times 0.25 + 10 \times 0.36\ (= 6.1)$ | B1 | |
| $1 - e^{-6.1}\left(1 + 6.1 + \frac{6.1^2}{2} + \frac{6.1^3}{3!}\right)$ | M1 | $1 - P(X \leqslant 3)$, any $\lambda$. Allow one end error |
| $= 0.857$ | A1 | Allow 0.858 |
| **Total: 3** | | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 61$ | B1 ft | Ft from (i) |
| $N(61', 61')$ | M1 | N with $\mu = \lambda$, any $\lambda$. May be implied |
| $\frac{59.5 - 61}{\sqrt{61'}}\ (= -0.192)$ | M1 | Standardise with their mean and variance. Allow no or wrong cc. not 61/100 |
| $\Phi(-0.192') = 1 - \Phi(0.192')$ | M1 | Correct area consistent with their working |
| $= 0.424$ | A1 | |
| **Total: 5** | | |
---4 Small drops of two liquids, $A$ and $B$, are randomly and independently distributed in the air. The average numbers of drops of $A$ and $B$ per cubic centimetre of air are 0.25 and 0.36 respectively.\\
(i) A sample of $10 \mathrm {~cm} ^ { 3 }$ of air is taken at random. Find the probability that the total number of drops of $A$ and $B$ in this sample is at least 4 .\\
(ii) A sample of $100 \mathrm {~cm} ^ { 3 }$ of air is taken at random. Use an approximating distribution to find the probability that the total number of drops of $A$ and $B$ in this sample is less than 60 .\\
\hfill \mbox{\textit{CAIE S2 2018 Q4 [8]}}