| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | Find quartiles or percentiles |
| Difficulty | Standard +0.3 This is a straightforward application of standard exponential distribution techniques. Part (i) involves routine CDF manipulation to find quartiles and apply the outlier rule (Q3 + 1.5×IQR). Part (ii) requires integrating a simple exponential PDF (standard result), substituting values into the CDF, and applying the memoryless property. All techniques are textbook exercises with no novel insight required, making it slightly easier than average for S3 level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Increasing curve through \((0,0)\), in first quadrant only | M1 | |
| Asymptotic behaviour | A1 | |
| Asymptote labelled; condone absence of axis labels | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For UQ: \(G(u) = 0.75\), \(\therefore \left(1 + \dfrac{u}{200}\right)^{-2} = \dfrac{1}{4}\), \(\therefore u = 200\) | M1 A1 | Use of \(G(x)\) for either quartile. c.a.o. |
| For LQ: \(G(l) = 0.25\), \(\therefore \left(1+\dfrac{l}{200}\right)^{-2} = \dfrac{3}{4}\), \(\therefore l = 200\left(\dfrac{2}{\sqrt{3}}-1\right) = 30.94\ldots\) | A1 | c.a.o. |
| \(\therefore \text{IQR} = 200 - 30.94 = 169(.06)\) | M1 | UQ \(-\) LQ |
| For an outlier \(x > \text{UQ} + 1.5 \times \text{IQR} = 200 + 1.5 \times 169 = 453(.58) \approx 454\) (nearest hour) | M1 E1 | UQ \(+ 1.5 \times\) IQR. Answer given; must be obtained genuinely |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F(x) = \int_0^x \dfrac{1}{200} e^{\frac{-t}{200}}\, dt\) | M1 | Correct integral including limits |
| \(= \left[-e^{\frac{-t}{200}}\right]_0^x = \left(-e^{\frac{-x}{200}}\right) - \left(-e^0\right) = 1 - e^{\frac{-x}{200}}\) | A1 E1 | Correctly integrated. Answer given; must be shown convincingly. Condone omission of \(x<0\) part |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 50) = 1 - F(50) = e^{\frac{-50}{200}} = e^{-0.25}\) | M1 E1 | Use of \(1-F(x)\). Answer given; must be convincing \((= 0.7788(0))\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 400) = e^{\frac{-400}{200}} = 0.1353(35)\) | B1 | Accept any form |
| \(P(X > 450) = e^{\frac{-450}{200}} = 0.1053(99)\) | B1 | Accept any form |
| \(P(X > 450 \mid X > 400) = \dfrac{P(X>450)}{P(X>400)}\) | M1 | Conditional probability. Not \(P(X>50) \times P(X>400)\) unless clearly justified |
| \(= \dfrac{e^{\frac{-450}{200}}}{e^{\frac{-400}{200}}} = e^{\frac{-50}{200}} = e^{-0.25}\ (= 0.7788)\) | A1 | Accept division of decimals, 3dp or better. Accept 0.778 or 0.779 |
# Question 3:
## Part (i)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| Increasing curve through $(0,0)$, in first quadrant only | M1 | |
| Asymptotic behaviour | A1 | |
| Asymptote labelled; condone absence of axis labels | A1 | |
## Part (i)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| For UQ: $G(u) = 0.75$, $\therefore \left(1 + \dfrac{u}{200}\right)^{-2} = \dfrac{1}{4}$, $\therefore u = 200$ | M1 A1 | Use of $G(x)$ for either quartile. c.a.o. |
| For LQ: $G(l) = 0.25$, $\therefore \left(1+\dfrac{l}{200}\right)^{-2} = \dfrac{3}{4}$, $\therefore l = 200\left(\dfrac{2}{\sqrt{3}}-1\right) = 30.94\ldots$ | A1 | c.a.o. |
| $\therefore \text{IQR} = 200 - 30.94 = 169(.06)$ | M1 | UQ $-$ LQ |
| For an outlier $x > \text{UQ} + 1.5 \times \text{IQR} = 200 + 1.5 \times 169 = 453(.58) \approx 454$ (nearest hour) | M1 E1 | UQ $+ 1.5 \times$ IQR. Answer given; must be obtained genuinely |
## Part (ii)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| $F(x) = \int_0^x \dfrac{1}{200} e^{\frac{-t}{200}}\, dt$ | M1 | Correct integral including limits |
| $= \left[-e^{\frac{-t}{200}}\right]_0^x = \left(-e^{\frac{-x}{200}}\right) - \left(-e^0\right) = 1 - e^{\frac{-x}{200}}$ | A1 E1 | Correctly integrated. Answer given; must be shown convincingly. Condone omission of $x<0$ part |
## Part (ii)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 50) = 1 - F(50) = e^{\frac{-50}{200}} = e^{-0.25}$ | M1 E1 | Use of $1-F(x)$. Answer given; must be convincing $(= 0.7788(0))$ |
## Part (ii)(C)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 400) = e^{\frac{-400}{200}} = 0.1353(35)$ | B1 | Accept any form |
| $P(X > 450) = e^{\frac{-450}{200}} = 0.1053(99)$ | B1 | Accept any form |
| $P(X > 450 \mid X > 400) = \dfrac{P(X>450)}{P(X>400)}$ | M1 | Conditional probability. Not $P(X>50) \times P(X>400)$ unless clearly justified |
| $= \dfrac{e^{\frac{-450}{200}}}{e^{\frac{-400}{200}}} = e^{\frac{-50}{200}} = e^{-0.25}\ (= 0.7788)$ | A1 | Accept division of decimals, 3dp or better. Accept 0.778 or 0.779 |
---3 The time, in hours, until an electronic component fails is represented by the random variable $X$. In this question two models for $X$ are proposed.
\begin{enumerate}[label=(\roman*)]
\item In one model, $X$ has cumulative distribution function
$$\mathrm { G } ( x ) = \begin{cases} 0 & x \leqslant 0 \\ 1 - \left( 1 + \frac { x } { 200 } \right) ^ { - 2 } & x > 0 \end{cases}$$
(A) Sketch $\mathrm { G } ( x )$.\\
(B) Find the interquartile range for this model. Hence show that a lifetime of more than 454 hours (to the nearest hour) would be classed as an outlier.
\item In the alternative model, $X$ has probability density function
$$\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { 200 } \mathrm { e } ^ { - \frac { 1 } { 200 } x } & x > 0 \\ 0 & \text { elsewhere. } \end{cases}$$
(A) For this model show that the cumulative distribution function of $X$ is
$$\mathrm { F } ( x ) = \begin{cases} 0 & x \leqslant 0 \\ 1 - \mathrm { e } ^ { - \frac { 1 } { 200 } x } & x > 0 \end{cases}$$
(B) Show that $\mathrm { P } ( X > 50 ) = \mathrm { e } ^ { - 0.25 }$.\\
(C) It is observed that a particular component is still working after 400 hours. Find the conditional probability that it will still be working after a further 50 hours (i.e. after a total of 450 hours) given that it is still working after 400 hours.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2011 Q3 [18]}}