| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF of transformed variable |
| Difficulty | Challenging +1.2 This is a standard transformation of variables question requiring finding the CDF through substitution and differentiation, followed by a routine expectation calculation. While it involves multiple steps and careful algebraic manipulation (including finding the range of Y and differentiating), the technique is a core S3/Further Stats topic with no novel insight required. Slightly above average difficulty due to the inverse transformation and need for precision, but well within standard Further Maths scope. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(x) = x^{3/2}\ (0 < x \leq 1)\) | B1 | |
| \(G(y) = P(Y \leq y) = P(1/\sqrt{X} \leq y) = P(X \geq 1/y^2)\) | M1 | |
| \(= 1 - F(1/y^2) = 1 - 1/y^3\ \text{ for } y \geq 1\) | M1, A1,B1 | B1 for \(y \geq 1\) seen anywhere |
| \((G(y) = 0 \text{ otherwise})\) | ||
| \(g(y) = G'(y) = 3/y^4\ \ y \geq 1\) AG, \((= 0 \text{ otherwise})\) | M1A1 | M1 for differentiating. Allow from \(-1/y^3\) |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\alpha)\ \int_1^{\infty} \frac{3}{y^2}\,dy = 3\) | M1A1 | Ft range from (i) for M1, but not \(\int_0^1 \frac{3}{y^2}\,dy\) |
| \((\beta)\ \int_0^1 \frac{3}{2\sqrt{x}}\,dx = 3\) | M1A1 | |
| [2] |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(x) = x^{3/2}\ (0 < x \leq 1)$ | B1 | |
| $G(y) = P(Y \leq y) = P(1/\sqrt{X} \leq y) = P(X \geq 1/y^2)$ | M1 | |
| $= 1 - F(1/y^2) = 1 - 1/y^3\ \text{ for } y \geq 1$ | M1, A1,B1 | B1 for $y \geq 1$ seen anywhere |
| $(G(y) = 0 \text{ otherwise})$ | | |
| $g(y) = G'(y) = 3/y^4\ \ y \geq 1$ AG, $(= 0 \text{ otherwise})$ | M1A1 | M1 for differentiating. Allow from $-1/y^3$ |
| | **[7]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\alpha)\ \int_1^{\infty} \frac{3}{y^2}\,dy = 3$ | M1A1 | Ft range from (i) for M1, but not $\int_0^1 \frac{3}{y^2}\,dy$ |
| $(\beta)\ \int_0^1 \frac{3}{2\sqrt{x}}\,dx = 3$ | M1A1 | |
| | **[2]** | |
---4 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 3 } { 2 } \sqrt { x } & 0 < x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$
The random variable $Y$ is given by $Y = \frac { 1 } { \sqrt { X } }$.\\
(i) Find the (cumulative) distribution function of $Y$, and hence show that its probability density function is given by
$$\mathrm { g } ( y ) = \frac { 3 } { y ^ { 4 } }$$
for a set of values of $y$ to be stated.\\
(ii) Find the value of $\mathrm { E } \left( Y ^ { 2 } \right)$.
\hfill \mbox{\textit{OCR S3 2013 Q4 [9]}}