| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | Derive CDF from PDF |
| Difficulty | Standard +0.3 This is a straightforward application of standard techniques for the exponential distribution. Part (i) requires integrating the PDF to find the CDF (a routine calculus exercise), and part (ii) involves finding the mean (λ=6) and median (6ln2) using standard formulas, then computing a probability using the CDF. While it requires multiple steps, each step follows directly from textbook methods with no problem-solving insight needed. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| State or find by integration \(F(x)\): \(F(x) = 1 - e^{-x/6}\) \((x \geq 0)\), \(0\) otherwise | M1 A1 | Total part: [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| State or find mean \(\mu\): \(\mu = 1/(1/6) = 6\) | B1 | |
| Find \(\pm P(m \leq X \leq \mu)\) [\(m = 4.16\) not reqd]: \(F(\mu) - \frac{1}{2} = 1 - e^{-1} - \frac{1}{2}\) | M1 A1 | |
| Required probability \(= 0.132\) | A1 | Total part: [4], Question total: [6] |
## Question 6(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| State or find by integration $F(x)$: $F(x) = 1 - e^{-x/6}$ $(x \geq 0)$, $0$ otherwise | M1 A1 | Total part: [2] |
## Question 6(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| State or find mean $\mu$: $\mu = 1/(1/6) = 6$ | B1 | |
| Find $\pm P(m \leq X \leq \mu)$ [$m = 4.16$ not reqd]: $F(\mu) - \frac{1}{2} = 1 - e^{-1} - \frac{1}{2}$ | M1 A1 | |
| Required probability $= 0.132$ | A1 | Total part: [4], Question total: [6] |
---6 The random variable $X$ has probability density function f given by
$$\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { 6 } \mathrm { e } ^ { - \frac { 1 } { 6 } x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$
Find\\
(i) the distribution function of $X$,\\
(ii) the probability that $X$ lies between the median and the mean.
\hfill \mbox{\textit{CAIE FP2 2012 Q6 [6]}}