| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| 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 structured multi-part question on transforming CDFs that follows a standard template taught in S3. Part (i) requires the routine transformation Y=1/X², part (ii) is straightforward differentiation to verify a given pdf, and part (iii) involves a standard expectation calculation. While it requires careful algebraic manipulation and understanding of the transformation method, it's a textbook-style question with clear signposting and no novel problem-solving required. |
| Spec | 5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(G(y) = P(Y \leq y) = P(X^2 \leq 1/y)\) [or \(P(X > 1/\sqrt{y})\)] \(= 1 - F(1/\sqrt{y})\) | M1 | May be implied by following line; Accept strict inequalities |
| \[G(y) = \begin{cases} 0 & y \leq 0 \\ y^2 & 0 \leq y \leq 1 \\ 1 & y > 1 \end{cases}\] | A1 | |
| A1 | ||
| A1 | 4 marks; Or \(F(x) = P(X \leq x) = P(Y \geq 1/x^2) = 1-(P(Y < 1/x^2)) = 1-G(y)\) ;etc; A1 A1 | |
| (ii) Differentiate their \(G(y)\) to obtain \(g(y) = 2y\) for \(0 < y \leq 1\) AG | M1 | |
| A1 | 2 marks; Only from \(G\) correctly | |
| (iii) \(\int_0^1 2y(\sqrt[5]{y})dy\) | M1 | Unsimplified, but with limits |
| \(= [6y^{7/5}/7]\) | B1 | OR: Find \(f(x)\), \(\int x^{-2/3}f(x)dx\); \(= 4(x^{14/3})/(14/3)]; \)\frac{6}{7}\(; B1A1; OR: Find \)H(z), Z = Y^{1/5}$ |
| \(= \frac{6}{7}\) | A1 | 3 marks |
(i) $G(y) = P(Y \leq y) = P(X^2 \leq 1/y)$ [or $P(X > 1/\sqrt{y})$] $= 1 - F(1/\sqrt{y})$ | M1 | May be implied by following line; Accept strict inequalities
$$G(y) = \begin{cases} 0 & y \leq 0 \\ y^2 & 0 \leq y \leq 1 \\ 1 & y > 1 \end{cases}$$ | A1 |
| A1 |
| A1 | 4 marks; Or $F(x) = P(X \leq x) = P(Y \geq 1/x^2) = 1-(P(Y < 1/x^2)) = 1-G(y)$ ;etc; A1 A1
(ii) Differentiate their $G(y)$ to obtain $g(y) = 2y$ for $0 < y \leq 1$ AG | M1 |
| A1 | 2 marks; Only from $G$ correctly
(iii) $\int_0^1 2y(\sqrt[5]{y})dy$ | M1 | Unsimplified, but with limits
$= [6y^{7/5}/7]$ | B1 | OR: Find $f(x)$, $\int x^{-2/3}f(x)dx$; $= 4(x^{14/3})/(14/3)]; $\frac{6}{7}$; B1A1; OR: Find $H(z), Z = Y^{1/5}$
$= \frac{6}{7}$ | A1 | 3 marks |7 The continuous random variable $X$ has (cumulative) distribution function given by
$$\mathrm { F } ( x ) = \begin{cases} 0 & x < 1 \\ 1 - \frac { 1 } { x ^ { 4 } } & x \geqslant 1 \end{cases}$$
(i) Find the (cumulative) distribution function, $\mathrm { G } ( y )$, of the random variable $Y$, where $Y = \frac { 1 } { X ^ { 2 } }$.\\
(ii) Hence show that the probability density function of $Y$ is given by
$$g ( y ) = \begin{cases} 2 y & 0 < y \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$
(iii) Find $\mathrm { E } ( \sqrt [ 3 ] { Y } )$.
\hfill \mbox{\textit{OCR S3 2007 Q7 [9]}}