| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2002 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Standard applied PDF calculations |
| Difficulty | Moderate -0.8 This is a straightforward S2 question requiring standard PDF calculations: finding E(X) and Var(X) using integration formulas, then solving F(x) = 0.98 for a percentile. The linear PDF makes integration simple, and all three parts follow routine procedures with no problem-solving insight required. Easier than average A-level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 2x - 2x^2 \, dx\) | M1, A1 (3 marks) | For sensible attempt to integrate \(xf(x)\); For correct integrand (any form) |
| \(= \left[x^2 - \frac{2x^3}{3}\right]_0^1 = 0.333\) | A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.0556\) | M1, M1\(^{\text{dep}}\), A1 (3 marks) | For sensible attempt to integrate \(x^2f(x)\); For their integral − (their mean)²; For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| 859 tonnes | M1, A1, M1, A1, B1 (5 marks) | For identifying both sides of equation; For correct equation in any form; For solving for \(x\) (must be sensible attempt); For correct answer; For applying concept of continuous rv. |
| Answer | Marks |
|---|---|
| M1, A1, M1, A1, B1 (5 marks) | For identifying \(x\) from a relevant diagram; For correct equation; For solving for \(x\); For correct answer; For applying concept of continuous rv. |
**(i)** $E(X) = \int_0^1 2x(1-x) \, dx$
$\int_0^1 2x - 2x^2 \, dx$ | M1, A1 (3 marks) | For sensible attempt to integrate $xf(x)$; For correct integrand (any form)
$= \left[x^2 - \frac{2x^3}{3}\right]_0^1 = 0.333$ | A1 | For correct answer
**(ii)** $\text{Var}(X) = \int_0^1 2x^2 - 2x^3 \, dx - (0.333)^2$
$= \left[\frac{2x^3}{3} - \frac{2x^4}{4}\right] - (0.333)^2$
$= 0.0556$ | M1, M1$^{\text{dep}}$, A1 (3 marks) | For sensible attempt to integrate $x^2f(x)$; For their integral − (their mean)²; For correct answer
**(iii)** $\int_0^{2(1-x)} dx = 0.98$
$[2x - x^2] = 0.98$
$x^2 - 2x + 0.98 = 0$
$x = 0.859$
859 tonnes | M1, A1, M1, A1, B1 (5 marks) | For identifying both sides of equation; For correct equation in any form; For solving for $x$ (must be sensible attempt); For correct answer; For applying concept of continuous rv.
**OR** (using diagram)
| M1, A1, M1, A1, B1 (5 marks) | For identifying $x$ from a relevant diagram; For correct equation; For solving for $x$; For correct answer; For applying concept of continuous rv.7 A factory is supplied with grain at the beginning of each week. The weekly demand, $X$ thousand tonnes, for grain from this factory is a continuous random variable having the probability density function given by
$$f ( x ) = \begin{cases} 2 ( 1 - x ) & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$
Find\\
(i) the mean value of $X$,\\
(ii) the variance of $X$,\\
(iii) the quantity of grain in tonnes that the factory should have in stock at the beginning of a week, in order to be $98 \%$ certain that the demand in that week will be met.
\hfill \mbox{\textit{CAIE S2 2002 Q7 [11]}}