| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2018 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Direct probability from given distribution |
| Difficulty | Standard +0.3 This is a straightforward application of expectation properties with independent variables. Students need to recognize E(X² + Y²) = E(X²) + E(Y²), calculate E(X²) from the given discrete distribution (simple weighted sum), and integrate to find E(Y²) using a standard power function. All techniques are routine for Further Maths statistics with no conceptual challenges or novel insights required. |
| Spec | 5.02b Expectation and variance: discrete random variables5.03c Calculate mean/variance: by integration5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(\boldsymbol { x }\) | 1 | 2 | 4 | 9 |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.2 | 0.4 | 0.35 | 0.05 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X^2) = 1^2 \times 0.2 + 2^2 \times 0.4 + 4^2 \times 0.35 + 9^2 \times 0.05 = 11.45\) or \(\frac{229}{20}\) | M1 | Evaluates \(E(X^2)\) by calculating \(1^2 \times 0.2 + 2^2 \times 0.4 + 4^2 \times 0.35 + 9^2 \times 0.05\) |
| \(E(Y^2) = \int_{0}^{4} y^2 \times \frac{1}{64}y^3 \, dy = \frac{1}{64}\left[\frac{1}{6}y^6\right]_0^4 = \left[\frac{1}{384}y^6\right]_0^4 = \frac{32}{3}\) | M1 | Evaluates \(E(Y^2)\) by integrating \(\int y^2 \times f(y) \, dy\); must see the integral |
| \(E(X^2 + Y^2) = E(X^2) + E(Y^2) = 11.45 + \frac{32}{3}\) | A1 | Finds \(E(X^2) = 11.45\) and \(E(Y^2) = \frac{32}{3}\) |
| \(= \frac{1327}{60}\) | R1 | Uses \(E(X^2 + Y^2) = E(X^2) + E(Y^2)\) to show result is \(\frac{1327}{60}\) AG; mark awarded for completely correct, clear solution with no slips |
| Total: 4 |
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X^2) = 1^2 \times 0.2 + 2^2 \times 0.4 + 4^2 \times 0.35 + 9^2 \times 0.05 = 11.45$ or $\frac{229}{20}$ | M1 | Evaluates $E(X^2)$ by calculating $1^2 \times 0.2 + 2^2 \times 0.4 + 4^2 \times 0.35 + 9^2 \times 0.05$ |
| $E(Y^2) = \int_{0}^{4} y^2 \times \frac{1}{64}y^3 \, dy = \frac{1}{64}\left[\frac{1}{6}y^6\right]_0^4 = \left[\frac{1}{384}y^6\right]_0^4 = \frac{32}{3}$ | M1 | Evaluates $E(Y^2)$ by integrating $\int y^2 \times f(y) \, dy$; must see the integral |
| $E(X^2 + Y^2) = E(X^2) + E(Y^2) = 11.45 + \frac{32}{3}$ | A1 | Finds $E(X^2) = 11.45$ and $E(Y^2) = \frac{32}{3}$ |
| $= \frac{1327}{60}$ | R1 | Uses $E(X^2 + Y^2) = E(X^2) + E(Y^2)$ to show result is $\frac{1327}{60}$ AG; mark awarded for completely correct, clear solution with no slips |
| **Total: 4** | | |3 The discrete random variable $X$ has the following probability distribution
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 1 & 2 & 4 & 9 \\
\hline
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & 0.2 & 0.4 & 0.35 & 0.05 \\
\hline
\end{tabular}
\end{center}
The continuous random variable $Y$ has the following probability density function
$$\mathrm { f } ( y ) = \begin{cases} \frac { 1 } { 64 } y ^ { 3 } & 0 \leq y \leq 4 \\ 0 & \text { otherwise } \end{cases}$$
Given that $X$ and $Y$ are independent, show that $\mathrm { E } \left( X ^ { 2 } + Y ^ { 2 } \right) = \frac { 1327 } { 60 }$\\
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2018 Q3 [4]}}