| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | Link Poisson to exponential |
| Difficulty | Standard +0.3 This is a straightforward application of the relationship between Poisson and exponential distributions. The 'show that' part requires converting the Poisson rate (1.6 per 100m → 0.016 per metre) and recognizing P(X > x) = P(0 flaws in distance x) = e^{-λx}. Parts (i) and (ii) are routine calculations: setting F(x) = 0.5 for the median and computing 1 - F(50). While this is Further Maths content, it's a standard textbook exercise requiring only direct application of formulas with minimal problem-solving, making it slightly easier than average overall. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.03a Continuous random variables: pdf and cdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks |
|---|---|
| (ii) | Relate F(x) to Poisson distribution (ignore x |
| Answer | Marks |
|---|---|
| Find P(X ≥ 50) (or > 50): 1 – F(50) = e –0⋅8 = 0⋅449 M1 A1 | 4 |
| Answer | Marks |
|---|---|
| 2 | [9] |
Question 8:
--- 8
(i)
(ii) ---
8
(i)
(ii) | Relate F(x) to Poisson distribution (ignore x
< 0): F(x) = 1 – P(X > x)
= 1 – P(no flaws in length x) M1
= 1 – e –(x/100)1⋅6 = 1 – e –0⋅016x A.G. M1 A1
Equate F(x) to 1 – e –λx or f(x) to e –λx to find λ = 1/0⋅016 = 62⋅5 B1
mean λ:
Formulate eqn for median m of X: 1 – e –0⋅016m = ½ M1
Find value of m: m = – ln ½ / 0⋅016 = 43⋅3 M1 A1
Find P(X ≥ 50) (or > 50): 1 – F(50) = e –0⋅8 = 0⋅449 M1 A1 | 4
3
2 | [9]The number of flaws in a randomly chosen 100 metre length of ribbon is modelled by a Poisson distribution with mean 1.6. The random variable $X$ metres is the distance between two successive flaws. Show that the distribution function of $X$ is given by
$$\text{F}(x) = \begin{cases}
1 - e^{-0.016x} & x \geqslant 0, \\
0 & x < 0,
\end{cases}$$
and deduce that $X$ has a negative exponential distribution, stating its mean.
[4]
Find
\begin{enumerate}[label=(\roman*)]
\item the median distance between two successive flaws, [3]
\item the probability that there is a distance of at least 50 metres between two successive flaws. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2012 Q8 [9]}}