| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Expectation of function of X |
| Difficulty | Standard +0.3 This is a straightforward application of expectation properties and integration for a continuous random variable. Finding E(2X^{-1} - 3) requires using linearity of expectation and computing E(X^{-1}), which is a standard technique in Further Maths statistics. The algebraic manipulation is routine, making this slightly easier than average for Further AS level. |
| Spec | 5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X>1) = \frac{4}{99}\int_1^3 12x - x^2 - x^3 \, dx\) | M1 | Forms integral with correct integrand (limits not needed) |
| \(= \frac{4}{99}\left[\frac{12x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4}\right]_1^3\) | A1 | Integrates correctly and applies correct limits correctly |
| \(= \frac{4}{99} \times \frac{58}{3} = \frac{232}{297}\) | A1 | Correct answer \(\frac{232}{297}\) or 0.781 (AWRT) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^{-1}) = \frac{4}{99}\int_0^3 x^{-1}(12x - x^2 - x^3)\, dx\) | M1 | Selects correct integral, can be unsimplified (limits not needed) |
| \(= \frac{4}{99}\int_0^3 (12 - x - x^2)\, dx\) | A1 | Obtains \(\frac{4}{99}\int_0^3(12-x-x^2)\,dx\) or \(\int_0^3\left(\frac{16}{33}-\frac{4}{99}x-\frac{4}{99}x^2\right)dx\) |
| \(= \frac{4}{99}\left[12x - \frac{x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{10}{11}\) | R1 | Shows \(E(X^{-1}) = \frac{10}{11}\); fully correct solution with correct notation required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(2X^{-1}-3) = 2E(X^{-1}) - 3 = 2\times\frac{10}{11} - 3\) | M1 | Applies \(E(aY+b) = aE(Y)+b\) or selects correct integral |
| \(= -\frac{13}{11}\) | A1 | Obtains \(-\frac{13}{11}\) or \(-1.18\) (AWRT) |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X>1) = \frac{4}{99}\int_1^3 12x - x^2 - x^3 \, dx$ | M1 | Forms integral with correct integrand (limits not needed) |
| $= \frac{4}{99}\left[\frac{12x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4}\right]_1^3$ | A1 | Integrates correctly and applies correct limits correctly |
| $= \frac{4}{99} \times \frac{58}{3} = \frac{232}{297}$ | A1 | Correct answer $\frac{232}{297}$ or 0.781 (AWRT) |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^{-1}) = \frac{4}{99}\int_0^3 x^{-1}(12x - x^2 - x^3)\, dx$ | M1 | Selects correct integral, can be unsimplified (limits not needed) |
| $= \frac{4}{99}\int_0^3 (12 - x - x^2)\, dx$ | A1 | Obtains $\frac{4}{99}\int_0^3(12-x-x^2)\,dx$ or $\int_0^3\left(\frac{16}{33}-\frac{4}{99}x-\frac{4}{99}x^2\right)dx$ |
| $= \frac{4}{99}\left[12x - \frac{x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{10}{11}$ | R1 | Shows $E(X^{-1}) = \frac{10}{11}$; fully correct solution with correct notation required |
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(2X^{-1}-3) = 2E(X^{-1}) - 3 = 2\times\frac{10}{11} - 3$ | M1 | Applies $E(aY+b) = aE(Y)+b$ or selects correct integral |
| $= -\frac{13}{11}$ | A1 | Obtains $-\frac{13}{11}$ or $-1.18$ (AWRT) |4
\begin{enumerate}[label=(\alph*)]
\item \(\text { Find } \mathrm { P } ( X > 1 )\) \\[0pt]
[3 marks] \\
4
\item \\[0pt]
[3 marks] \\
□\\
4
\item Find $\mathrm { E } \left( 2 X ^ { - 1 } - 3 \right)$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2019 Q4 [8]}}