| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Sample mean distribution of Poisson |
| Difficulty | Standard +0.8 This question requires knowledge that the sum of independent Poisson distributions is Poisson, conditional probability with discrete distributions, and applying the Central Limit Theorem to approximate the sample mean distribution. Part (ii) involves non-trivial conditional probability calculation, and part (iii) requires recognizing when to use normal approximation—these go beyond routine application and require synthesis of multiple statistical concepts. |
| Spec | 5.02n Sum of Poisson variables: is Poisson5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-5.6} \times \frac{5.6^3}{3!}\) | M1 | Allow any \(\lambda\) |
| \(= 0.108\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=2\ \&\ Y=1) = e^{-2.1} \times \frac{2.1^2}{2} \times e^{-3.5} \times 3.5\) | M1 | |
| \((0.2700 \times 0.10569 = 0.028538)\) | ||
| \(\frac{P(X=2\ \&\ Y=1)}{P(X+Y=3)}\) attempted \(= \frac{0.028538}{0.108234}\) | M1 | For attempt at fraction with their (i) as denominator or \(\frac{2.1^2}{2} \times 3.5 \div \frac{5.6^3}{3}\) M2 |
| \(= 0.264\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(X) = 2.1\) | B1 | soi |
| \(\bar{X} \sim N\left(2.1, \frac{2.1}{100}\right)\) or \(N(210, 210)\) | B1 | soi B1 for \(N(2.1, ...)\) |
| B1 | B1 for \(\frac{2.1}{100}\) oe. Standardise with their values. Allow with or without cc or with incorrect cc | |
| \(\dfrac{2.2-2.1}{\sqrt{\frac{2.1}{100}}}\) oe \((220-210)/\sqrt{210}\) \((= 0.690)\) | M1 | or \(\dfrac{2.2+0.5\pm100-2.1}{\sqrt{\frac{2.1}{100}}}\) or \((220.5-210)/\sqrt{210}\) \((= 0.725)\); no mixed methods |
| \(1 - \phi(\text{`}0.690\text{'})\) | M1 | Correct area consistent with their working or \(1 - \phi(\text{`}0.725\text{'})\) |
| \(= 0.245\) (3 sf) | A1 | \(= 0.234\) (3 sf) |
| 6 |
## Question 7(i):
| $e^{-5.6} \times \frac{5.6^3}{3!}$ | M1 | Allow any $\lambda$ |
|---|---|---|
| $= 0.108$ (3 sf) | A1 | |
---
## Question 7(ii):
| $P(X=2\ \&\ Y=1) = e^{-2.1} \times \frac{2.1^2}{2} \times e^{-3.5} \times 3.5$ | M1 | |
|---|---|---|
| $(0.2700 \times 0.10569 = 0.028538)$ | | |
| $\frac{P(X=2\ \&\ Y=1)}{P(X+Y=3)}$ attempted $= \frac{0.028538}{0.108234}$ | M1 | For attempt at fraction with their (i) as denominator or $\frac{2.1^2}{2} \times 3.5 \div \frac{5.6^3}{3}$ M2 |
| $= 0.264$ (3 sf) | A1 | |
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(X) = 2.1$ | **B1** | soi |
| $\bar{X} \sim N\left(2.1, \frac{2.1}{100}\right)$ or $N(210, 210)$ | **B1** | soi B1 for $N(2.1, ...)$ |
| | **B1** | B1 for $\frac{2.1}{100}$ oe. Standardise with their values. Allow with or without cc or with incorrect cc |
| $\dfrac{2.2-2.1}{\sqrt{\frac{2.1}{100}}}$ oe $(220-210)/\sqrt{210}$ $(= 0.690)$ | **M1** | or $\dfrac{2.2+0.5\pm100-2.1}{\sqrt{\frac{2.1}{100}}}$ or $(220.5-210)/\sqrt{210}$ $(= 0.725)$; no mixed methods |
| $1 - \phi(\text{`}0.690\text{'})$ | **M1** | Correct area consistent with their working or $1 - \phi(\text{`}0.725\text{'})$ |
| $= 0.245$ (3 sf) | **A1** | $= 0.234$ (3 sf) |
| | **6** | |7 The independent random variables $X$ and $Y$ have the distributions $\operatorname { Po } ( 2.1 )$ and $\operatorname { Po } ( 3.5 )$ respectively.\\
(i) Find $\mathrm { P } ( X + Y = 3 )$.\\
(ii) Given that $X + Y = 3$, find $\mathrm { P } ( X = 2 )$.\\
(iii) A random sample of 100 values of $X$ is taken. Find the probability that the sample mean is more than 2.2.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S2 2018 Q7 [11]}}