| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 18 |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Challenging +1.8 This is a challenging Pre-U question requiring transformation of random variables using the Jacobian method, then finding median and expectation from a non-standard pdf. The lens law transformation is non-trivial, requiring careful manipulation to find the inverse function and its derivative. However, the steps are systematic once the method is recognized, and the integration for part (ii) is manageable with standard techniques. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(f_v(v) = f_u(g^{-1}(v)) \times | g^{-1}{}'(v) | \) — M1A1 (Formula; mod sign needed for A1) |
| \(= \dfrac{1}{40} \times \left | \dfrac{-1600}{(v-40)^2}\right | = \dfrac{40}{(v-40)^2}\) (AG) — A1 (Mod sign needed for A1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \left[40\ln | v-40 | - \dfrac{1600}{v-40}\right]_{60}^{80}\) — M1A1 |
**Question 6(i) and 6(ii)**
**(i)**
$f_u(u) = \dfrac{1}{40}$ — B1 (PDF of $u$)
$U = g^{-1}(V) = \dfrac{40V}{V - 40}$ — M1 ($U$ in terms of $V$)
$f_v(v) = f_u(g^{-1}(v)) \times |g^{-1}{}'(v)|$ — M1A1 (Formula; mod sign needed for A1)
$= \dfrac{1}{40} \times \left|\dfrac{-1600}{(v-40)^2}\right| = \dfrac{40}{(v-40)^2}$ (AG) — A1 (Mod sign needed for A1)
**[5]**
Or:
$F_u(u) = \dfrac{u - 80}{40}$; $U = \dfrac{40V}{V-40}$ — B1, M1
$\mathrm{P}(V < v) = \mathrm{P}\!\left(U > \dfrac{40V}{V-40}\right) = 1 - \mathrm{P}\!\left(U < \dfrac{40V}{V-40}\right)$ — M1 (Turn $F_U(u)$ into $F_V(v)$, allow no $1-$)
$= 2 - \dfrac{40}{v - 40}$ so $f_v(v) = \dfrac{40}{(v-40)^2}$ (AG) — A1A1 (Correct $F_V(v)$; correctly obtain AG)
**[5]**
**(ii)(a)**
$F(v) = 2 - \dfrac{40}{(v-40)}$, $\quad 60 \leq v \leq 80$ — M1 (Or by integration of pdf from 60 to median $= 0.5$; M1 needs limits or $c$)
$2 - \dfrac{40}{v-40} = \dfrac{1}{2} \Rightarrow v = \dfrac{200}{3}$ (OE) — A1
**[2]**
**(ii)(b)**
$\mathrm{E}(V) = \int_{60}^{80} \dfrac{40v}{(v-40)^2}\,\mathrm{d}v = \int_{60}^{80} \dfrac{40}{v-40} + \dfrac{1600}{(v-40)^2}\,\mathrm{d}v$ — M1, M1A1
Or by $x = v - 40$: $40\!\left[\ln x - \dfrac{40}{x}\right]_{20}^{40}$ — M1A1
$= \left[40\ln|v-40| - \dfrac{1600}{v-40}\right]_{60}^{80}$ — M1A1
$= 40\ln 2 + 40 \quad (= 67.7)$ — A1
**[6]**6 The object distance, $U \mathrm {~cm}$, and the image distance, $V \mathrm {~cm}$, for a convex lens of focal length 40 cm are related by the lens law
$$\frac { 1 } { U } + \frac { 1 } { V } = \frac { 1 } { 40 } .$$
The random variable $U$ is uniformly distributed over the interval $80 \leqslant u \leqslant 120$.\\
(i) Show that the probability density function of $V$ is given by
$$f ( v ) = \begin{cases} \frac { 40 } { ( v - 40 ) ^ { 2 } } & 60 \leqslant v \leqslant 80 \\ 0 & \text { otherwise } \end{cases}$$
(ii) Find
\begin{enumerate}[label=(\alph*)]
\item the median value of $V$,
\item the expected value of $V$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q6 [18]}}