| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Sketch or interpret PDF graph |
| Difficulty | Moderate -0.3 This is a straightforward piecewise PDF question requiring standard techniques: sketching a simple linear function, identifying the median by symmetry, computing E(T) via integration of two simple pieces, and applying variance transformation rules. All steps are routine for Further Maths students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks |
|---|---|
| Part 1: Draws horizontal line \((0, \frac{1}{3})\) to \((\frac{3}{2}, \frac{1}{3})\) and straight line joining \((\frac{3}{2}, \frac{1}{3})\) to \(t\)-axis | M1 |
| Part 2: Correct shape, accurate and fully labelled (endpoints at \(0\), \(\frac{3}{2}\), \(\frac{9}{2}\) on \(t\)-axis, height \(\frac{1}{3}\)) | A1 |
| Answer | Marks |
|---|---|
| Area under \(f(t)\) for \(0 \leq t \leq \frac{3}{2}\) is \(\frac{1}{2}\); median of \(t\) is \(\frac{3}{2}\) | B1 |
| Answer | Marks |
|---|---|
| Part 1: \(E(T) = \int_0^{\frac{3}{2}} \frac{1}{3}t \, dt + \int_{\frac{3}{2}}^{\frac{9}{2}} \frac{t(9-2t)}{18} \, dt\) | M1 |
| Part 2: \(= \left[\frac{1}{6}t^2\right]_0^{\frac{3}{2}} + \left[\frac{t^2}{4} - \frac{t^3}{27}\right]_{\frac{3}{2}}^{\frac{9}{2}} = \frac{13}{8}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Part 1: \(\text{Var}(T) = E(T^2) - (E(T))^2\) | M1 | |
| Part 2: \(\text{Var}(T) = \frac{15}{4} - \left(\frac{13}{8}\right)^2 = \frac{71}{64}\) | A1 | |
| Part 3: \(\text{Var}(4T-5) = 16\,\text{Var}(T) = \frac{71}{4}\) | A1F | Follow through 'their' \(\text{Var}(T)\) |
## Question 6(a)(i):
**Part 1:** Draws horizontal line $(0, \frac{1}{3})$ to $(\frac{3}{2}, \frac{1}{3})$ and straight line joining $(\frac{3}{2}, \frac{1}{3})$ to $t$-axis | M1 |
**Part 2:** Correct shape, accurate and fully labelled (endpoints at $0$, $\frac{3}{2}$, $\frac{9}{2}$ on $t$-axis, height $\frac{1}{3}$) | A1 |
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## Question 6(a)(ii):
Area under $f(t)$ for $0 \leq t \leq \frac{3}{2}$ is $\frac{1}{2}$; median of $t$ is $\frac{3}{2}$ | B1 |
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## Question 6(b)(i):
**Part 1:** $E(T) = \int_0^{\frac{3}{2}} \frac{1}{3}t \, dt + \int_{\frac{3}{2}}^{\frac{9}{2}} \frac{t(9-2t)}{18} \, dt$ | M1 |
**Part 2:** $= \left[\frac{1}{6}t^2\right]_0^{\frac{3}{2}} + \left[\frac{t^2}{4} - \frac{t^3}{27}\right]_{\frac{3}{2}}^{\frac{9}{2}} = \frac{13}{8}$ | A1 |
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## Question 6(b)(ii):
**Part 1:** $\text{Var}(T) = E(T^2) - (E(T))^2$ | M1 |
**Part 2:** $\text{Var}(T) = \frac{15}{4} - \left(\frac{13}{8}\right)^2 = \frac{71}{64}$ | A1 |
**Part 3:** $\text{Var}(4T-5) = 16\,\text{Var}(T) = \frac{71}{4}$ | A1F | Follow through 'their' $\text{Var}(T)$
---6 The continuous random variable $T$ has probability density function defined by
$$\mathrm { f } ( t ) = \left\{ \begin{array} { c c }
\frac { 1 } { 3 } & 0 \leq t \leq \frac { 3 } { 2 } \\
\frac { 9 - 2 t } { 18 } & \frac { 3 } { 2 } \leq t \leq \frac { 9 } { 2 } \\
0 & \text { otherwise }
\end{array} \right.$$
6
\begin{enumerate}[label=(\alph*)]
\item (i) Sketch this probability density function below.\\
\includegraphics[max width=\textwidth, alt={}, center]{6ccf7d1d-5a7b-47d1-b38e-c7e762204746-07_1009_1041_1073_520}
6 (a) (ii) State the median of $T$.
6
\item (i) Find $\mathrm { E } ( T )$\\[0pt]
[2 marks]\\
6 (b) (ii) Given that $\mathrm { E } \left( T ^ { 2 } \right) = \frac { 15 } { 4 }$, find $\operatorname { Var } ( 4 T - 5 )$ [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics Q6 [8]}}