| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find parameter from median |
| Difficulty | Standard +0.3 This is a straightforward continuous probability distribution question requiring standard techniques: integrating to find k using the median condition (part a), then applying the same integral for a probability calculation (part b). Part (c) requires minimal conceptual insight about adjusting the domain. The integration of 1/x is routine A-level material, and the multi-step nature is typical rather than challenging. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_{1}^{200} \frac{k}{x}\,dx = 0.5\) | M1 | Forms integral with correct upper limit using \(P(X \leq 200) = 0.5\); condone missing \(dx\) |
| \(\left[k \ln x\right]_{1}^{200} = 0.5\) | M1 | Integrates correctly; condone missing limits |
| \(k = \dfrac{0.5}{\ln 200} = \dfrac{1}{2\ln 200}\) | R1 | Shows substitution of limits and solves; mark awarded for completely correct, clear solution with correct notation and no slips |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{1}{2\ln 200}\int_{1}^{2000} \frac{1}{x}\,dx = 0.717\) | M1 | States correct integral to find \(P(X < 2000)\); condone missing \(dx\) |
| \(0.717\) (AWRT) | A1 | Correct probability |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Set \(a = 2000\) and change value of \(k\) so that \(P(X < 2000) = 1\) | E1 | Explains model is restricted to values from 1 to 2000; implied by set \(a = 2000\) |
| Model with pdf unchanged for \((1, 2000)\) and probability \(1 -\) their (b) for 2000; or same pdf with different/increased \(k\) or different pdf so \(P(X < 2000) = 1\); implied by \(k = 1/\ln 2000\) or AWRT \(0.132\) | E1 |
# Question 5:
## Part 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_{1}^{200} \frac{k}{x}\,dx = 0.5$ | M1 | Forms integral with correct upper limit using $P(X \leq 200) = 0.5$; condone missing $dx$ |
| $\left[k \ln x\right]_{1}^{200} = 0.5$ | M1 | Integrates correctly; condone missing limits |
| $k = \dfrac{0.5}{\ln 200} = \dfrac{1}{2\ln 200}$ | R1 | Shows substitution of limits and solves; mark awarded for completely correct, clear solution with correct notation and no slips |
## Part 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{1}{2\ln 200}\int_{1}^{2000} \frac{1}{x}\,dx = 0.717$ | M1 | States correct integral to find $P(X < 2000)$; condone missing $dx$ |
| $0.717$ (AWRT) | A1 | Correct probability |
## Part 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Set $a = 2000$ and change value of $k$ so that $P(X < 2000) = 1$ | E1 | Explains model is restricted to values from 1 to 2000; implied by set $a = 2000$ |
| Model with pdf unchanged for $(1, 2000)$ and probability $1 -$ their (b) for 2000; or same pdf with different/increased $k$ or different pdf so $P(X < 2000) = 1$; implied by $k = 1/\ln 2000$ or AWRT $0.132$ | E1 | |
---5 An insurance company models the claims it pays out in pounds $( \pounds )$ with a random variable $X$ which has probability density function
$$f ( x ) = \begin{cases} \frac { k } { x } & 1 < x < a \\ 0 & \text { otherwise } \end{cases}$$
5
\begin{enumerate}[label=(\alph*)]
\item The median claim is $\pounds 200$\\
Show that $k = \frac { 1 } { 2 \ln 200 }$\\
5
\item Find $\mathrm { P } ( X < 2000 )$, giving your answer to three significant figures.\\
5
\item The insurance company finds that the maximum possible claim is $\pounds 2000$ and they decide to refine their probability density function.
Suggest how this could be done.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2019 Q5 [7]}}