| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.3 This is a straightforward probability density function question requiring standard techniques: integrating to find k using the normalization condition, then computing E(X) using integration by parts. Both are routine S2 procedures with no conceptual challenges, though the integration by parts adds slight computational work beyond the most basic questions. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int ke^{-x}dx = 1\) | M1 | Int = 1, ignore limits |
| \(\left[-ke^{-x}\right]_0^1 = 1\) | A1 | Correct integral & limits, & = 1 |
| \((= -ke^{-1} - (-ke^0))\) | ||
| \(= k \times \frac{e-1}{e} = 1\) or \(k(e-1) = e\) | ||
| \(k = \frac{e}{e-1}\) | A1 | [3] Correctly obtained, no errors seen |
| AG | ||
| (ii) \(\frac{e}{e-1} \int xe^{-x}dx\) | M1 | Attempt \(\int xf(x)dx\), ignore limits |
| \(= \frac{e}{e-1}\left[\{x(-e^{-x})\}_0^1 - \int(-e^{-x})dx\right]\) | M1* | Attempt integration by parts the correct way round, ignore limits |
| \(= \frac{e}{e-1}\left(-xe^{-x}\Big | _0^1 - \left[-e^{-x}\right]_0^1\right)\) | M1dep* |
| \(\left(= \frac{e}{e-1}(-e^{-1} - 0 - (e^{-1} - 1))\right)\) | ||
| \(= \frac{e}{e-1}\left(1 - \frac{2}{e}\right)\) or \(\frac{e-2}{e-1}\) or equivalent | A1 | [4] Accept \(k\) instead of \(\frac{e}{e-1}\) throughout except ans |
**(i)** $\int ke^{-x}dx = 1$ | M1 | Int = 1, ignore limits
$\left[-ke^{-x}\right]_0^1 = 1$ | A1 | Correct integral & limits, & = 1
$(= -ke^{-1} - (-ke^0))$ | |
$= k \times \frac{e-1}{e} = 1$ or $k(e-1) = e$ | |
$k = \frac{e}{e-1}$ | A1 | [3] Correctly obtained, no errors seen
| AG |
**(ii)** $\frac{e}{e-1} \int xe^{-x}dx$ | M1 | Attempt $\int xf(x)dx$, ignore limits
$= \frac{e}{e-1}\left[\{x(-e^{-x})\}_0^1 - \int(-e^{-x})dx\right]$ | M1* | Attempt integration by parts the correct way round, ignore limits
$= \frac{e}{e-1}\left(-xe^{-x}\Big|_0^1 - \left[-e^{-x}\right]_0^1\right)$ | M1dep* | Attempt second integral of the form $\pm k^-dx$, ignore limits
$\left(= \frac{e}{e-1}(-e^{-1} - 0 - (e^{-1} - 1))\right)$ | |
$= \frac{e}{e-1}\left(1 - \frac{2}{e}\right)$ or $\frac{e-2}{e-1}$ or equivalent | A1 | [4] Accept $k$ instead of $\frac{e}{e-1}$ throughout except ansThe random variable $X$ has probability density function given by
$$f(x) = \begin{cases}
ke^{-x} & 0 \leqslant x \leqslant 1, \\
0 & \text{otherwise}.
\end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Show that $k = \frac{e}{e-1}$. [3]
\item Find E($X$) in terms of $e$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2011 Q4 [7]}}