| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2018 |
| Session | September |
| Marks | 9 |
| Topic | Exponential Distribution |
| Type | Find parameter from given information |
| Difficulty | Standard +0.3 This is a straightforward application of standard exponential distribution properties. Part (i) uses the direct relationship between mean and λ (mean = 1/λ), part (ii) is a routine probability calculation using the exponential CDF, and part (iii) requires sketching a truncated distribution. All parts involve recall and basic application rather than problem-solving or novel insight, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| DR: \(\int_0^x \lambda e^{-\lambda x}dx\) | M1 | 1.1a |
| \(= [-xe^{-\lambda x}]_0^x - \int_0^x -e^{-\lambda x}dx\) | M1 | 1.1 |
| \(= \lim_{k \to \infty}\left[-xe^{-\lambda x} - \frac{1}{\lambda}e^{-\lambda x}\right]_0^k\) | E1 | 2.4 |
| \(= \frac{1}{\lambda}\) so \(\lambda = \frac{1}{5.0} = 0.2\) | A1 [4] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{20}^x 0.2e^{-0.2x}dx\) | M1 | 3.1b |
| \(= 0.0183156\ldots\) | A1 | 3.4 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(x)\) [graph showing correct shape for \(y = 2e^{-\lambda x}\), asymptotic to x-axis only] | B1 [1] | 1.1 |
| (b) [Any curve truncated at 15, More than half above their curve from part (a)] | M1, A1 [2] | 3.5c, 3.3 |
## (i)
DR: $\int_0^x \lambda e^{-\lambda x}dx$ | M1 | 1.1a | Use $\int_0^x x2e^{-\lambda x}dx$
$= [-xe^{-\lambda x}]_0^x - \int_0^x -e^{-\lambda x}dx$ | M1 | 1.1 | Parts used explicitly, $e^{-\lambda x}$ integrated
$= \lim_{k \to \infty}\left[-xe^{-\lambda x} - \frac{1}{\lambda}e^{-\lambda x}\right]_0^k$ | E1 | 2.4 | or equivalent consideration of limits | accept $= 0 - \left[\frac{1}{0.2}e^{-0.2x}\right]_0^{\infty}$
$= \frac{1}{\lambda}$ so $\lambda = \frac{1}{5.0} = 0.2$ | A1 [4] | 1.1 | Correctly obtain $\frac{1}{\lambda}$ and $\lambda = 0.2$, www
## (ii)
$\int_{20}^x 0.2e^{-0.2x}dx$ | M1 | 3.1b | Or $1 - \int_0^{20} 0.2e^{-0.2x}dx$
$= 0.0183156\ldots$ | A1 | 3.4 | BC Awrt 0.0183
| [2] |
## (iii)
(a) $f(x)$ [graph showing correct shape for $y = 2e^{-\lambda x}$, asymptotic to x-axis only] | B1 [1] | 1.1 | Correct shape for $y = 2e^{-\lambda x}$, asymptotic to x-axis only
(b) [Any curve truncated at 15, More than half above their curve from part (a)] | M1, A1 [2] | 3.5c, 3.3 | Don't need vertical line i.e., awareness that total area is the same
---4 A survey is carried out into the length of time for which customers wait for a response on a telephone helpline. A statistician who is analysing the results of the survey starts by modelling the waiting time, $x$ minutes, by an exponential distribution with probability density function (PDF)
$$\mathrm { f } ( x ) = \begin{cases} \lambda \mathrm { e } ^ { - \lambda x } & x \geqslant 0 \\ 0 & x < 0 \end{cases}$$
\section*{(i) In this question you must show detailed reasoning.}
The mean waiting time is found to be 5.0 minutes. Show that $\lambda = 0.2$.\\
(ii) Use the model to calculate the probability that a customer has to wait longer than 20 minutes for a response.
In practice it is found that no customer waits for more than 15 minutes for a response. The statistician constructs an improved model to incorporate this fact.\\
(iii) On the diagram in the Printed Answer Booklet, sketch the following, labelling the curves clearly:
\begin{enumerate}[label=(\alph*)]
\item the PDF of the model using the exponential distribution,
\item a possible PDF for the improved model.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2018 Q4 [9]}}