| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Composite/applied transformation |
| Difficulty | Challenging +1.2 This is a multi-step transformation problem requiring students to derive the CDF and PDF of a transformed random variable. While it involves several techniques (geometric reasoning, function of a random variable, differentiation), the question provides significant scaffolding through parts (i)-(iii), and the transformation from uniform X to A follows a standard template taught in S3. The algebraic manipulation is moderate but routine for this level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles5.07a Non-parametric tests: when to use |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = X(10 - X)\) | B1 | from base \(\times\) height |
| Use CTS | M1 | or quadratic formula. Allow verification. |
| \(A = 25 - (X-5)^2\) AG | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f_x(x) = \frac{1}{2}\) | B1 | Ignore range. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F_X(x) = \frac{1}{2}x\) | B1 | Only if (ii) correct |
| \((F_A(a) =) \ P(A \leq a) = P[X(10-X) \leq a]\) | M1 | |
| \(= F_X(5 - \sqrt{25-a})\) | A1 | Fully justified. \(X \text{(or } x) \geq 5 + \sqrt{25-a}\) is impossible |
| \(= \frac{1}{2}(5 - \sqrt{25-a})\) AG | A1 | |
| \(0 \leq A \leq 16\) AG explained | B1 | e.g. \(x=2 \Rightarrow a=16\). \(F_A(16)=1\) is not enough |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f_A(a) = \frac{1}{4}(25-a)^{-\frac{1}{2}}\) | M1, A1 | M1 for attempt at differentiation |
## Question 9:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = X(10 - X)$ | B1 | from base $\times$ height |
| Use CTS | M1 | or quadratic formula. Allow verification. |
| $A = 25 - (X-5)^2$ AG | A1 | |
| **[3]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f_x(x) = \frac{1}{2}$ | B1 | Ignore range. |
## Question (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F_X(x) = \frac{1}{2}x$ | B1 | Only if (ii) correct |
| $(F_A(a) =) \ P(A \leq a) = P[X(10-X) \leq a]$ | M1 | |
| $= F_X(5 - \sqrt{25-a})$ | A1 | Fully justified. $X \text{(or } x) \geq 5 + \sqrt{25-a}$ is impossible |
| $= \frac{1}{2}(5 - \sqrt{25-a})$ AG | A1 | |
| $0 \leq A \leq 16$ AG explained | B1 | e.g. $x=2 \Rightarrow a=16$. $F_A(16)=1$ is not enough |
**Total: [5]**
---
## Question (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f_A(a) = \frac{1}{4}(25-a)^{-\frac{1}{2}}$ | M1, A1 | M1 for attempt at differentiation |
**Total: [2]**9 A rectangle of area $A \mathrm {~m} ^ { 2 }$ has a perimeter of 20 m and each of the two shorter sides are of length $X \mathrm {~m}$, where $X$ is uniformly distributed between 0 and 2 .\\
(i) Write down an expression for $A$ in terms of $X$, and hence show that $A = 25 - ( X - 5 ) ^ { 2 }$.\\
(ii) Write down the probability density function of $X$.\\
(iii) Show that the cumulative distribution function of $A$ is
$$\mathrm { F } ( a ) = \left\{ \begin{array} { l r }
0 & a < 0 , \\
\frac { 1 } { 2 } ( 5 - \sqrt { 25 - a } ) & 0 \leqslant a \leqslant 16 , \\
1 & a > 16 .
\end{array} \right.$$
(iv) Find the probability density function of $A$.
\section*{END OF QUESTION PAPER}
\section*{OCR}
\hfill \mbox{\textit{OCR S3 2014 Q9 [11]}}