| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Find quantiles from CDF |
| Difficulty | Standard +0.8 This S3 question requires finding a median from a non-standard CDF (part i), deriving a new CDF through transformation Y=1/X² and then finding its PDF (part ii), and computing an expectation using the relationship between X and the CDF (part iii). While the techniques are standard for S3, the transformation in part (ii) requires careful handling of the inverse relationship and range, and part (iii) requires recognizing how to use the CDF structure efficiently. This is moderately challenging for Further Maths statistics but still follows established methods. |
| Spec | 5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Solve \(\frac{4}{3}(1-\frac{1}{m^2}) = \frac{1}{2}\) | M1 | |
| Giving \(m = \sqrt{\frac{5}{3}}\) | A1 | 2 Or equivalent. 1.26, 1.265, \(\sqrt{10}/5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(G(y) = P(Y \leq y)\) or \(<\) | M1 | |
| \(= P(X \geq 1/\sqrt{y})\) | A1 | |
| \(= 1 - F(1/\sqrt{y})\) | M1 | |
| \(= 1 - ^1/_{4}(1-y)\) or \((4y)-1/3\) | A1 | |
| \(1 \leq 1/\sqrt{y} \leq 2\) \(\frac{1}{4} \leq y \leq 1\) | B1 | |
| \(g(y)=\begin{cases} 4/3 & 1/4 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}\) | B1 | 6 Ft G(y) |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER: \(E(2-2Y)\) | M1 | |
| \(= 2 - 2x^7/8\) | A1 | |
| A1 | ||
| OR \(2 \cdot \int_1^{16(3^x)}dx\) OR \(\int_2^{(2-2/x^2)}/3x^3)dx\) | M1 | |
| \(= 2 + [4/(3x^3)]\) | A1 | |
| \(= 3/4\) | A1 | 3(11) CAO AEF; From 2 - \(\int xf(x)dx\); \(\sqrt{f(x)}\); CAO AEF |
**5(i)**
Solve $\frac{4}{3}(1-\frac{1}{m^2}) = \frac{1}{2}$ | M1 |
Giving $m = \sqrt{\frac{5}{3}}$ | A1 | 2 Or equivalent. 1.26, 1.265, $\sqrt{10}/5$
**5(ii)**
$G(y) = P(Y \leq y)$ or $<$ | M1 |
$= P(X \geq 1/\sqrt{y})$ | A1 |
$= 1 - F(1/\sqrt{y})$ | M1 |
$= 1 - ^1/_{4}(1-y)$ or $(4y)-1/3$ | A1 |
$1 \leq 1/\sqrt{y} \leq 2$ $\frac{1}{4} \leq y \leq 1$ | B1 |
$g(y)=\begin{cases} 4/3 & 1/4 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$ | B1 | 6 Ft G(y)
**5(iii)**
EITHER: $E(2-2Y)$ | M1 |
$= 2 - 2x^7/8$ | A1 |
| A1 |
OR $2 \cdot \int_1^{16(3^x)}dx$ OR $\int_2^{(2-2/x^2)}/3x^3)dx$ | M1 |
$= 2 + [4/(3x^3)]$ | A1 |
$= 3/4$ | A1 | 3(11) CAO AEF; From 2 - $\int xf(x)dx$; $\sqrt{f(x)}$; CAO AEF
---5 The continuous random variable $X$ has (cumulative) distribution function given by
$$\mathrm { F } ( x ) = \begin{cases} 0 & x < 1 , \\ \frac { 4 } { 3 } \left( 1 - \frac { 1 } { x ^ { 2 } } \right) & 1 \leqslant x \leqslant 2 , \\ 1 & x > 2 . \end{cases}$$
(i) Find the median value of $X$.\\
(ii) Find the (cumulative) distribution function of $Y$, where $Y = \frac { 1 } { X ^ { 2 } }$, and hence find the probability density function of $Y$.\\
(iii) Evaluate $\mathrm { E } \left( 2 - \frac { 2 } { X ^ { 2 } } \right)$.
\hfill \mbox{\textit{OCR S3 2011 Q5 [11]}}