| Exam Board | Edexcel |
|---|---|
| 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 | Piecewise PDF with multiple regions |
| Difficulty | Challenging +1.2 This S2 question requires understanding of PDF properties (integration to 1), mode/median concepts, and linear equations, but follows a structured multi-part format with clear guidance. The constraint that mode equals median provides a helpful starting point, and the parts build logically. While requiring several techniques (finding k from normalization, using geometric properties for the linear section, and qualitative reasoning about skewness), each step is relatively standard for S2 level with no particularly novel insights required. |
| 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 |
| Mode \(= 3\) from graph | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^3 kx^2\, dx = 0.5 \Rightarrow \left[\frac{kx^3}{3}\right]_0^3 = 0.5\) | M1 A1 | 1st M1 for attempt to integrate \(f(x)\) (need \(x^3\)). 1st A1 for correct integration, ignore limits |
| \(\frac{27k}{3} - 0 = 0.5 \Rightarrow k = \frac{1}{18}\) (using median \(= 3\)) | M1d A1 | 2nd M1 dependent on previous M mark, use of correct limits set equal to \(0.5\). 2nd A1 for \(k = \frac{1}{18}\) or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Height of triangle \(= \frac{1}{18} \times 3^2 = \frac{1}{2}\) | B1ft | B1 for correct height using their \(k\), ie \(9k\) |
| Area of triangle \(= \frac{1}{2} \times (a-3) \times \frac{1}{2} = \frac{1}{2}\), so \(a = 5\) | M1 | Or correct gradient of line ie \(\frac{9k}{(3-a)}\) |
| \(a = 5\) cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| From graph distribution is negative skew (left tail is longer) | B1 | |
| \(\mu <\) median for negative skew so \(E(X) < 3\) | B1d | NB \(E(X) = 2\frac{23}{24}\) |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mode $= 3$ from graph | B1 | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^3 kx^2\, dx = 0.5 \Rightarrow \left[\frac{kx^3}{3}\right]_0^3 = 0.5$ | M1 A1 | 1st M1 for attempt to integrate $f(x)$ (need $x^3$). 1st A1 for correct integration, ignore limits |
| $\frac{27k}{3} - 0 = 0.5 \Rightarrow k = \frac{1}{18}$ (using median $= 3$) | M1d A1 | 2nd M1 dependent on previous M mark, use of correct limits set equal to $0.5$. 2nd A1 for $k = \frac{1}{18}$ or exact equivalent |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Height of triangle $= \frac{1}{18} \times 3^2 = \frac{1}{2}$ | B1ft | B1 for correct height using their $k$, ie $9k$ |
| Area of triangle $= \frac{1}{2} \times (a-3) \times \frac{1}{2} = \frac{1}{2}$, so $a = 5$ | M1 | Or correct gradient of line ie $\frac{9k}{(3-a)}$ |
| $a = 5$ cao | A1 | |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| From graph distribution is negative skew (left tail is longer) | B1 | |
| $\mu <$ median for negative skew so $E(X) < 3$ | B1d | NB $E(X) = 2\frac{23}{24}$ |
# Mark Scheme Extraction3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6e4af10e-ee8d-493f-bd72-34b231003d97-05_455_1026_242_484}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the probability density function $\mathrm { f } ( x )$ of the random variable $X$.\\
For $0 \leqslant x \leqslant 3 , \mathrm { f } ( x )$ is represented by a curve $O B$ with equation $\mathrm { f } ( x ) = k x ^ { 2 }$, where $k$ is a constant.
For $3 \leqslant x \leqslant a$, where $a$ is a constant, $\mathrm { f } ( x )$ is represented by a straight line passing through $B$ and the point ( $a , 0$ ).
For all other values of $x , \mathrm { f } ( x ) = 0$.\\
Given that the mode of $X =$ the median of $X$, find
\begin{enumerate}[label=(\alph*)]
\item the mode,
\item the value of $k$,
\item the value of $a$.
Without calculating $\mathrm { E } ( X )$ and with reference to the skewness of the distribution
\item state, giving your reason, whether $\mathrm { E } ( X ) < 3 , \mathrm { E } ( X ) = 3$ or $\mathrm { E } ( X ) > 3$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2011 Q3 [10]}}