| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a straightforward S3 question requiring standard techniques: integrating a piecewise pdf to find the upper quartile (Q3) by solving F(x) = 0.75, and computing E(X) and E(X²) using standard expectation formulas. The piecewise nature adds mild complexity but the integrals are routine (power functions), and the algebra is manageable. Slightly easier than average since it's purely procedural with no conceptual insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: \(\int_{q_3}^{4} \frac{1}{12}x\,dx = \frac{1}{4}\) or \(\int_{1}^{2} \frac{4}{3x^3}\,dx + \int_{2}^{q_3} \frac{1}{12}x\,dx = \frac{3}{4}\) | M1* | |
| \([x^2/24]\) OR \([-2/(3x^2)] + [x^2/24]\) | A1 | Either |
| \((16 - q_3^2)/24 = 1/4\) or \(1/3 + q_3^2/24 = 3/4\) | dep *M1 | Form equation and attempt to solve |
| \(q_3 = \sqrt{10}\) | A1 | 4 |
| If they find \(F(x)\): M1A1, M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X^2) = \int_{1}^{2} \frac{4}{3x}\,dx + \int_{2}^{4} \frac{x^3}{12}\,dx\) | ||
| \(E(X) = \int_{1}^{2} \frac{4}{3x^2}\,dx + \int_{2}^{4} \frac{x^2}{12}\,dx\) | M1 | Either correct |
| \(\left[\frac{4}{3}\ln x\right]_1^2 + \left[\frac{x^4}{48}\right]_2^4\) | A1 | |
| \(\left[\frac{-4}{3x}\right]_1^2 + \left[\frac{x^3}{36}\right]_2^4\) | A1 | |
| \(a = E(X^2)/E(X)\) | M1 | |
| \(a = 2.6659\), \(2.67\) | A1 | 5 |
# Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| EITHER: $\int_{q_3}^{4} \frac{1}{12}x\,dx = \frac{1}{4}$ or $\int_{1}^{2} \frac{4}{3x^3}\,dx + \int_{2}^{q_3} \frac{1}{12}x\,dx = \frac{3}{4}$ | M1* | |
| $[x^2/24]$ OR $[-2/(3x^2)] + [x^2/24]$ | A1 | Either |
| $(16 - q_3^2)/24 = 1/4$ or $1/3 + q_3^2/24 = 3/4$ | dep *M1 | Form equation and attempt to solve |
| $q_3 = \sqrt{10}$ | A1 | **4** | Accept to 3 SF |
| If they find $F(x)$: M1A1, M1A1 | | |
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# Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X^2) = \int_{1}^{2} \frac{4}{3x}\,dx + \int_{2}^{4} \frac{x^3}{12}\,dx$ | | |
| $E(X) = \int_{1}^{2} \frac{4}{3x^2}\,dx + \int_{2}^{4} \frac{x^2}{12}\,dx$ | M1 | Either correct |
| $\left[\frac{4}{3}\ln x\right]_1^2 + \left[\frac{x^4}{48}\right]_2^4$ | A1 | |
| $\left[\frac{-4}{3x}\right]_1^2 + \left[\frac{x^3}{36}\right]_2^4$ | A1 | |
| $a = E(X^2)/E(X)$ | M1 | |
| $a = 2.6659$, $2.67$ | A1 | **5** | Or exact value, $(3\ln 2)/5 + 9/4$ or equiv. |
---4 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 4 } { 3 x ^ { 3 } } & 1 \leqslant x < 2 \\ \frac { 1 } { 12 } x & 2 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
(i) Find the upper quartile of $X$.\\
(ii) Find the value of $a$ for which $\mathrm { E } \left( X ^ { 2 } \right) = a \mathrm { E } ( X )$.
\hfill \mbox{\textit{OCR S3 2006 Q4 [9]}}