| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Mixed continuous-discrete problems |
| Difficulty | Challenging +1.2 This question requires finding E(X³+Y) = E(X³) + E(Y) using independence. Students must differentiate the CDF to get the PDF, then integrate x³f(x) from 0 to 2, and calculate E(Y) from the table. While it involves multiple steps (differentiation, integration, expectation calculation), these are all standard A-level Further Maths techniques with no novel insight required. The algebra is moderately involved but straightforward. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(y\) | 2 | 7 | 13 | 19 |
| \(\mathrm { P } ( Y = y )\) | 0.5 | 0.1 | 0.1 | 0.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \frac{1}{124}(24x^2 + 24x + 6) = \frac{3}{62}(4x^2 + 4x + 1)\) | M1 | Differentiates the cumulative distribution function of \(X\) to obtain \(f(x)\) in the form \(ax^2 + bx + c\) |
| \(E(X^3) = \frac{3}{62}\int_0^2 x^3(4x^2 + 4x + 1)\, dx\) | M1 | Uses correct integral of the form \(\int x^3 f(x)\, dx\) with any limits to find \(E(X^3)\) |
| \(E(X^3) = \frac{542}{155}\) or AWRT 3.5 | A1 | Obtains \(E(X^3) = \frac{542}{155}\) |
| \(E(Y) = 2 \times 0.5 + 7 \times 0.1 + 13 \times 0.1 + 19 \times 0.3 = \frac{87}{10}\) | M1 | Obtains \(E(Y)\) by evaluating \(2 \times 0.5 + 7 \times 0.1 + 13 \times 0.1 + 19 \times 0.3\) |
| \(E(X^3 + Y) = E(X^3) + E(Y) = \frac{542}{155} + \frac{87}{10}\) | M1 | Uses formula \(E(X^3 + Y) = E(X^3) + E(Y)\) |
| \(E(X^3 + Y) = \frac{3781}{310}\) | R1 | Completes a rigorous method to correctly show that \(E(X^3 + Y) = \frac{3781}{310}\) |
## Question 9:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \frac{1}{124}(24x^2 + 24x + 6) = \frac{3}{62}(4x^2 + 4x + 1)$ | M1 | Differentiates the cumulative distribution function of $X$ to obtain $f(x)$ in the form $ax^2 + bx + c$ |
| $E(X^3) = \frac{3}{62}\int_0^2 x^3(4x^2 + 4x + 1)\, dx$ | M1 | Uses correct integral of the form $\int x^3 f(x)\, dx$ with any limits to find $E(X^3)$ |
| $E(X^3) = \frac{542}{155}$ or AWRT 3.5 | A1 | Obtains $E(X^3) = \frac{542}{155}$ |
| $E(Y) = 2 \times 0.5 + 7 \times 0.1 + 13 \times 0.1 + 19 \times 0.3 = \frac{87}{10}$ | M1 | Obtains $E(Y)$ by evaluating $2 \times 0.5 + 7 \times 0.1 + 13 \times 0.1 + 19 \times 0.3$ |
| $E(X^3 + Y) = E(X^3) + E(Y) = \frac{542}{155} + \frac{87}{10}$ | M1 | Uses formula $E(X^3 + Y) = E(X^3) + E(Y)$ |
| $E(X^3 + Y) = \frac{3781}{310}$ | R1 | Completes a rigorous method to correctly show that $E(X^3 + Y) = \frac{3781}{310}$ |
**Total: 6 marks**9 The continuous random variable $X$ has the cumulative distribution function shown below.
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c }
0 & x < 0 \\
\frac { 1 } { 62 } \left( 4 x ^ { 3 } + 6 x ^ { 2 } + 3 x \right) & 0 \leq x \leq 2 \\
1 & x > 2
\end{array} \right.$$
The discrete random variable $Y$ has the probability distribution shown below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$y$ & 2 & 7 & 13 & 19 \\
\hline
$\mathrm { P } ( Y = y )$ & 0.5 & 0.1 & 0.1 & 0.3 \\
\hline
\end{tabular}
\end{center}
The random variables $X$ and $Y$ are independent.\\
Find the exact value of $\mathrm { E } \left( X ^ { 3 } + Y \right)$.\\
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2020 Q9 [6]}}