| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.8 This is a straightforward S2 probability density function question requiring only standard techniques: integrating to find the constant (given answer), evaluating a probability integral, and computing an expectation. All three parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int(a + x/3)dx = 1\) | M1 | Equating to 1 and attempting to integrate |
| \(\left[ax + \frac{x^2}{6}\right] = 1\) | A1 | Correct integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 1/2\) AG | A1 (3 marks) | Given answer legit obtained |
| (ii) \(P(X > 1.8) = \int_{1.8}^{2}(1/2 + x/3)dx = \left[\frac{x}{2} + \frac{x^3}{6}\right]_{1.8} = 0.227\) | M1, A1, A1 (2 marks) | For integrating and using limits 1.8 and 2 or 0 and 1.8 and subt from 1; For correct answer |
| (iii) \(E(X) = \int_1^2 (x/2 + x^3/3)dx = [x^2/4 + x^3/9]^2_1 = [1+8/9] - [1/4+1/9] = 55/36(1.53)\) | M1, A1, A1 (3 marks) | For attempting to evaluate integral \(xf(x)\) between limits; For correct integration; For correct answer |
**(i)** $\int(a + x/3)dx = 1$ | M1 | Equating to 1 and attempting to integrate
$\left[ax + \frac{x^2}{6}\right] = 1$ | A1 | Correct integration
$[2a + 2/3] - [a + 1/6] = 1$
$a = 1/2$ AG | A1 (3 marks) | Given answer legit obtained
**(ii)** $P(X > 1.8) = \int_{1.8}^{2}(1/2 + x/3)dx = \left[\frac{x}{2} + \frac{x^3}{6}\right]_{1.8} = 0.227$ | M1, A1, A1 (2 marks) | For integrating and using limits 1.8 and 2 or 0 and 1.8 and subt from 1; For correct answer
**(iii)** $E(X) = \int_1^2 (x/2 + x^3/3)dx = [x^2/4 + x^3/9]^2_1 = [1+8/9] - [1/4+1/9] = 55/36(1.53)$ | M1, A1, A1 (3 marks) | For attempting to evaluate integral $xf(x)$ between limits; For correct integration; For correct answer
---5 A continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} a + \frac { 1 } { 3 } x & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $a$ is a constant.\\
(i) Show that the value of $a$ is $\frac { 1 } { 2 }$.\\
(ii) Find $\mathrm { P } ( X > 1.8 )$.\\
(iii) Find $\mathrm { E } ( X )$.
\hfill \mbox{\textit{CAIE S2 2005 Q5 [8]}}