| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| 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 straightforward continuous probability distribution question requiring standard techniques: integrating to find k (given answer to verify), sketching an exponential curve, and solving P(T > t) = 0.8 using integration. All steps are routine S2 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int_0^5 ke^{0.2t} dt = 1\) | M1 | Equating to 1 and attempting to integrate |
| \(\left[\frac{k}{0.2} e^{1.0}\right] - \left[\frac{k}{0.2} e^0\right] = 1\) | A1 | Correct integrand and limits |
| \(\frac{k}{0.2}(e - 1) = 1\) | ||
| \(k = \frac{1}{5(e-1)}\) AG | A1 [3] | Correct answer legitimately obtained |
| (ii) | B1 [2] | Correct curve shape |
| B1 | Correct horizontal lines (need to see a 5) | |
| (iii) \(\int_0^T ke^{0.2t} dt = 0.2\) | M1 | Equation relating \(T\) and 0.2 or 0.8 |
| \([5ke^{0.2t}] - [5k] = 0.2\) | A1 | Correct equation (can be in '\(k\)') |
| \(e^{0.2T} = \frac{0.2}{5k} + 1 = 1.344\) | ||
| \(T = 1.48\) (seconds) | A1 [3] | Correct answer |
**(i)** $\int_0^5 ke^{0.2t} dt = 1$ | M1 | Equating to 1 and attempting to integrate
$\left[\frac{k}{0.2} e^{1.0}\right] - \left[\frac{k}{0.2} e^0\right] = 1$ | A1 | Correct integrand and limits
$\frac{k}{0.2}(e - 1) = 1$ | |
$k = \frac{1}{5(e-1)}$ AG | A1 [3] | Correct answer legitimately obtained
**(ii)** | B1 [2] | Correct curve shape
| B1 | Correct horizontal lines (need to see a 5)
**(iii)** $\int_0^T ke^{0.2t} dt = 0.2$ | M1 | Equation relating $T$ and 0.2 or 0.8
$[5ke^{0.2t}] - [5k] = 0.2$ | A1 | Correct equation (can be in '$k$')
$e^{0.2T} = \frac{0.2}{5k} + 1 = 1.344$ | |
$T = 1.48$ (seconds) | A1 [3] | Correct answer5 The random variable $T$ denotes the time in seconds for which a firework burns before exploding. The probability density function of $T$ is given by
$$\mathrm { f } ( t ) = \begin{cases} k \mathrm { e } ^ { 0.2 t } & 0 \leqslant t \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 1 } { 5 ( \mathrm { e } - 1 ) }$.\\
(ii) Sketch the probability density function.\\
(iii) $80 \%$ of fireworks burn for longer than a certain time before they explode. Find this time.
\hfill \mbox{\textit{CAIE S2 2010 Q5 [8]}}