| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | State distribution and mean |
| Difficulty | Moderate -0.8 This is a straightforward question requiring recognition of the exponential distribution from its CDF and direct application of standard formulas. All parts involve routine calculations: identifying λ=0.6, finding mean=1/λ, computing P(X>4) from the CDF, and solving for the median. No problem-solving or novel insight required—pure recall and basic manipulation. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Negative exponential | B1 | Identify distribution of \(X\) |
| \(5/3\) or \(1.67\) | B1 | State mean of \(X\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 4) = 1 - F(4) = e^{-2.4} = 0.0907\) | M1 A1 | Find \(P(X > 4)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F(m)\) [or \(1 - F(m)\)] \(= \frac{1}{2}\) | M1 | State or use equation for median \(m\) of \(X\) |
| \(e^{-0.6m} = \frac{1}{2}\), \(m = (5/3)\ln 2\) or \(1.16\) | M1 A1 | Find value of \(m\) |
## Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Negative exponential | B1 | Identify distribution of $X$ |
| $5/3$ or $1.67$ | B1 | State mean of $X$ |
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 4) = 1 - F(4) = e^{-2.4} = 0.0907$ | M1 A1 | Find $P(X > 4)$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(m)$ [or $1 - F(m)$] $= \frac{1}{2}$ | M1 | State or use equation for median $m$ of $X$ |
| $e^{-0.6m} = \frac{1}{2}$, $m = (5/3)\ln 2$ or $1.16$ | M1 A1 | Find value of $m$ |
**Total: 7 marks**
---6 The random variable $X$ has distribution function F given by
$$\mathrm { F } ( x ) = \begin{cases} 1 - \mathrm { e } ^ { - 0.6 x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$
Identify the distribution of $X$ and state its mean.
Find\\
(i) $\mathrm { P } ( X > 4 )$,\\
(ii) the median of $X$.
\hfill \mbox{\textit{CAIE FP2 2013 Q6 [7]}}