| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Geometric applications |
| Difficulty | Standard +0.8 This Further Statistics 2 question requires students to work with transformations of continuous uniform distributions, find probabilities involving non-linear transformations (area = πx²), reason about medians of transformed variables, and perform algebraic integration of a trigonometric expression (½x²sin(x)). While the individual techniques are A-level standard, the combination of probability theory with geometric applications and the need for integration by parts makes this moderately challenging, though still within reach for well-prepared Further Maths students. |
| 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 |
| \(P(\pi X^2 > 10) \Rightarrow P\left(X > \sqrt{\frac{10}{\pi}}\right)\) | M1 | Reduce the problem to a probability about \(X\) |
| \(= \dfrac{\pi - \sqrt{\frac{10}{\pi}}}{\pi}\) | M1 | For use of the uniform distribution (a correct expression ft their value 1.784..) |
| \(= 0.43209\ldots =\) awrt \(\mathbf{0.432}\) | A1 | For awrt 0.432 |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{area} > \text{median}) = 0.5\); since (a) \(< 0.5\) therefore median \(\leq 10\) | B1 | For statement that median \(< 10\) supported by argument about answer to (a) being \(< 0.5\) |
| (1 mark) | ||
| ALT: Median area is given by \(\pi \times \left(\dfrac{\pi}{2}\right)^2 = 7.751\ldots < 10\) so median \(< 10\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of triangle \(= 0.5x^2 \sin x\) | M1 | For a correct expression for area in terms of \(x\) |
| \(E(\text{area}) = \int_{[0]}^{[\pi]} \frac{1}{\pi} \cdot \frac{1}{2} x^2 \sin x \, dx\) | M1 | For realisation that need to use \(E(g(X))\) formula and a correct expression (ignore limits) |
| \(= \dfrac{1}{2\pi} \int_{[0]}^{[\pi]} x^2 \, d(-\cos x) = \dfrac{1}{2\pi}\left\{\left[-x^2 \cos x\right]_{[0]}^{[\pi]} - \int_{[0]}^{[\pi]} -2x\cos x \, dx\right\}\) | M1 | For attempt to use integration by parts |
| \(= \left\{\left[\dfrac{-x^2 \cos x}{2\pi}\right]_{[0]}^{[\pi]}\right\} + \left[\dfrac{x \sin x}{\pi}\right]_{[0]}^{[\pi]} - \dfrac{1}{\pi}\int_{[0]}^{[\pi]} \sin x \, dx\) | M1 | For a \(2^{\text{nd}}\) use of integration by parts |
| A1 | For correct integration (ignore limits) | |
| \(= \dfrac{\pi^2}{2\pi} - 0 + 0 - 0 + \left(-\dfrac{1}{\pi}\right) - \left(\dfrac{1}{\pi}\right) = \dfrac{\pi}{2} - \dfrac{2}{\pi}\) | M1 | For clear use of the correct limits |
| \(= \dfrac{\pi}{2} - \dfrac{2}{\pi}\) | A1 | For \(\dfrac{\pi}{2} - \dfrac{2}{\pi}\) |
| (7 marks) | ||
| Total: 11 marks | [(c) is an extended problem and also involves work from pure for the integration] |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\pi X^2 > 10) \Rightarrow P\left(X > \sqrt{\frac{10}{\pi}}\right)$ | M1 | Reduce the problem to a probability about $X$ |
| $= \dfrac{\pi - \sqrt{\frac{10}{\pi}}}{\pi}$ | M1 | For use of the uniform distribution (a correct expression ft their value 1.784..) |
| $= 0.43209\ldots =$ awrt $\mathbf{0.432}$ | A1 | For awrt 0.432 |
| **(3 marks)** | | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{area} > \text{median}) = 0.5$; since (a) $< 0.5$ therefore **median $\leq 10$** | B1 | For statement that median $< 10$ supported by argument about answer to (a) being $< 0.5$ |
| **(1 mark)** | | |
| **ALT:** Median area is given by $\pi \times \left(\dfrac{\pi}{2}\right)^2 = 7.751\ldots < 10$ so median $< 10$ | | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of triangle $= 0.5x^2 \sin x$ | M1 | For a correct expression for area in terms of $x$ |
| $E(\text{area}) = \int_{[0]}^{[\pi]} \frac{1}{\pi} \cdot \frac{1}{2} x^2 \sin x \, dx$ | M1 | For realisation that need to use $E(g(X))$ formula and a correct expression (ignore limits) |
| $= \dfrac{1}{2\pi} \int_{[0]}^{[\pi]} x^2 \, d(-\cos x) = \dfrac{1}{2\pi}\left\{\left[-x^2 \cos x\right]_{[0]}^{[\pi]} - \int_{[0]}^{[\pi]} -2x\cos x \, dx\right\}$ | M1 | For attempt to use integration by parts |
| $= \left\{\left[\dfrac{-x^2 \cos x}{2\pi}\right]_{[0]}^{[\pi]}\right\} + \left[\dfrac{x \sin x}{\pi}\right]_{[0]}^{[\pi]} - \dfrac{1}{\pi}\int_{[0]}^{[\pi]} \sin x \, dx$ | M1 | For a $2^{\text{nd}}$ use of integration by parts |
| | A1 | For correct integration (ignore limits) |
| $= \dfrac{\pi^2}{2\pi} - 0 + 0 - 0 + \left(-\dfrac{1}{\pi}\right) - \left(\dfrac{1}{\pi}\right) = \dfrac{\pi}{2} - \dfrac{2}{\pi}$ | M1 | For clear use of the correct limits |
| $= \dfrac{\pi}{2} - \dfrac{2}{\pi}$ | A1 | For $\dfrac{\pi}{2} - \dfrac{2}{\pi}$ |
| **(7 marks)** | | |
| **Total: 11 marks** | | [(c) is an extended problem and also involves work from pure for the integration] |8 A circle, centre $O$, has radius $x \mathrm {~cm}$, where $x$ is an observation from the random variable $X$ which has a rectangular distribution on $[ 0 , \pi ]$
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the area of the circle is greater than $10 \mathrm {~cm} ^ { 2 }$
\item State, giving a reason, whether the median area of the circle is greater or less than $10 \mathrm {~cm} ^ { 2 }$
The triangle $O A B$ is drawn inside the circle with $O A$ and $O B$ as radii of length $x \mathrm {~cm}$ and angle $A O B x$ radians.
\item Use algebraic integration to find the expected value of the area of triangle $O A B$. Give your answer as an exact value.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2020 Q8 [11]}}