| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Geometric/graphical PDF with k |
| Difficulty | Moderate -0.5 This is a straightforward continuous probability distribution question requiring basic integration and PDF properties. Part (a)(i) uses the area-under-curve = 1 property to find k (simple triangle area), part (a)(ii) is a standard E(X) integral, and part (b) uses the median definition with another triangle area calculation. All techniques are routine for S2 level 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 pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 4 \times a = 1\) | M1 | For use of area \(= 1\), or let \(f(x) = kx\) and attempt \(\int_0^4 kx\, dx = 1\) |
| \(\left[a = \frac{1}{2}\right]\) \(f(x) = \frac{1}{8}x\) | A1 | \(k\left[\frac{x^2}{2}\right]_0^4 = 1\); \(8k=1\); \(k = \frac{1}{8}\). \(f(x) = \frac{1}{8}x\) or \(k = \frac{1}{8}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^4 x \times \frac{1}{8}x \, dx\) | M1 | Attempt to integrate \(x \times their\) \(f(x)\). Ignore limits accept in terms of \(k\). |
| \(\left[\frac{x^3}{24}\right]_0^4\) | A1ft | Their integral and correct limits accept in terms of \(k\). |
| \(= \frac{8}{3}\) or \(2.67\) (3 sf) | A1 | Note: Final answer of \(64k/3\) scores \(2/3\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{a-1}{a} = \frac{1}{\sqrt{2}}\) | M1 | Or attempt \(\int_0^1 g(w)\,dw = \frac{1}{2}\) i.e. \(\int_0^1\left(\frac{2}{a} - \frac{2}{a^2}w\right)dw = \frac{1}{2}\), or integral from \(1\) to \(a\). \(g(w)\) must be linear of form \(g(w) = mw (+c)\). Or area attempt: attempt to calculate heights using their linear equation (\(h_1 = 2/a\) and \(h_2 = -2/a^2 + 2/a\)) and use in either area trapezium \(= 0.5\), or area trapezium \(=\) area small triangle or area small triangle \(= 0.5\). Area trapezium \(= 1/2 \times 1\,(2/a + -2/a^2 + 2/a)\). Area triangle \(= 1/2(a-1)(-2/a^2 + 2/a)\). Note: alternative expression for \(h_1 = (a-2)/(a-1)\). |
| \(a\sqrt{2} - \sqrt{2} = a\) | A1 | Or \(a^2 - 4a + 2 = 0\). Any correct equation in \(a\), \(a\) not in denominator. |
| \(a = 2 + \sqrt{2} = 3.41\) | A1 |
## Question 7(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 4 \times a = 1$ | M1 | For use of area $= 1$, or let $f(x) = kx$ and attempt $\int_0^4 kx\, dx = 1$ |
| $\left[a = \frac{1}{2}\right]$ $f(x) = \frac{1}{8}x$ | A1 | $k\left[\frac{x^2}{2}\right]_0^4 = 1$; $8k=1$; $k = \frac{1}{8}$. $f(x) = \frac{1}{8}x$ or $k = \frac{1}{8}$ |
## Question 7(a)(ii):
$\int_0^4 x \times \frac{1}{8}x \, dx$ | **M1** | Attempt to integrate $x \times their$ $f(x)$. Ignore limits accept in terms of $k$.
$\left[\frac{x^3}{24}\right]_0^4$ | **A1ft** | Their integral and correct limits accept in terms of $k$.
$= \frac{8}{3}$ or $2.67$ (3 sf) | **A1** | Note: Final answer of $64k/3$ scores $2/3$.
**Total: 3 marks**
---
## Question 7(b):
$\frac{a-1}{a} = \frac{1}{\sqrt{2}}$ | **M1** | Or attempt $\int_0^1 g(w)\,dw = \frac{1}{2}$ i.e. $\int_0^1\left(\frac{2}{a} - \frac{2}{a^2}w\right)dw = \frac{1}{2}$, or integral from $1$ to $a$. $g(w)$ must be linear of form $g(w) = mw (+c)$. Or area attempt: attempt to calculate heights using their linear equation ($h_1 = 2/a$ and $h_2 = -2/a^2 + 2/a$) and use in either area trapezium $= 0.5$, or area trapezium $=$ area small triangle or area small triangle $= 0.5$. Area trapezium $= 1/2 \times 1\,(2/a + -2/a^2 + 2/a)$. Area triangle $= 1/2(a-1)(-2/a^2 + 2/a)$. Note: alternative expression for $h_1 = (a-2)/(a-1)$.
$a\sqrt{2} - \sqrt{2} = a$ | **A1** | Or $a^2 - 4a + 2 = 0$. Any correct equation in $a$, $a$ not in denominator.
$a = 2 + \sqrt{2} = 3.41$ | **A1** |
**Total: 3 marks**7
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{593c1ece-82a2-4dcd-8041-f39c98adf631-12_357_738_267_731}
The diagram shows the graph of the probability density function, f , of a random variable $X$ which takes values between 0 and 4 only. Between these two values the graph is a straight line.
\begin{enumerate}[label=(\roman*)]
\item Show that $\mathrm { f } ( x ) = k x$ for $0 \leqslant x \leqslant 4$, where $k$ is a constant to be determined.
\item Hence, or otherwise, find $\mathrm { E } ( X )$.
\end{enumerate}\item \\
\includegraphics[max width=\textwidth, alt={}, center]{593c1ece-82a2-4dcd-8041-f39c98adf631-13_383_752_269_731}
The diagram shows the graph of the probability density function, g , of a random variable $W$ which takes values between 0 and $a$ only, where $a > 0$. Between these two values the graph is a straight line.
Given that the median of $W$ is 1 , find the value of $a$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q7 [8]}}