| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find or specify CDF |
| Difficulty | Standard +0.3 This is a standard Further Maths Statistics question testing routine techniques: integrating to find k, finding CDF by integration, solving F(m)=0.5 for median, and computing E(X) with integration by parts. All steps are textbook procedures with straightforward trigonometric integration, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\frac{\pi}{6}} k\sin 2x \, dx = 1\) | M1 | Uses correct equation; condone missing \(dx\) |
| \(\left[-\frac{k}{2}\cos 2x\right]_0^{\frac{\pi}{6}} = 1\) | ||
| \(-\frac{k}{2}\cos\frac{\pi}{3} + \frac{k}{2} = 1\) | A1 | Obtains \(-\frac{k}{2}\cos 2x\) OE |
| \(\frac{k}{4} = 1\) | M1 | Substitutes limits into their integrated function and subtracts either way round |
| \(k = 4\) | R1 | Completes reasoned argument to show that \(k = 4\) by solving the correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(x) = \int_0^x 4\sin 2y \, dy\) | M1 | Uses integral of the form \(\int f(y)\,dy\) to find \(F(x)\) for the interval \(0 \leq x \leq \frac{\pi}{6}\); PI by \(2 - 2\cos 2x\) or \(-2\cos 2y\) or \(-2\cos 2x\); condone \(\int f(x)\,dx\) |
| \(= [-2\cos 2y]_0^x = 2 - 2\cos 2x\) | ||
| \[F(x) = \begin{cases} 0 & x < 0 \\ 2 - 2\cos 2x & 0 \leq x \leq \frac{\pi}{6} \\ 1 & x > \frac{\pi}{6} \end{cases}\] | A1 | Obtains correct \(F(x)\) for the interval \(0 \leq x \leq \frac{\pi}{6}\) |
| A1 | Obtains correct \(F(x)\) for all intervals; \(0\) and \(\frac{\pi}{6}\) must each be in one interval only |
| Answer | Marks | Guidance |
|---|---|---|
| \(2 - 2\cos 2m = 0.5\) | M1 | Sets their integrated expression for \(F(m)\) equal to \(0.5\); PI |
| \(\cos 2m = 0.75\) | ||
| \(m = 0.361\) | A1 | Obtains correct median; AWRT \(0.361\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Mean} = \int_0^{\frac{\pi}{6}} 4x\sin 2x \, dx\) | M1 | Uses correct integral of the form \(\int xf(x)\,dx\) with any limits |
| \(= \left[-2x\cos 2x\right]_0^{\frac{\pi}{6}} - \int_0^{\frac{\pi}{6}} -2\cos 2x \, dx\) | M1 | Integrates by parts with \(u = x\) and \(v' = \cos 2x\) to reach the form \([\pm Ax\cos 2x] - \int \pm A\cos 2x \, dx\) |
| \(= \left[-2x\cos 2x + \sin 2x\right]_0^{\frac{\pi}{6}}\) | A1 | Obtains the correct integrated function; may not be seen as a single expression |
| \(= \left(-\frac{\pi}{3} \times \frac{1}{2} + \frac{\sqrt{3}}{2}\right) - 0\) | M1 | Substitutes the correct limits into their integrated function and subtracts either way round |
| \(= \frac{1}{6}(3\sqrt{3} - \pi)\) | R1 | Completes reasoned argument to show that the mean is \(\frac{1}{6}(3\sqrt{3}-\pi)\) by correctly simplifying the correct expression |
## Question 8(a):
$\int_0^{\frac{\pi}{6}} k\sin 2x \, dx = 1$ | M1 | Uses correct equation; condone missing $dx$
$\left[-\frac{k}{2}\cos 2x\right]_0^{\frac{\pi}{6}} = 1$ | | |
$-\frac{k}{2}\cos\frac{\pi}{3} + \frac{k}{2} = 1$ | A1 | Obtains $-\frac{k}{2}\cos 2x$ **OE**
$\frac{k}{4} = 1$ | M1 | Substitutes limits into their integrated function and subtracts either way round
$k = 4$ | R1 | Completes reasoned argument to show that $k = 4$ by solving the correct equation
---
## Question 8(b):
$F(x) = \int_0^x 4\sin 2y \, dy$ | M1 | Uses integral of the form $\int f(y)\,dy$ to find $F(x)$ for the interval $0 \leq x \leq \frac{\pi}{6}$; **PI** by $2 - 2\cos 2x$ or $-2\cos 2y$ or $-2\cos 2x$; condone $\int f(x)\,dx$
$= [-2\cos 2y]_0^x = 2 - 2\cos 2x$ | | |
$$F(x) = \begin{cases} 0 & x < 0 \\ 2 - 2\cos 2x & 0 \leq x \leq \frac{\pi}{6} \\ 1 & x > \frac{\pi}{6} \end{cases}$$ | A1 | Obtains correct $F(x)$ for the interval $0 \leq x \leq \frac{\pi}{6}$
| A1 | Obtains correct $F(x)$ for all intervals; $0$ and $\frac{\pi}{6}$ must each be in one interval only
---
## Question 8(c):
$2 - 2\cos 2m = 0.5$ | M1 | Sets their integrated expression for $F(m)$ equal to $0.5$; **PI**
$\cos 2m = 0.75$ | |
$m = 0.361$ | A1 | Obtains correct median; **AWRT** $0.361$
---
## Question 8(d):
$\text{Mean} = \int_0^{\frac{\pi}{6}} 4x\sin 2x \, dx$ | M1 | Uses correct integral of the form $\int xf(x)\,dx$ with any limits
$= \left[-2x\cos 2x\right]_0^{\frac{\pi}{6}} - \int_0^{\frac{\pi}{6}} -2\cos 2x \, dx$ | M1 | Integrates by parts with $u = x$ and $v' = \cos 2x$ to reach the form $[\pm Ax\cos 2x] - \int \pm A\cos 2x \, dx$
$= \left[-2x\cos 2x + \sin 2x\right]_0^{\frac{\pi}{6}}$ | A1 | Obtains the correct integrated function; may not be seen as a single expression
$= \left(-\frac{\pi}{3} \times \frac{1}{2} + \frac{\sqrt{3}}{2}\right) - 0$ | M1 | Substitutes the correct limits into their integrated function and subtracts either way round
$= \frac{1}{6}(3\sqrt{3} - \pi)$ | R1 | Completes reasoned argument to show that the mean is $\frac{1}{6}(3\sqrt{3}-\pi)$ by correctly simplifying the correct expression8 The continuous random variable $X$ has probability density function
$$f ( x ) = \begin{cases} k \sin 2 x & 0 \leq x \leq \frac { \pi } { 6 } \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.
8
\begin{enumerate}[label=(\alph*)]
\item Show that $k = 4$\\
8
\item Find the cumulative distribution function $\mathrm { F } ( x )$\\
8
\item Find the median of $X$, giving your answer to three significant figures.
8
\item Find the mean of $X$ giving your answer in the form $\frac { 1 } { a } ( b \sqrt { 3 } - \pi )$ where $a$ and $b$ are integers.\\
\includegraphics[max width=\textwidth, alt={}, center]{1e2fdd33-afa4-486f-a9e2-1d425ed14eee-14_2492_1721_217_150}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2023 Q8 [14]}}