| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Symmetry property of PDF |
| Difficulty | Standard +0.3 This question tests recognition of PDF symmetry properties to deduce E(X) and uses basic probability axioms. Part (a) requires spotting that the PDF is symmetric about x=1.5 (making E(X)=1.5 without integration), part (b) is trivial probability reasoning (P(3≤X≤4)=0 since support is [0,3]), and part (c) uses complement rule with given probability. All parts are conceptually straightforward with minimal calculation, slightly easier than a typical S2 question. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = 1.5\) | B1 | |
| \(\frac{2}{9}\int_0^3 (3x^3 - x^4)\,dx\) | M1 | Attempt integration of \(x^2f(x)\), ignore limits |
| \(= \frac{2}{9}\left[\frac{3x^4}{4} - \frac{x^5}{5}\right]_0^3\) | M1 | Substitute correct limits into correct integral |
| \(= \frac{2}{9}\left[\frac{243}{4} - \frac{243}{5}\right] = 2.7\) | ||
| \(\text{Var}(X) = 2.7 - 1.5^2 = 0.45\) | A1FT | FT their \(E(X)\), but no FT for negative variance |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(1 - \frac{13}{27}\right) \div 2\) | M1 | or \(\frac{2}{9}\int_2^3 (3x - x^2)\,dx\) or equivalent |
| \(= \frac{7}{27}\) or \(0.259\) | A1 | As final answer |
| Total: 2 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = 1.5$ | B1 | |
| $\frac{2}{9}\int_0^3 (3x^3 - x^4)\,dx$ | M1 | Attempt integration of $x^2f(x)$, ignore limits |
| $= \frac{2}{9}\left[\frac{3x^4}{4} - \frac{x^5}{5}\right]_0^3$ | M1 | Substitute correct limits into correct integral |
| $= \frac{2}{9}\left[\frac{243}{4} - \frac{243}{5}\right] = 2.7$ | | |
| $\text{Var}(X) = 2.7 - 1.5^2 = 0.45$ | A1FT | FT their $E(X)$, but no FT for negative variance |
| **Total: 4** | | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5$ | B1 | |
| **Total: 1** | | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(1 - \frac{13}{27}\right) \div 2$ | M1 | or $\frac{2}{9}\int_2^3 (3x - x^2)\,dx$ or equivalent |
| $= \frac{7}{27}$ or $0.259$ | A1 | As final answer |
| **Total: 2** | | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{43403c12-93e6-44e4-b15e-e3c4363be5f9-08_254_634_260_717}
The diagram shows the graph of the probability density function, f , of a random variable $X$, where
$$f ( x ) = \begin{cases} \frac { 2 } { 9 } \left( 3 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
(a) State the value of $\mathrm { E } ( X )$ and find $\operatorname { Var } ( X )$.\\
(b) State the value of $\mathrm { P } ( 1.5 \leqslant X \leqslant 4 )$.\\
(c) Given that $\mathrm { P } ( 1 \leqslant X \leqslant 2 ) = \frac { 13 } { 27 }$, find $\mathrm { P } ( X > 2 )$.\\
\hfill \mbox{\textit{CAIE S2 2020 Q5 [7]}}