| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Composite/applied transformation |
| Difficulty | Challenging +1.2 This is a standard S3 transformation question requiring derivation of a pdf through the CDF method, followed by a routine median calculation. While it involves multiple steps (finding area formula, deriving CDF, differentiating for pdf, finding median), these are all textbook techniques for continuous transformations with no novel insight required. The geometric setup is straightforward and the algebra is manageable. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(X = \frac{1}{4}S^2\) | B1 | |
| \(F(s) = \int_1^s \frac{8}{3s^3}\,ds = \left[-\frac{4}{3s^2}\right]_1^s\) | M1 | |
| \(= \frac{4}{3}(1 - 1/s^2)\) | A1 | Ignore range here |
| \(G(x) = P(X \leq x) = P(S \leq 2\sqrt{x}) = F(2\sqrt{x})\) | M1 | SR: B1 for \(G(x)=F(2\sqrt{x})\) without justification and with correct result ft F |
| \(= \frac{4}{3} - \frac{1}{3x}\) | A1 ft | |
| \(g(x) = \begin{cases} \frac{1}{3x^2} & \frac{1}{4} \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}\) | M1 | For \(G'(a)\) |
| B1 7 | For range |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| EITHER: \(G(m) = \frac{1}{2}\) | M1 | ft \(G(x)\) in (i) |
| \(\Rightarrow \frac{4}{3} - \frac{1}{3x} = \frac{1}{2}\) | A1 ft | CAO |
| \(\Rightarrow m = \frac{2}{5}\) | A1 | |
| OR: \(\int_{1/4}^m \frac{1}{3x^2}\,dx = \frac{1}{2}\) | M1 | Allow wrong \(\frac{1}{4}\) |
| \(\Rightarrow \left[-\frac{1}{3x}\right]_{1/4}^m = \frac{1}{2}\) | A1 | Allow wrong \(\frac{1}{4}\) |
| \(\Rightarrow m = \frac{2}{5}\) | A1 3 | CAO |
# Question 8:
## Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $X = \frac{1}{4}S^2$ | B1 | |
| $F(s) = \int_1^s \frac{8}{3s^3}\,ds = \left[-\frac{4}{3s^2}\right]_1^s$ | M1 | |
| $= \frac{4}{3}(1 - 1/s^2)$ | A1 | Ignore range here |
| $G(x) = P(X \leq x) = P(S \leq 2\sqrt{x}) = F(2\sqrt{x})$ | M1 | SR: B1 for $G(x)=F(2\sqrt{x})$ without justification and with correct result ft F |
| $= \frac{4}{3} - \frac{1}{3x}$ | A1 ft | |
| $g(x) = \begin{cases} \frac{1}{3x^2} & \frac{1}{4} \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$ | M1 | For $G'(a)$ |
| | B1 **7** | For range |
## Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| EITHER: $G(m) = \frac{1}{2}$ | M1 | ft $G(x)$ in (i) |
| $\Rightarrow \frac{4}{3} - \frac{1}{3x} = \frac{1}{2}$ | A1 ft | CAO |
| $\Rightarrow m = \frac{2}{5}$ | A1 | |
| OR: $\int_{1/4}^m \frac{1}{3x^2}\,dx = \frac{1}{2}$ | M1 | Allow wrong $\frac{1}{4}$ |
| $\Rightarrow \left[-\frac{1}{3x}\right]_{1/4}^m = \frac{1}{2}$ | A1 | Allow wrong $\frac{1}{4}$ |
| $\Rightarrow m = \frac{2}{5}$ | A1 **3** | CAO |8 The continuous random variable $S$ has probability density function given by
$$f ( s ) = \begin{cases} \frac { 8 } { 3 s ^ { 3 } } & 1 \leqslant s \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
An isosceles triangle has equal sides of length $S$, and the angle between them is $30 ^ { \circ }$ (see diagram).\\
(i) Find the (cumulative) distribution function of the area $X$ of the triangle, and hence show that the probability density function of $X$ is $\frac { 1 } { 3 x ^ { 2 } }$ over an interval to be stated.\\
(ii) Find the median value of $X$.
www.ocr.org.uk after the live examination series.\\
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\hfill \mbox{\textit{OCR S3 2010 Q8 [10]}}