| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2003 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate probability P(X in interval) |
| Difficulty | Moderate -0.8 This is a straightforward continuous probability distribution question requiring standard techniques: integration to find a probability, calculating E(X) using the standard formula, and solving for the median. All three parts are routine applications of well-practiced methods with a simple linear pdf, making this easier than average for A-level statistics. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(X > 1.5) = \left[x - \frac{x^2}{4}\right]_{1.5}^2\) | \(M1\) | For substituting 2 and 1.5 in their \(\int f(x)dx\) (or area method \(\frac{1}{2}\) their base \(\times\) their height) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.0625\) | \(A1\) | For correct answer |
| (ii) \(E(X) = \int_0^2 \left(x - \frac{1}{2}x^2\right)dx = \left[\frac{x^2}{2} - \frac{x^3}{6}\right]_0^2\) | \(M1\) | For evaluating their \(\int xf(x)dx\) |
| \(= \frac{2}{3}\) | \(A1\) | For correct answer |
| (iii) \(m - \frac{m^2}{4} = 0.5\) | \(M1\) | For equating their \(\int f(x)dx\) to 0.5 |
| \(m = 0.586\) \((2 - \sqrt{2})\) | \(M1, A1\) | For solving the related quadratic |
| For correct answer |
**(i)** $P(X > 1.5) = \left[x - \frac{x^2}{4}\right]_{1.5}^2$ | $M1$ | For substituting 2 and 1.5 in their $\int f(x)dx$ (or area method $\frac{1}{2}$ their base $\times$ their height)
or $1 - \left[x - \frac{x^2}{4}\right]_0^{1.5}$
$= 0.0625$ | $A1$ | For correct answer
**(ii)** $E(X) = \int_0^2 \left(x - \frac{1}{2}x^2\right)dx = \left[\frac{x^2}{2} - \frac{x^3}{6}\right]_0^2$ | $M1$ | For evaluating their $\int xf(x)dx$
$= \frac{2}{3}$ | $A1$ | For correct answer
**(iii)** $m - \frac{m^2}{4} = 0.5$ | $M1$ | For equating their $\int f(x)dx$ to 0.5
$m = 0.586$ $(2 - \sqrt{2})$ | $M1, A1$ | For solving the related quadratic
| | For correct answer4 A random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} 1 - \frac { 1 } { 2 } x & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
(i) Find $\mathrm { P } ( X > 1.5 )$.\\
(ii) Find the mean of $X$.\\
(iii) Find the median of $X$.
\hfill \mbox{\textit{CAIE S2 2003 Q4 [7]}}