| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Explain why not valid PDF |
| Difficulty | Moderate -0.5 Part (a) tests basic PDF properties (non-negativity and total area = 1) which is fundamental recall. Part (b) involves standard integration techniques for E(X), median, and probability calculations with a given PDF. While multi-step, these are routine S2 procedures without requiring novel insight or complex problem-solving. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) g: Area \(\neq 1\) or \(> 1\) | B1 | |
| h: pdf cannot be neg | B1 | |
| [2] | ||
| (b) (i) \(\int_{16}^{\infty} \frac{30}{x}\) | M1 | Attempt integ xf(x), ignore limits |
| \(= [30\ln x]_{16}^{\infty}\) | A1 | Correct integrand and limits |
| \(= 30(\ln 5 - \ln 10)\) | A1 | or \(30\ln(5/10)\) |
| \((= 30\ln 1.5\) AG\()\) | ||
| [3] | ||
| (ii) \(\int_{10}^{m} \frac{30}{x^2} dx = 0.5\) | M1 | Integ \(f(x) = 0.5\), limits 10 to unknown |
| \(\left[-30x^{-1}\right]_{10}^{m} = 0.5\) | A1 | Correct integrand, limits and \(= 0.5\) |
| \(-\frac{30}{m} - \left(-\frac{30}{10}\right) = 0.5\) | ||
| \(m = 12\) | A1 | |
| \(30\ln 1.5\) | M1 | |
| \(\int_{1}^{1.2^{1.5}} \frac{30}{x^2} dx\) | ||
| \(= 0.0337\) (3 sfs) | A1 | |
| [5] |
**(a)** g: Area $\neq 1$ or $> 1$ | B1 |
h: pdf cannot be neg | B1 |
| [2] |
**(b) (i)** $\int_{16}^{\infty} \frac{30}{x}$ | M1 | Attempt integ xf(x), ignore limits
$= [30\ln x]_{16}^{\infty}$ | A1 | Correct integrand and limits
$= 30(\ln 5 - \ln 10)$ | A1 | or $30\ln(5/10)$
$(= 30\ln 1.5$ AG$)$ | | |
| [3] |
**(ii)** $\int_{10}^{m} \frac{30}{x^2} dx = 0.5$ | M1 | Integ $f(x) = 0.5$, limits 10 to unknown
$\left[-30x^{-1}\right]_{10}^{m} = 0.5$ | A1 | Correct integrand, limits and $= 0.5$
$-\frac{30}{m} - \left(-\frac{30}{10}\right) = 0.5$ | | |
$m = 12$ | A1 |
$30\ln 1.5$ | M1 |
$\int_{1}^{1.2^{1.5}} \frac{30}{x^2} dx$ | | |
$= 0.0337$ (3 sfs) | A1 |
| [5] |4
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{7c9a87ac-69c6-4850-82aa-8235bba581e8-2_611_712_1466_358}\\
\includegraphics[max width=\textwidth, alt={}, center]{7c9a87ac-69c6-4850-82aa-8235bba581e8-2_618_716_1464_1155}
The diagrams show the graphs of two functions, $g$ and $h$. For each of the functions $g$ and $h$, give a reason why it cannot be a probability density function.
\item The distance, in kilometres, travelled in a given time by a cyclist is represented by the continuous random variable $X$ with probability density function given by
$$f ( x ) = \begin{cases} \frac { 30 } { x ^ { 2 } } & 10 \leqslant x \leqslant 15 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Show that $\mathrm { E } ( X ) = 30 \ln 1.5$.
\item Find the median of $X$. Find also the probability that $X$ lies between the median and the mean.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2011 Q4 [10]}}