| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Sketch or interpret PDF graph |
| Difficulty | Moderate -0.3 This is a straightforward PDF question requiring students to read the mode from a graph and determine a piecewise linear function from visual information. Part (a) is trivial observation, while part (b) requires finding equations of line segments and ensuring total area equals 1, which is standard bookwork for this topic with minimal problem-solving demand. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mode \(= 1.75\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2k}{2} = 1 \Rightarrow k = 1\); maximum point of p.d.f. is \((1.75, 1)\) | M1 | Uses area under p.d.f. \(= 1\) to find \(y\)-coordinate of maximum |
| \(y = \frac{k}{1.75}x = \frac{1}{1.75}x = \frac{4}{7}x\) | M1 | Finds equation of one line using \(k\) and \(y\)-coordinate of maximum; may be in terms of \(k\) |
| \(y = -\frac{k}{0.25}(x-2) = -\frac{1}{0.25}(x-2) = -4x+8\) | A1F | Finds equations of both lines; follow through their \(k\) |
| \(f(x) = \begin{cases} \frac{4}{7}x & 0 \leq x < 1.75 \\ -4x+8 & 1.75 \leq x < 2 \\ 0 & \text{otherwise} \end{cases}\) | A1 | Fully defined p.d.f.; CAO |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mode $= 1.75$ | B1 | |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2k}{2} = 1 \Rightarrow k = 1$; maximum point of p.d.f. is $(1.75, 1)$ | M1 | Uses area under p.d.f. $= 1$ to find $y$-coordinate of maximum |
| $y = \frac{k}{1.75}x = \frac{1}{1.75}x = \frac{4}{7}x$ | M1 | Finds equation of one line using $k$ and $y$-coordinate of maximum; may be in terms of $k$ |
| $y = -\frac{k}{0.25}(x-2) = -\frac{1}{0.25}(x-2) = -4x+8$ | A1F | Finds equations of both lines; follow through their $k$ |
| $f(x) = \begin{cases} \frac{4}{7}x & 0 \leq x < 1.75 \\ -4x+8 & 1.75 \leq x < 2 \\ 0 & \text{otherwise} \end{cases}$ | A1 | Fully defined p.d.f.; CAO |
---5 The diagram shows a graph of the probability density function of the random variable $X$.\\
\includegraphics[max width=\textwidth, alt={}, center]{313cd5ce-07ff-4781-a134-565b8b221145-05_574_1086_406_479}
5
\begin{enumerate}[label=(\alph*)]
\item State the mode of $X$.\\
5
\item Find the probability density function of $X$.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2018 Q5 [5]}}