| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | PDF with multiple constants |
| Difficulty | Standard +0.3 This is a standard S3 PDF question requiring integration to find constants using continuity and the total probability condition, followed by calculating an expectation. The piecewise structure and E(1/X) calculation add mild complexity, but the techniques are routine for this module with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (Continuity at \(x=1\) gives) \(a = b\) | B1 | \(a=b\) seen or implied, eg by \(a=1.2,\ b=1.2\), even from two separate areas eg 0.5, 0.5 |
| \(\int_0^1 ax\,dx + \int_1^2 b(2-x)^2\,dx = 1\) | M1 | |
| \(\frac{a}{2},\left[\frac{-b}{3}(2-x)^3\right]_1^2\ (=1)\) oe both seen | B1 | Allow \(\left[\frac{b}{3}(2-x)^3\right]_1^2\). Allow without limits \(\ldots + b\left[4x - 2x^2 + \frac{x^3}{3}\right]_1^2\) |
| \(a/2 + b/3 = 1\) | B1 | cwo |
| solving gives \(a = b = 6/5\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \frac{6}{5}\,dx + \int_1^2 \frac{6}{5}\!\left(\frac{4}{x} - 4 + x\right)dx\) aef | M1 | From \(\int \frac{f(x)}{x}\,dx\), ft \(a,b\) for M1A1 |
| \(= [6x/5 + [(6/5)(4\ln x - 4x + x^2/2)]\) | A1ft | allow \(a + b(4\ln 2 - \frac{5}{2})\). Either their \(a/b\) or letters |
| \(= (24/5)\ln 2 - 9/5\) AEF or \(1.53\) (3SF) | A1 | |
| [3] |
# Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| (Continuity at $x=1$ gives) $a = b$ | B1 | $a=b$ seen or implied, eg by $a=1.2,\ b=1.2$, even from two separate areas eg 0.5, 0.5 |
| $\int_0^1 ax\,dx + \int_1^2 b(2-x)^2\,dx = 1$ | M1 | |
| $\frac{a}{2},\left[\frac{-b}{3}(2-x)^3\right]_1^2\ (=1)$ oe both seen | B1 | Allow $\left[\frac{b}{3}(2-x)^3\right]_1^2$. Allow without limits $\ldots + b\left[4x - 2x^2 + \frac{x^3}{3}\right]_1^2$ |
| $a/2 + b/3 = 1$ | B1 | cwo |
| solving gives $a = b = 6/5$ | A1 | |
| | **[5]** | |
---
# Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \frac{6}{5}\,dx + \int_1^2 \frac{6}{5}\!\left(\frac{4}{x} - 4 + x\right)dx$ aef | M1 | From $\int \frac{f(x)}{x}\,dx$, ft $a,b$ for M1A1 |
| $= [6x/5 + [(6/5)(4\ln x - 4x + x^2/2)]$ | A1ft | allow $a + b(4\ln 2 - \frac{5}{2})$. Either their $a/b$ or letters |
| $= (24/5)\ln 2 - 9/5$ AEF or $1.53$ (3SF) | A1 | |
| | **[3]** | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{c4adc528-ae3f-4ea7-9420-d3e1068a85fe-2_524_796_1105_623}
The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} a x & 0 < x \leqslant 1 \\ b ( 2 - x ) ^ { 2 } & 1 < x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $a$ and $b$ are constants. The graph is shown in the above diagram.\\
(i) Find the values of $a$ and $b$.\\
(ii) Find the value of $\mathrm { E } \left( \frac { 1 } { X } \right)$.
\hfill \mbox{\textit{OCR S3 2013 Q3 [8]}}