| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Expectation of reciprocals and nonlinear functions |
| Difficulty | Challenging +1.8 This Further Maths statistics question requires integration of a piecewise continuous distribution with a discrete component, finding expectations of nonlinear functions (√|T|), and careful handling of absolute values across negative and positive domains. While the integration itself is standard, the combination of mixed distribution types and the specific nonlinear transformation elevates this beyond routine A-level work to require genuine problem-solving and careful algebraic manipulation. |
| Spec | 5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(c + 225k - k(15-12)^2 = 1\) | M1 | Identifies that total probability must sum to 1 |
| \(c + 216k = 1\); \(1-c = 216k\) | M1 | Substitutes \(t=12\) to form an equation in \(c\) and \(k\) |
| Rearranges to required result | R1 | Rearranges equation with correct rigorous workings to show required result AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(T) = (-2 \times c) + \int_0^{12} t \times f(t)\,dt\); \(f(t) = \dfrac{d}{dt}(225k - k(15-t)^2) = 2k(15-t)\) | M1 | Correct expression for \(E(T)\) (PI) |
| \(k = \dfrac{1}{240}\); \(E(T) = -0.2 + 2k\int_0^{12}15t - t^2\,dt\); \(0.9 = 216k\) so \(k = \dfrac{9}{2160} = \dfrac{1}{240}\); \(E(T) = -0.2 + \dfrac{2}{240}\left[\dfrac{15}{2}t^2 - \dfrac{1}{3}t^3\right]_0^{12}\) | M1 | Uses \(k=\dfrac{1}{240}\) and integrates with correct limits |
| \(= -0.2 + \dfrac{1}{120}\times 504 = -0.2 + 4.2\); \(E(T) = 4\) | A1 | Obtains correct value for \(E(T)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(\sqrt{\ | T\ | }) = 0.1\times\sqrt{2} + \int_0^{12}2k\times\sqrt{t}\times(15-t)\,dt\) |
| \(\dfrac{1}{120}\left(10\times 12^{\frac{3}{2}} - \dfrac{2}{5}\times 12^{\frac{5}{2}}\right) = \dfrac{12^{\frac{3}{2}}}{120}\left(10-\dfrac{24}{5}\right)\) | M1 | Integrates and uses limits |
| \(= \dfrac{12\times\sqrt{12}}{120}\times\dfrac{26}{5} = \dfrac{13\sqrt{12}}{25} = \dfrac{26\sqrt{3}}{25}\) | R1 | Completes clear correct rigorous workings to show required result AG |
| \(E(\sqrt{\ | T\ | }) = 0.1\times\sqrt{2} + \dfrac{26\sqrt{3}}{25} = \dfrac{\sqrt{2}}{10}+\dfrac{26\sqrt{3}}{25} = \dfrac{5\sqrt{2}+52\sqrt{3}}{50}\) |
## Question 6(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $c + 225k - k(15-12)^2 = 1$ | M1 | Identifies that total probability must sum to 1 |
| $c + 216k = 1$; $1-c = 216k$ | M1 | Substitutes $t=12$ to form an equation in $c$ and $k$ |
| Rearranges to required result | R1 | Rearranges equation with correct rigorous workings to show required result **AG** |
---
## Question 6(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(T) = (-2 \times c) + \int_0^{12} t \times f(t)\,dt$; $f(t) = \dfrac{d}{dt}(225k - k(15-t)^2) = 2k(15-t)$ | M1 | Correct expression for $E(T)$ (PI) |
| $k = \dfrac{1}{240}$; $E(T) = -0.2 + 2k\int_0^{12}15t - t^2\,dt$; $0.9 = 216k$ so $k = \dfrac{9}{2160} = \dfrac{1}{240}$; $E(T) = -0.2 + \dfrac{2}{240}\left[\dfrac{15}{2}t^2 - \dfrac{1}{3}t^3\right]_0^{12}$ | M1 | Uses $k=\dfrac{1}{240}$ and integrates with correct limits |
| $= -0.2 + \dfrac{1}{120}\times 504 = -0.2 + 4.2$; $E(T) = 4$ | A1 | Obtains correct value for $E(T)$ |
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(\sqrt{\|T\|}) = 0.1\times\sqrt{2} + \int_0^{12}2k\times\sqrt{t}\times(15-t)\,dt$ | M1 | Correct expression for $E(\sqrt{\|T\|})$ |
| $\dfrac{1}{120}\left(10\times 12^{\frac{3}{2}} - \dfrac{2}{5}\times 12^{\frac{5}{2}}\right) = \dfrac{12^{\frac{3}{2}}}{120}\left(10-\dfrac{24}{5}\right)$ | M1 | Integrates and uses limits |
| $= \dfrac{12\times\sqrt{12}}{120}\times\dfrac{26}{5} = \dfrac{13\sqrt{12}}{25} = \dfrac{26\sqrt{3}}{25}$ | R1 | Completes clear correct rigorous workings to show required result **AG** |
| $E(\sqrt{\|T\|}) = 0.1\times\sqrt{2} + \dfrac{26\sqrt{3}}{25} = \dfrac{\sqrt{2}}{10}+\dfrac{26\sqrt{3}}{25} = \dfrac{5\sqrt{2}+52\sqrt{3}}{50}$ | | Only award if completely correct, clear, easy to follow, and contains no slips |
---6 The random variable $T$ can take the value $T = - 2$ or any value in the range $0 \leq T < 12$
The distribution of $T$ is given by $\mathrm { P } ( T = - 2 ) = c , \mathrm { P } ( 0 \leq T \leq t ) = 225 k - k ( 15 - t ) ^ { 2 }$
6
\begin{enumerate}[label=(\alph*)]
\item (i) Show that $1 - c = 216 k$\\[0pt]
[3 marks]
6 (a) (ii) Given that $c = 0.1$, find the value of $\mathrm { E } ( T )$\\[0pt]
[3 marks]\\
6
\item Show that $\mathrm { E } ( \sqrt { | T | } ) = \frac { 5 \sqrt { 2 } + 52 \sqrt { 3 } } { 50 }$\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics Q6 [9]}}