| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | PDF from CDF |
| Difficulty | Standard +0.3 This is a straightforward application of differentiation to find PDF from CDF, followed by standard variance calculation using E(X²) - [E(X)]². The piecewise structure requires careful bookkeeping but involves only routine calculus techniques (differentiating polynomials, integrating polynomials). The 'show that' in part (b) provides the target answer, reducing computational anxiety. Slightly easier than average due to mechanical nature and clear structure. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \begin{cases} \frac{1}{10} & 1 < x \leq 6 \\ \frac{1}{45}x & 6 < x \leq 9 \\ 0 & \text{otherwise} \end{cases}\) | M1 (AO1.1a) | Differentiates to find either \(\frac{1}{10}\) or \(\frac{1}{45}x\) or equivalent |
| Fully defined \(f(x)\) | A1 (AO1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \int_1^6 \frac{1}{10}x\ dx + \int_6^9 \frac{1}{45}x^2\ dx = 5.55\) | M1 (AO1.1a) | Uses integrals of \(xf(x)\) to find \(E(X)\) |
| \(E(X^2) = \int_1^6 \frac{1}{10}x^2\ dx + \int_6^9 \frac{1}{45}x^3\ dx = \frac{437}{12}\) | M1 (AO1.1a) | Uses integrals of \(x^2f(x)\) to find \(E(X^2)\) |
| \(E(X) = 5.55\) or \(E(X^2) = \frac{437}{12}\) | A1 (AO1.1b) | Accept AWRT 36.4 for \(E(X^2)\); PI |
| \(\text{Var}(X) = \frac{437}{12} - (5.55)^2 = \frac{6737}{1200}\) | R1 (AO2.1) | Uses \(\text{Var}(X) = E(X^2) - (E(X))^2\) |
# Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \begin{cases} \frac{1}{10} & 1 < x \leq 6 \\ \frac{1}{45}x & 6 < x \leq 9 \\ 0 & \text{otherwise} \end{cases}$ | M1 (AO1.1a) | Differentiates to find either $\frac{1}{10}$ or $\frac{1}{45}x$ or equivalent |
| Fully defined $f(x)$ | A1 (AO1.1b) | |
---
# Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \int_1^6 \frac{1}{10}x\ dx + \int_6^9 \frac{1}{45}x^2\ dx = 5.55$ | M1 (AO1.1a) | Uses integrals of $xf(x)$ to find $E(X)$ |
| $E(X^2) = \int_1^6 \frac{1}{10}x^2\ dx + \int_6^9 \frac{1}{45}x^3\ dx = \frac{437}{12}$ | M1 (AO1.1a) | Uses integrals of $x^2f(x)$ to find $E(X^2)$ |
| $E(X) = 5.55$ or $E(X^2) = \frac{437}{12}$ | A1 (AO1.1b) | Accept AWRT 36.4 for $E(X^2)$; PI |
| $\text{Var}(X) = \frac{437}{12} - (5.55)^2 = \frac{6737}{1200}$ | R1 (AO2.1) | Uses $\text{Var}(X) = E(X^2) - (E(X))^2$ |
---5 The continuous random variable $X$ has cumulative distribution function
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c l }
0 & x \leq 1 \\
\frac { 1 } { 10 } x - \frac { 1 } { 10 } & 1 < x \leq 6 \\
\frac { 1 } { 90 } x ^ { 2 } + \frac { 1 } { 10 } & 6 < x \leq 9 \\
1 & x > 9
\end{array} \right.$$
5
\begin{enumerate}[label=(\alph*)]
\item Find the probability density function $\mathrm { f } ( x )$\\
5
\item Show that $\operatorname { Var } ( X ) = \frac { 6737 } { 1200 }$\\
\includegraphics[max width=\textwidth, alt={}, center]{3ef4c3fd-cbf0-4ac0-a072-a07d763fd50a-07_2488_1716_219_153}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2021 Q5 [6]}}