| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 13 |
| Topic | Exponential Distribution |
| Type | Derive CDF from PDF |
| Difficulty | Standard +0.8 This is a multi-part probability question requiring integration by substitution (recognizing the derivative structure), finding median/mode/percentiles from a non-standard pdf, and sketching. While the integration is straightforward once the substitution u=t²/5 is spotted, students must work with an unfamiliar exponential distribution and apply multiple statistical concepts. The modal value calculation (part iii) requires differentiation and further integration, making this more demanding than routine A-level statistics questions but accessible to strong Further Maths students. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Correct position, including origin as a point on sketch | B1 | |
| Correct form at infinity | B1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(t) = \int f(t)dt = -e^{-\frac{t^2}{9}} + c\) | M1 | |
| \(F(0) = 0 \Rightarrow c = 1\) \(\therefore F(t) = 1 - e^{-\frac{t^2}{9}}\) | M1 | |
| \(F(\text{median}) = 0.5 \Rightarrow \text{median} = 2.50\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(t) = \frac{2}{9}e^{-\frac{t^2}{9}} - \left(\frac{t}{9}\right)^2 e^{-\frac{t^2}{9}}\) | M1 | |
| \(\frac{2}{9}e^{-\frac{t^2}{9}}\left(1-\frac{t^2}{2}\right) = 0 \Rightarrow \text{mode} = \frac{3}{\sqrt{2}}\) (= 2.12) | M1, A1 | |
| \(P(T > \text{mode}) = 1 - F\left(\frac{3}{\sqrt{2}}\right) = e^{-\frac{1}{2}}\) (= 0.607) | M1, A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T > N) = 0.05 \Rightarrow e^{-\frac{N^2}{9}} = \frac{1}{20}\) | M1 | |
| \(\Rightarrow \frac{N^2}{9} = \ln 20\) (Any base for logs) | M1 | |
| \(\Rightarrow N = \sqrt{9\ln 20} = 5.19\) | A1 | 3 |
### (i)
Correct position, including origin as a point on sketch | B1 |
Correct form at infinity | B1 | **2**
### (ii)
$F(t) = \int f(t)dt = -e^{-\frac{t^2}{9}} + c$ | M1 |
$F(0) = 0 \Rightarrow c = 1$ $\therefore F(t) = 1 - e^{-\frac{t^2}{9}}$ | M1 |
$F(\text{median}) = 0.5 \Rightarrow \text{median} = 2.50$ | A1 | **3**
### (iii)
$f(t) = \frac{2}{9}e^{-\frac{t^2}{9}} - \left(\frac{t}{9}\right)^2 e^{-\frac{t^2}{9}}$ | M1 |
$\frac{2}{9}e^{-\frac{t^2}{9}}\left(1-\frac{t^2}{2}\right) = 0 \Rightarrow \text{mode} = \frac{3}{\sqrt{2}}$ (= 2.12) | M1, A1 |
$P(T > \text{mode}) = 1 - F\left(\frac{3}{\sqrt{2}}\right) = e^{-\frac{1}{2}}$ (= 0.607) | M1, A1 | **5**
### (iv)
$P(T > N) = 0.05 \Rightarrow e^{-\frac{N^2}{9}} = \frac{1}{20}$ | M1 |
$\Rightarrow \frac{N^2}{9} = \ln 20$ (Any base for logs) | M1 |
$\Rightarrow N = \sqrt{9\ln 20} = 5.19$ | A1 | **3**The length of time, in years, that a salesman keeps his company car may be modelled by the continuous random variable $T$. The probability density function of $T$ is given by
$$f(t) = \begin{cases}
\frac{2}{5}t e^{-\frac{t^2}{5}} & t \geq 0, \\
0 & \text{otherwise}.
\end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Sketch the graph of $f(t)$. [2]
\item Find the cumulative distribution function $F(t)$ and hence find the median value of $T$. [3]
\item Find the probability that $T$ is greater than the modal value of $T$. [5]
\item The probability that a randomly chosen salesman keeps his car longer than $N$ years is $0.05$. Find the value of $N$ correct to $3$ significant figures. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q12 [13]}}