| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Standard +0.3 This is a straightforward S3 question requiring integration of a piecewise PDF to find a constant (using ∫f(t)dt = 1) and then solving for a quartile. Both parts use standard techniques with no conceptual challenges—slightly easier than average due to the simple exponential function and clear structure. |
| Spec | 5.03a Continuous random variables: pdf and cdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^2 ae^{-1}\,dt + \int_2^\infty ae^{-\frac{1}{2}t}\,dt = 1\) | M1 | |
| \([ae^{-1}t] + [-2ae^{-\frac{1}{2}t}]\) | A1 | |
| \(\Rightarrow a = \frac{1}{4}e\) AG | A1 3 | Properly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_{q_3}^\infty \frac{1}{4}e^{1-\frac{1}{2}t}\,dt = \frac{1}{4}\) | M1 | Or \(\int_0^2 \frac{1}{4}\,dt + \int_2^{q_3}\frac{1}{4}e^{1-t/2}\,dt = \frac{3}{4}\) |
| \([-\frac{1}{2}/e^{1-\frac{1}{2}t}]\) | B1 | AEF |
| \(-\frac{1}{2}q_3 + 1 = -\ln 2\) | M1 | For taking logs (not \(\ln(-)\)) |
| \(\Rightarrow q_3 = 2(\ln 2 + 1)\) or \(3.39\) | A1 4 [7] | AEF |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^2 ae^{-1}\,dt + \int_2^\infty ae^{-\frac{1}{2}t}\,dt = 1$ | M1 | |
| $[ae^{-1}t] + [-2ae^{-\frac{1}{2}t}]$ | A1 | |
| $\Rightarrow a = \frac{1}{4}e$ AG | A1 **3** | Properly obtained |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_{q_3}^\infty \frac{1}{4}e^{1-\frac{1}{2}t}\,dt = \frac{1}{4}$ | M1 | Or $\int_0^2 \frac{1}{4}\,dt + \int_2^{q_3}\frac{1}{4}e^{1-t/2}\,dt = \frac{3}{4}$ |
| $[-\frac{1}{2}/e^{1-\frac{1}{2}t}]$ | B1 | AEF |
| $-\frac{1}{2}q_3 + 1 = -\ln 2$ | M1 | For taking logs (not $\ln(-)$) |
| $\Rightarrow q_3 = 2(\ln 2 + 1)$ or $3.39$ | A1 **4** [7] | AEF |
---3 The continuous random variable $T$ has probability density function given by
$$\mathrm { f } ( t ) = \begin{cases} 0 & t < 0 , \\ \frac { a } { \mathrm { e } } & 0 \leqslant t < 2 , \\ a \mathrm { e } ^ { - \frac { 1 } { 2 } t } & t \geqslant 2 , \end{cases}$$
where $a$ is a positive constant.\\
(i) Show that $a = \frac { 1 } { 4 } \mathrm { e }$.\\
(ii) Find the upper quartile of $T$.
\hfill \mbox{\textit{OCR S3 2011 Q3 [7]}}