| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a straightforward Further Maths Statistics question requiring standard techniques: using continuity of CDF to find k, setting F(x)=0.5 for the median, and integrating xf(x) for the mean. All steps are routine applications of definitions with simple exponential algebra. Slightly above average difficulty due to algebraic manipulation of logarithms and the integration by parts likely needed for part (c), but no novel insight required. |
| Spec | 5.02b Expectation and variance: discrete random variables5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| 8 | Show that the mean of \(X\) is \(p - \frac { q } { \ln 2 }\), where \(p\) and \(q\) are integers to be found. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{5k} - 1 = 1\) | M1 (1.1a) | Forms a correct equation |
| \(e^{5k} = 2 \Rightarrow 5k = \ln 2 \Rightarrow k = \frac{1}{5}\ln 2\) | R1 (2.1) | Completes reasoned argument to obtain correct exact value of \(k = \frac{1}{5}\ln 2\). Minimum evidence required as shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{\frac{m}{5}\ln 2} - 1 = 0.5\) | M1 (1.1a) | Forms a correct equation |
| \(e^{\frac{m}{5}\ln 2} = 1.5 \Rightarrow \frac{m}{5}\ln 2 = \ln 1.5\) | M1 (1.1a) | Uses logarithms to correctly rearrange into form \(pm = q\) where \(p\) and \(q\) are constants and \(m\) is their median |
| \(\frac{m}{5}\ln 2 = \ln 3 - \ln 2 \Rightarrow m = 5\frac{\ln 3}{\ln 2} - 5\) | R1 (2.1) | Completes reasoned argument to obtain correct exact value of median \(= 5\frac{\ln 3}{\ln 2} - 5\). \(\frac{1}{5}\ln 2\) may be left as \(k\) until final line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \left(\frac{1}{5}\ln 2\right)e^{\frac{x}{5}\ln 2}\); \(E(X) = \frac{1}{5}\ln 2\int_0^5 x e^{\frac{x}{5}\ln 2}\,dx\) | M1 (3.1a) | Differentiates \(F(x)\) to find pdf and obtains function of form \(Ae^{kx}\) where \(A\) is non-zero constant |
| \(= \left[xe^{\frac{x}{5}\ln 2}\right]_0^5 - \int_0^5 e^{\frac{x}{5}\ln 2}\,dx\) | M1 (1.1a) | Forms integral of form \(A\int xe^{kx}\,dx\) where \(A\) is non-zero constant. Condone missing \(dx\) |
| \(= \left[xe^{\frac{x}{5}\ln 2} - \frac{5}{\ln 2}e^{\frac{x}{5}\ln 2}\right]_0^5\) | M1 (3.1a) | Uses integration by parts the correct way round to rearrange integral. Condone one slip provided intent is clear |
| Correct integrated function, possibly in terms of \(k\) and \(x\) | A1 (1.1b) | Components of integrated function may be seen on different lines |
| \(= \left(10 - \frac{10}{\ln 2}\right) - \left(-\frac{5}{\ln 2}\right)\) | M1 (1.1a) | Substitutes limits of 0 and 5 and subtracts correct way round |
| \(E(X) = 10 - \frac{5}{\ln 2}\) | R1 (2.1) | Completes reasoned argument. \(\frac{1}{5}\ln 2\) may be left as \(k\) until final line |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{5k} - 1 = 1$ | M1 (1.1a) | Forms a correct equation |
| $e^{5k} = 2 \Rightarrow 5k = \ln 2 \Rightarrow k = \frac{1}{5}\ln 2$ | R1 (2.1) | Completes reasoned argument to obtain correct exact value of $k = \frac{1}{5}\ln 2$. Minimum evidence required as shown |
---
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{\frac{m}{5}\ln 2} - 1 = 0.5$ | M1 (1.1a) | Forms a correct equation |
| $e^{\frac{m}{5}\ln 2} = 1.5 \Rightarrow \frac{m}{5}\ln 2 = \ln 1.5$ | M1 (1.1a) | Uses logarithms to correctly rearrange into form $pm = q$ where $p$ and $q$ are constants and $m$ is their median |
| $\frac{m}{5}\ln 2 = \ln 3 - \ln 2 \Rightarrow m = 5\frac{\ln 3}{\ln 2} - 5$ | R1 (2.1) | Completes reasoned argument to obtain correct exact value of median $= 5\frac{\ln 3}{\ln 2} - 5$. $\frac{1}{5}\ln 2$ may be left as $k$ until final line |
---
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \left(\frac{1}{5}\ln 2\right)e^{\frac{x}{5}\ln 2}$; $E(X) = \frac{1}{5}\ln 2\int_0^5 x e^{\frac{x}{5}\ln 2}\,dx$ | M1 (3.1a) | Differentiates $F(x)$ to find pdf and obtains function of form $Ae^{kx}$ where $A$ is non-zero constant |
| $= \left[xe^{\frac{x}{5}\ln 2}\right]_0^5 - \int_0^5 e^{\frac{x}{5}\ln 2}\,dx$ | M1 (1.1a) | Forms integral of form $A\int xe^{kx}\,dx$ where $A$ is non-zero constant. Condone missing $dx$ |
| $= \left[xe^{\frac{x}{5}\ln 2} - \frac{5}{\ln 2}e^{\frac{x}{5}\ln 2}\right]_0^5$ | M1 (3.1a) | Uses integration by parts the correct way round to rearrange integral. Condone one slip provided intent is clear |
| Correct integrated function, possibly in terms of $k$ and $x$ | A1 (1.1b) | Components of integrated function may be seen on different lines |
| $= \left(10 - \frac{10}{\ln 2}\right) - \left(-\frac{5}{\ln 2}\right)$ | M1 (1.1a) | Substitutes limits of 0 and 5 and subtracts correct way round |
| $E(X) = 10 - \frac{5}{\ln 2}$ | R1 (2.1) | Completes reasoned argument. $\frac{1}{5}\ln 2$ may be left as $k$ until final line |8 The continuous random variable $X$ has cumulative distribution function $\mathrm { F } ( x )$ where
$$\mathrm { F } ( x ) = \begin{cases} 0 & x = 0 \\ \mathrm { e } ^ { k x } - 1 & 0 \leq x \leq 5 \\ 1 & x > 5 \end{cases}$$
8
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 5 } \ln 2$\\[0pt]
[2 marks]\\
8
\item Show that the median of $X$ is $a \frac { \ln b } { \ln 2 } - c$, where $a , b$ and $c$ are integers to be found.\\
\begin{center}
\begin{tabular}{|l|l|}
\hline
8
\item & Show that the mean of $X$ is $p - \frac { q } { \ln 2 }$, where $p$ and $q$ are integers to be found. \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2022 Q8 [11]}}