| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a straightforward application of integration to find cumulative probabilities and percentiles from a piecewise pdf. Part (a) requires computing a definite integral over two pieces, part (b) involves solving for the lower quartile (similar process), and part (c) is a standard expectation calculation. While the piecewise function and algebraic expressions look complex, the techniques are routine for Further Maths statistics—no novel insight or particularly challenging manipulation is required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03d E(g(X)): general expectation formula |
| 5 (d) | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X < 1.8) = \int_0^1 x^3\,dx + \dfrac{9}{1696}\int_1^{1.8} x^3(x^2+1)\,dx\) | M1 | Uses at least one of \(\int x^3\,dx\) or \(\dfrac{9}{1696}\int x^3(x^2+1)\,dx\); condone missing \(dx\); PI |
| \(= \dfrac{1}{4} + 0.042\) | A1 | Forms both correct integrals with correct limits with no other integrals: \(\int_0^1 x^3\,dx\) and \(\dfrac{9}{1696}\int_1^{1.8} x^3(x^2+1)\,dx\); PI; or forms \(\dfrac{9}{1696}\int_{1.8}^{3} x^3(x^2+1)\,dx\) OE with no other integrals |
| \(= 0.292\) | A1 | AWRT 0.292 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Lower quartile \(= 1\) | B1 | Obtains the correct value of the lower quartile |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E\!\left(\dfrac{1}{X^2}\right) = \int_0^1 \dfrac{x^3}{x^2}\,dx + \int_1^3 \dfrac{9x^3(x^2+1)}{1696x^2}\,dx\) | M1 | Uses at least one of the correct integrals for \(\int \dfrac{f(x)}{x^2}\,dx\) with any limits; condone missing \(dx\) |
| \(= \int_0^1 x\,dx + \dfrac{9}{1696}\int_1^3 x^3 + x\,dx\) | A1 | Obtains the correct value for one of the integrals OE |
| \(= \dfrac{1}{2} + \dfrac{27}{212}\) | A1 | Obtains the correct value for both of the integrals OE |
| \(= \dfrac{133}{212}\) | R1 | Shows that \(E\!\left(\dfrac{1}{X^2}\right) = \dfrac{133}{212}\) by first showing \(\int_0^1 \dfrac{x^3}{x^2}\,dx = \dfrac{1}{2}\) OE and \(\int_1^3 \dfrac{9x^3(x^2+1)}{1696x^2}\,dx = \dfrac{27}{212}\) OE and adding them together |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y) = 3\) | B1 | Obtains the correct value of \(E(Y)\); may be unsimplified |
| \(E\!\left(\dfrac{1}{X^2} + Y\right) = E\!\left(\dfrac{1}{X^2}\right) + E(Y) = \dfrac{133}{212} + 3\) | M1 | Uses the formula \(E\!\left(\dfrac{1}{X^2}+Y\right) = E\!\left(\dfrac{1}{X^2}\right)+E(Y)\) to obtain \(\dfrac{133}{212}\) + their \(E(Y)\) |
| \(E\!\left(\dfrac{1}{X^2} + Y\right) = \dfrac{769}{212}\) | A1 | Obtains the correct exact value |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X < 1.8) = \int_0^1 x^3\,dx + \dfrac{9}{1696}\int_1^{1.8} x^3(x^2+1)\,dx$ | M1 | Uses at least one of $\int x^3\,dx$ or $\dfrac{9}{1696}\int x^3(x^2+1)\,dx$; condone missing $dx$; PI |
| $= \dfrac{1}{4} + 0.042$ | A1 | Forms both correct integrals with correct limits with no other integrals: $\int_0^1 x^3\,dx$ and $\dfrac{9}{1696}\int_1^{1.8} x^3(x^2+1)\,dx$; PI; or forms $\dfrac{9}{1696}\int_{1.8}^{3} x^3(x^2+1)\,dx$ OE with no other integrals |
| $= 0.292$ | A1 | AWRT 0.292 |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Lower quartile $= 1$ | B1 | Obtains the correct value of the lower quartile |
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E\!\left(\dfrac{1}{X^2}\right) = \int_0^1 \dfrac{x^3}{x^2}\,dx + \int_1^3 \dfrac{9x^3(x^2+1)}{1696x^2}\,dx$ | M1 | Uses at least one of the correct integrals for $\int \dfrac{f(x)}{x^2}\,dx$ with any limits; condone missing $dx$ |
| $= \int_0^1 x\,dx + \dfrac{9}{1696}\int_1^3 x^3 + x\,dx$ | A1 | Obtains the correct value for one of the integrals OE |
| $= \dfrac{1}{2} + \dfrac{27}{212}$ | A1 | Obtains the correct value for both of the integrals OE |
| $= \dfrac{133}{212}$ | R1 | Shows that $E\!\left(\dfrac{1}{X^2}\right) = \dfrac{133}{212}$ by first showing $\int_0^1 \dfrac{x^3}{x^2}\,dx = \dfrac{1}{2}$ OE and $\int_1^3 \dfrac{9x^3(x^2+1)}{1696x^2}\,dx = \dfrac{27}{212}$ OE and adding them together |
## Question 5(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y) = 3$ | B1 | Obtains the correct value of $E(Y)$; may be unsimplified |
| $E\!\left(\dfrac{1}{X^2} + Y\right) = E\!\left(\dfrac{1}{X^2}\right) + E(Y) = \dfrac{133}{212} + 3$ | M1 | Uses the formula $E\!\left(\dfrac{1}{X^2}+Y\right) = E\!\left(\dfrac{1}{X^2}\right)+E(Y)$ to obtain $\dfrac{133}{212}$ + their $E(Y)$ |
| $E\!\left(\dfrac{1}{X^2} + Y\right) = \dfrac{769}{212}$ | A1 | Obtains the correct exact value |
---5 The continuous random variable $X$ has probability density function
$$f ( x ) = \begin{cases} x ^ { 3 } & 0 < x \leq 1 \\ \frac { 9 } { 1696 } x ^ { 3 } \left( x ^ { 2 } + 1 \right) & 1 < x \leq 3 \\ 0 & \text { otherwise } \end{cases}$$
5
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X < 1.8 )$, giving your answer to three decimal places.\\[0pt]
[3 marks]\\
5
\item Find the lower quartile of $X$\\
\begin{center}
\begin{tabular}{|l|l|}
\hline
5 (d) & 5
\item Show that $\mathrm { E } \left( \frac { 1 } { X ^ { 2 } } \right) = \frac { 133 } { 212 }$ \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2022 Q5 [11]}}