| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a straightforward application of standard techniques: integrating an exponential PDF to find the CDF (routine integration), using F(m)=0.5 for the median (direct substitution), and applying the transformation formula for PDFs (standard Further Maths content). All three parts follow textbook procedures with no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(x) = \int f(x)\,\mathrm{d}x = -e^{-2x} + c\) | M1 | Find or state distribution function \(F(x)\) for \(x \geq 0\) |
| \(F(x) = 0\ (x < 0)\), \(1 - e^{-2x}\ (x \geq 0)\) | A1 | Use \(F(0) = 0\) to find \(F(x)\); Total: [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - e^{-2m} = \frac{1}{2}\), \(e^{2m} = 2\) | M1 | Find median value \(m\) from \(F(m) = \frac{1}{2}\) |
| \(m = \frac{1}{2}\ln 2\) or \(0.347\) | M1 A1 | Total: [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G(y) = P(Y < y) = P(e^X < y) = P(X < \ln y) = F(\ln y)\) | M1 A1 | Find or state \(G(y)\) from \(Y = e^X\) for \(x \geq 0\); allow \(<\) or \(\leq\) throughout |
| \(= 1 - e^{-2\ln y} [= 1 - 1/y^2]\) | ||
| \(g(y) = 2/y^3\) | A1 | Find \(g(y)\) by differentiation |
| for \(y \geq 1\) [\(g(y) = 0\) for \(y < 1\)] | A1 | State corresponding range of \(y\); Total: [4] |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = \int f(x)\,\mathrm{d}x = -e^{-2x} + c$ | M1 | Find or state distribution function $F(x)$ for $x \geq 0$ |
| $F(x) = 0\ (x < 0)$, $1 - e^{-2x}\ (x \geq 0)$ | A1 | Use $F(0) = 0$ to find $F(x)$; Total: [2] |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - e^{-2m} = \frac{1}{2}$, $e^{2m} = 2$ | M1 | Find median value $m$ from $F(m) = \frac{1}{2}$ |
| $m = \frac{1}{2}\ln 2$ or $0.347$ | M1 A1 | Total: [3] |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G(y) = P(Y < y) = P(e^X < y) = P(X < \ln y) = F(\ln y)$ | M1 A1 | Find or state $G(y)$ from $Y = e^X$ for $x \geq 0$; allow $<$ or $\leq$ throughout |
| $= 1 - e^{-2\ln y} [= 1 - 1/y^2]$ | | |
| $g(y) = 2/y^3$ | A1 | Find $g(y)$ by differentiation |
| for $y \geq 1$ [$g(y) = 0$ for $y < 1$] | A1 | State corresponding range of $y$; Total: [4] |
---8 The random variable $X$ has probability density function f given by
$$\mathrm { f } ( x ) = \begin{cases} 2 \mathrm { e } ^ { - 2 x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$
(i) Find the distribution function of $X$.\\
(ii) Find the median value of $X$.
The random variable $Y$ is defined by $Y = \mathrm { e } ^ { X }$.\\
(iii) Find the probability density function of $Y$.
\hfill \mbox{\textit{CAIE FP2 2016 Q8 [9]}}