| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question requiring routine integration techniques: finding the constant c by integrating to 1, calculating a probability by integrating over an interval, and finding the mean using E(T). All three parts follow textbook procedures with straightforward polynomial integration, making it slightly easier than average for 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 |
|---|---|---|
| (i) \(\int_0^4 c(2s-t^2)dt = 1\) | M1 | For equating to 1 and a sensible attempt to integrate |
| \(c\left[\frac{2st^2}{2} - \frac{t^4}{4}\right]_0^4 = 1\) | A1 | For correct integration and correct limits |
| \(c\left[\frac{625}{2} - \frac{625}{4}\right] = 1 \Rightarrow c = \frac{4}{625}\) | A1 | For given answer correctly obtained |
| 3 | ||
| (ii) \(\int_2^4 c(2s-t^2)dt = \left[\frac{2st^2}{2} - \frac{t^4}{4}\right]_2^4 = c[36] - c[46] = \frac{72}{125}\) (0.576) | M1* M1*dep A1 | For attempting to integrate \(f(t)\) between 2 and 4 (or attempt 2 and 4); For subtracting their value when \(t = 2\) from their value when \(t = 4\); For correct answer |
| 3 | ||
| (iii) \(\int_0^5 ct^2(2s-t^2)dt = \left[\frac{4}{625} \times \frac{2s^3}{3} - \frac{4s^5}{625 \times 5}\right]_0^5 = \frac{8}{3}\) | M1* A1 M1*dep A1 | For attempting to integrate \(tf(t)\), no limits needed; For correct integrand can have \(c\) (or their \(c\)); For subtracting their value when \(t=0\) from their value when \(t=5\); For correct answer |
| 4 |
**(i)** $\int_0^4 c(2s-t^2)dt = 1$ | M1 | For equating to 1 and a sensible attempt to integrate
$c\left[\frac{2st^2}{2} - \frac{t^4}{4}\right]_0^4 = 1$ | A1 | For correct integration and correct limits
$c\left[\frac{625}{2} - \frac{625}{4}\right] = 1 \Rightarrow c = \frac{4}{625}$ | A1 | For given answer correctly obtained
| | **3** | |
**(ii)** $\int_2^4 c(2s-t^2)dt = \left[\frac{2st^2}{2} - \frac{t^4}{4}\right]_2^4 = c[36] - c[46] = \frac{72}{125}$ (0.576) | M1* M1*dep A1 | For attempting to integrate $f(t)$ between 2 and 4 (or attempt 2 and 4); For subtracting their value when $t = 2$ from their value when $t = 4$; For correct answer
| | **3** | |
**(iii)** $\int_0^5 ct^2(2s-t^2)dt = \left[\frac{4}{625} \times \frac{2s^3}{3} - \frac{4s^5}{625 \times 5}\right]_0^5 = \frac{8}{3}$ | M1* A1 M1*dep A1 | For attempting to integrate $tf(t)$, no limits needed; For correct integrand can have $c$ (or their $c$); For subtracting their value when $t=0$ from their value when $t=5$; For correct answer
| | **4** | |7 The queuing time, $T$ minutes, for a person queuing at a supermarket checkout has probability density function given by
$$f ( t ) = \begin{cases} c t \left( 25 - t ^ { 2 } \right) & 0 \leqslant t \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
where $c$ is a constant.\\
(i) Show that the value of $c$ is $\frac { 4 } { 625 }$.\\
(ii) Find the probability that a person will have to queue for between 2 and 4 minutes.\\
(iii) Find the mean queuing time.
\hfill \mbox{\textit{CAIE S2 2004 Q7 [10]}}