| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | Link Poisson to exponential |
| Difficulty | Standard +0.8 This question requires understanding the conceptual link between Poisson and exponential distributions, then applying standard exponential distribution techniques. Part (i) demands genuine insight into why P(X>x) relates to the Poisson parameter, which is non-trivial. Parts (ii)-(iii) are routine calculations once the connection is established, but the multi-step nature and the need to bridge two distributions elevates this above average difficulty. |
| 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 cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X > x) = P(\text{zero flaws in } x \text{ m})\) | B1 | Relate \(P(X > x)\) to number of flaws (AEF) |
| \(= P_0(0.8x) = e^{-0.8x}\) A.G. | B1 | Relate this to Poisson distribution (AEF) |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - P_0(0.8 \times 4) = 1 - e^{-3.2}\) | Find \(P(\text{number of flaws} \geq 1)\) | |
| \(= 1 - 0.0408 = 0.959\) | M1 A1 | M0 if "\(1-\)" omitted |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(x) = P(X \leq x) = 1 - P(X > x) = 1 - e^{-0.8x}\) | B1 | Find or state distribution function \(F(x)\) |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = dF/dx = 0.8e^{-0.8x}\) | M1 A1 | Find or state probability density function \(f(x)\) |
| Total: 2 marks | S.R. Deduct A1 if (a), (b) interchanged |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(Q) = 1 - e^{-0.8Q} = \frac{1}{4}\) *or* \(\frac{3}{4}\) | M1 | Formulate equation for either quartile value \(Q\) |
| \(Q_1 = 1.2\ln 4/3\ [= 0.360]\) | A1 | Find lower quartile \(Q_1\) (AEF) |
| \(Q_3 = 1.2\ln 4\ [= 1.733]\) | A1 | Find upper quartile \(Q_3\) (AEF) |
| \(Q_3 - Q_1\ [= 1.2\ln 3] = 1.37\) | A1 | Find interquartile range (allow \(Q_1 - Q_3\)) |
| Total: 4 marks |
## Question 9(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X > x) = P(\text{zero flaws in } x \text{ m})$ | B1 | Relate $P(X > x)$ to number of flaws (AEF) |
| $= P_0(0.8x) = e^{-0.8x}$ **A.G.** | B1 | Relate this to Poisson distribution (AEF) |
| **Total: 2 marks** | | |
## Question 9(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - P_0(0.8 \times 4) = 1 - e^{-3.2}$ | | Find $P(\text{number of flaws} \geq 1)$ |
| $= 1 - 0.0408 = 0.959$ | M1 A1 | M0 if "$1-$" omitted |
| **Total: 2 marks** | | |
## Question 9(iii)(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = P(X \leq x) = 1 - P(X > x) = 1 - e^{-0.8x}$ | B1 | Find or state distribution function $F(x)$ |
| **Total: 1 mark** | | |
## Question 9(iii)(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = dF/dx = 0.8e^{-0.8x}$ | M1 A1 | Find or state probability density function $f(x)$ |
| **Total: 2 marks** | | S.R. Deduct A1 if (a), (b) interchanged |
## Question 9(iii)(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(Q) = 1 - e^{-0.8Q} = \frac{1}{4}$ *or* $\frac{3}{4}$ | M1 | Formulate equation for either quartile value $Q$ |
| $Q_1 = 1.2\ln 4/3\ [= 0.360]$ | A1 | Find lower quartile $Q_1$ (AEF) |
| $Q_3 = 1.2\ln 4\ [= 1.733]$ | A1 | Find upper quartile $Q_3$ (AEF) |
| $Q_3 - Q_1\ [= 1.2\ln 3] = 1.37$ | A1 | Find interquartile range (allow $Q_1 - Q_3$) |
| **Total: 4 marks** | | |9 Cotton cloth is sold from long rolls of cloth. The number of flaws on a randomly chosen piece of cloth of length $a$ metres has a Poisson distribution with mean $0.8 a$. The random variable $X$ is the length of cloth, in metres, between two successive flaws.\\
(i) Explain why, for $x \geqslant 0 , \mathrm { P } ( X > x ) = \mathrm { e } ^ { - 0.8 x }$.\\
(ii) Find the probability that there is at least one flaw in a 4 metre length of cloth.\\
(iii) Find
\begin{enumerate}[label=(\alph*)]
\item the distribution function of $X$,
\item the probability density function of $X$,
\item the interquartile range of $X$.
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2015 Q9 [11]}}