| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF of transformed variable |
| Difficulty | Challenging +1.2 This S3 question requires finding the CDF of a transformed variable Y=1/X through careful manipulation of inequalities and substitution, then using the symmetry result to find expectations. While it involves multiple steps and requires understanding the relationship between CDFs under transformation, the techniques are standard for Further Maths S3 level. The algebraic manipulation is moderately demanding but follows established methods, making it above average difficulty but not exceptionally challenging. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G(y) = P(Y \leq y)\) | M1 | |
| \(= P(X \geq 1/y)\) | A1 | |
| \(= 1 - F(1/y)\) | M1 | |
| \(= (2y-1)/(y+1)\) | A1 | |
| For \(\frac{1}{2} \leq 1/y \leq 2 \Rightarrow \frac{1}{2} \leq y \leq 2\) | B1 | Seen |
| \(X\) and \(Y\) have identical distributions | B1 6 | |
| Alternative method: | SR: CDF not used | |
| \(y\) decreases with \(x\) | ||
| Use \(g(y) = f(x(y))\ | dx/dy\ | \) |
| \(f(x) = 3/(x+1)^2\) | M1A1 | |
| \(\ | dx/dy\ | = 1/y^2\) |
| \(g(y) = [3/(y^{-1}+1)^2][1/y^2] = 3/(y+1)^2\); for \(\frac{1}{2}\leq y \leq 2\) | M1A1B1 | |
| So \(X\) and \(Y\) have identical distributions | B1 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = F'(x) = 3/(x+1)^2,\quad \frac{1}{2}\leq x \leq 2\) | M1A1 | Must have range of \(x\); AEF Not if awarded in (i) |
| \(E(X+1) = \int_{1/2}^{2} \frac{3}{x+1}\,dx\) | M1 | |
| \(= 3\ln 2\ (2.08)\) | A1 | |
| \(E(1/X) = E(X)\) | M1 | |
| \(= 3\ln 2 - 1\ (1.08)\) | A1 6 [12] |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G(y) = P(Y \leq y)$ | M1 | |
| $= P(X \geq 1/y)$ | A1 | |
| $= 1 - F(1/y)$ | M1 | |
| $= (2y-1)/(y+1)$ | A1 | |
| For $\frac{1}{2} \leq 1/y \leq 2 \Rightarrow \frac{1}{2} \leq y \leq 2$ | B1 | Seen |
| $X$ and $Y$ have identical distributions | B1 **6** | |
| **Alternative method:** | | SR: CDF not used |
| $y$ decreases with $x$ | | |
| Use $g(y) = f(x(y))\|dx/dy\|$ | M1 | |
| $f(x) = 3/(x+1)^2$ | M1A1 | |
| $\|dx/dy\| = 1/y^2$ | B1 | |
| $g(y) = [3/(y^{-1}+1)^2][1/y^2] = 3/(y+1)^2$; for $\frac{1}{2}\leq y \leq 2$ | M1A1B1 | |
| So $X$ and $Y$ have identical distributions | B1 **8** | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = F'(x) = 3/(x+1)^2,\quad \frac{1}{2}\leq x \leq 2$ | M1A1 | Must have range of $x$; AEF Not if awarded in (i) |
| $E(X+1) = \int_{1/2}^{2} \frac{3}{x+1}\,dx$ | M1 | |
| $= 3\ln 2\ (2.08)$ | A1 | |
| $E(1/X) = E(X)$ | M1 | |
| $= 3\ln 2 - 1\ (1.08)$ | A1 **6** [12] | |
---6 The continuous random variable $X$ has (cumulative) distribution function given by
$$\mathrm { F } ( x ) = \begin{cases} 0 & x < \frac { 1 } { 2 } \\ \frac { 2 x - 1 } { x + 1 } & \frac { 1 } { 2 } \leqslant x \leqslant 2 \\ 1 & x > 2 . \end{cases}$$
(i) Given that $Y = \frac { 1 } { X }$, find the (cumulative) distribution function of $Y$, and deduce that $Y$ and $X$ have identical distributions.\\
(ii) Find $\mathrm { E } ( X + 1 )$ and deduce the value of $\mathrm { E } \left( \frac { 1 } { X } \right)$.
\hfill \mbox{\textit{OCR S3 2011 Q6 [12]}}