| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Standard +0.8 This is a Further Statistics 2 question requiring integration of trigonometric functions to verify k, finding E(X) using integration by parts, equating pdf values to find σ, and computing probabilities involving both distributions. While the individual techniques are standard for FS2, the multi-step nature, integration complexity, and need to work with both continuous distributions elevates this above average A-level difficulty but remains within expected FS2 scope. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int(1-\cos x)\,dx = [x - \sin x]\) | M1 | Attempt to integrate \((1-\cos x)\) — one correct term |
| Use of correct limits and \(\int f(x)\,dx = 1 \Rightarrow 2\pi - 0 - 0 = 1\) | M1 | For use of correct limits and correct method for \(k\) |
| so \(k = \frac{1}{2\pi}\) | A1*cso | Use of \(\int f(x)\,dx=1\) seen and no incorrect working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = \pi\) (symmetry) so \(\mu = \pi\) so \(f(\mu) = \frac{1}{2\pi}(1-\cos\pi) = \frac{1}{\pi}\) | B1 | For correctly deducing the value of \(f(\mu)\) |
| \(\frac{1}{\sigma\sqrt{2\pi}} = \frac{1}{\pi}\); so \(\sigma = \sqrt{\frac{\pi}{2}}\) | M1; A1 | M1 for a correct equation for \(\sigma\); A1 for \(\sqrt{\frac{\pi}{2}}\) or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P\!\left(\frac{\pi}{2} < X < \frac{3\pi}{2}\right) = \frac{1}{2\pi}\left[x-\sin x\right]_{\pi/2}^{3\pi/2} = \frac{1}{2\pi}\left[\left(\frac{3\pi}{2}-{-1}\right)-\left(\frac{\pi}{2}-1\right)\right]\) | M1 | For a correct attempt to find prob — some correct integration and use of limits |
| \(= \frac{2+\pi}{2\pi}\ (= 0.81830\ldots)\) | A1 | For a correct answer (exact or \(0.818\ldots\) or better) |
| \(P\!\left(\frac{\pi}{2} < Y < \frac{3\pi}{2}\right) = 0.7899\ldots\) | B1 | For a correct probability from calculator i.e. \(0.7899\) or better, accept \(0.79\) |
| So error is \(0.81830\ldots - 0.7899\ldots = 0.0284\) | A1 | For \(0.0284\) or better |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int(1-\cos x)\,dx = [x - \sin x]$ | M1 | Attempt to integrate $(1-\cos x)$ — one correct term |
| Use of correct limits and $\int f(x)\,dx = 1 \Rightarrow 2\pi - 0 - 0 = 1$ | M1 | For use of correct limits and correct method for $k$ |
| so $k = \frac{1}{2\pi}$ | A1*cso | Use of $\int f(x)\,dx=1$ seen and no incorrect working seen |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \pi$ (symmetry) so $\mu = \pi$ so $f(\mu) = \frac{1}{2\pi}(1-\cos\pi) = \frac{1}{\pi}$ | B1 | For correctly deducing the value of $f(\mu)$ |
| $\frac{1}{\sigma\sqrt{2\pi}} = \frac{1}{\pi}$; so $\sigma = \sqrt{\frac{\pi}{2}}$ | M1; A1 | M1 for a correct equation for $\sigma$; A1 for $\sqrt{\frac{\pi}{2}}$ or exact equivalent |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P\!\left(\frac{\pi}{2} < X < \frac{3\pi}{2}\right) = \frac{1}{2\pi}\left[x-\sin x\right]_{\pi/2}^{3\pi/2} = \frac{1}{2\pi}\left[\left(\frac{3\pi}{2}-{-1}\right)-\left(\frac{\pi}{2}-1\right)\right]$ | M1 | For a correct attempt to find prob — some correct integration and use of limits |
| $= \frac{2+\pi}{2\pi}\ (= 0.81830\ldots)$ | A1 | For a correct answer (exact or $0.818\ldots$ or better) |
| $P\!\left(\frac{\pi}{2} < Y < \frac{3\pi}{2}\right) = 0.7899\ldots$ | B1 | For a correct probability from calculator i.e. $0.7899$ or better, accept $0.79$ |
| So error is $0.81830\ldots - 0.7899\ldots = 0.0284$ | A1 | For $0.0284$ or better |
---5
\begin{figure}[h]
\begin{center}
\includegraphics[width=\textwidth]{54bf68ab-7934-432a-890f-20093082ab07-12_446_1105_242_479}
\caption{Figure 1}
\end{center}
\end{figure}
The random variable $X$ has probability density function $\mathrm { f } ( x )$ and Figure 1 shows a sketch of $\mathrm { f } ( x )$ where
$$f ( x ) = \left\{ \begin{array} { c c }
k ( 1 - \cos x ) & 0 \leqslant x \leqslant 2 \pi \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 2 \pi }$
The random variable $Y \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$ and $\mathrm { E } ( Y ) = \mathrm { E } ( X )$\\
The probability density function of $Y$ is $g ( y )$, where
$$g ( y ) = \frac { 1 } { \sigma \sqrt { 2 \pi } } e ^ { - \frac { 1 } { 2 } \left( \frac { y - \mu } { \sigma } \right) ^ { 2 } } \quad - \infty < y < \infty$$
Given that $\mathrm { g } ( \mu ) = \mathrm { f } ( \mu )$
\item find the exact value of $\sigma$
\item Calculate the error in using $\mathrm { P } \left( \frac { \pi } { 2 } < Y < \frac { 3 \pi } { 2 } \right)$ as an approximation to $\mathrm { P } \left( \frac { \pi } { 2 } < X < \frac { 3 \pi } { 2 } \right)$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2020 Q5 [10]}}