| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Verify CDF properties |
| Difficulty | Challenging +1.2 This is a multi-part S3 question requiring verification of CDF properties (routine), solving sin^4(t) = 0.25 for the quartile (straightforward), finding a transformed distribution (standard technique but requires careful working with the transformation Y = sin T), and computing an expectation using the derived pdf. While it involves several steps and the transformation requires some care, these are all standard S3 techniques without requiring novel insight or particularly complex reasoning. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F(0)=0\), \(F(\pi/2)=1\) | B1 | Consider both end-points |
| Increasing | B1 2 | Consider F between end-points, can be asserted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sin^4(Q_1) = \frac{1}{4}\) | M1 | |
| \(\sin(Q_1) = 1/\sqrt{2}\) | A1 | Can be implied. Allow decimal approximations |
| \(Q_1 = \pi/4\) | A1 3 | Or 0.785(4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(G(y) = P(Y \leq y) = P(T \leq \sin^{-1}y) = F(\sin^{-1}y) = y^4\) | M1, A1, A1 | Ignore other ranges |
| \(g(y) = \begin{cases} 4y^3 & 0 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}\) | M1 | Differentiate \(G(y)\) |
| A1 5 | Function and range stated, allow if range given in G |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_0^1 \frac{4}{1+2y}\,dy = \big[2\ln(1+2y)\big]_0^1\) | M1, A1 | Attempt \(\int\frac{g(y)}{y^3+2y^4}dy\); \(\int_0^1\frac{4}{1+2y}dy\) |
| \(= 2\ln 3\) | A1 3 | Or 2.2, 2.197 or better |
## Question 6:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $F(0)=0$, $F(\pi/2)=1$ | B1 | Consider both end-points |
| Increasing | B1 **2** | Consider F between end-points, can be asserted |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sin^4(Q_1) = \frac{1}{4}$ | M1 | |
| $\sin(Q_1) = 1/\sqrt{2}$ | A1 | Can be implied. Allow decimal approximations |
| $Q_1 = \pi/4$ | A1 **3** | Or 0.785(4) |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $G(y) = P(Y \leq y) = P(T \leq \sin^{-1}y) = F(\sin^{-1}y) = y^4$ | M1, A1, A1 | Ignore other ranges |
| $g(y) = \begin{cases} 4y^3 & 0 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$ | M1 | Differentiate $G(y)$ |
| | A1 **5** | Function and range stated, allow if range given in G |
### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_0^1 \frac{4}{1+2y}\,dy = \big[2\ln(1+2y)\big]_0^1$ | M1, A1 | Attempt $\int\frac{g(y)}{y^3+2y^4}dy$; $\int_0^1\frac{4}{1+2y}dy$ |
| $= 2\ln 3$ | A1 **3** | Or 2.2, 2.197 or better |
---6 The function $\mathrm { F } ( t )$ is defined as follows.
$$\mathrm { F } ( t ) = \begin{cases} 0 & t < 0 \\ \sin ^ { 4 } t & 0 \leqslant t \leqslant \frac { 1 } { 2 } \pi \\ 1 & t > \frac { 1 } { 2 } \pi \end{cases}$$
(i) Verify that F is a (cumulative) distribution function.
The continuous random variable $T$ has (cumulative) distribution function $\mathrm { F } ( t )$.\\
(ii) Find the lower quartile of $T$.\\
(iii) Find the (cumulative) distribution function of $Y$, where $Y = \sin T$, and obtain the probability density function of $Y$.\\
(iv) Find the expected value of $\frac { 1 } { Y ^ { 3 } + 2 Y ^ { 4 } }$.
\hfill \mbox{\textit{OCR S3 2009 Q6 [13]}}