| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a straightforward S3 question involving standard techniques: integrating a simple polynomial PDF to get CDF (part i), using the transformation method for derived distributions with clear substitution (part ii), and computing an expectation using the derived PDF (part iii). All steps are routine applications of A-level Further Maths Statistics methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| B1 | For \(t^3\) | |
| B1 2 | For rest |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 A1 A1 A1 | Accept \(<\) | |
| M1 A1 | With attempt at differentiation; Only from G obtained correctly | |
| B1 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | For integrating \((1+2h^{-1})g(x)\), with limits from (ii) | |
| B1 | Limits not required | |
| A1 | ||
| M1 | ||
| B1 | Limits not required | |
| A1 | ||
| M1 B1 A1 | Limits not required | |
| 3 (12) |
## Part (i)
$$F(t) = \begin{cases} 0 & t \leq 0 \\ t^3 & 0 < t \leq 1 \\ 1 & \text{otherwise} \end{cases}$$
| Answer | Marks | Guidance |
|--------|-------|----------|
| | B1 | For $t^3$ |
| | B1 **2** | For rest |
## Part (ii)
$G(h) = P(H \leq h)$
$= P(T \geq 1/h^{1/3})$
$= 1 - F((1/h^{1/3})$
$= 1 - 1/h$
$g(h) = G'(h) = 1/h^2$
$h \geq 1,(0$ otherwise$)$
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 A1 A1 A1 | Accept $<$ |
| | M1 A1 | With attempt at differentiation; Only from G obtained correctly |
| | B1 **7** | |
## Part (iii)
EITHER: $\int_1^e (h^{-2} + 2h^{-3})dh$
$= \left[-h^{-1} - h^{-2}\right]_1^e$
$= \quad 2$
OR: $1 + 2\int_1^e \frac{1}{h^3}dh$
$= 1 + 2\left[-\frac{1}{2h^2}\right]_1^e$
$= \quad 2$
OR: $E(1+2T^7) = 1 + \int_0^e 8t^7 dt$
$= 1+[t^8]_0^e$
$= 2$
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | For integrating $(1+2h^{-1})g(x)$, with limits from (ii) |
| | B1 | Limits not required |
| | A1 | |
| | M1 | |
| | B1 | Limits not required |
| | A1 | |
| | M1 B1 A1 | Limits not required |
| | **3 (12)** | |7 The continuous random variable $T$ has probability density function given by
$$f ( t ) = \begin{cases} 4 t ^ { 3 } & 0 < t \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$
(i) Obtain the cumulative distribution function of $T$.\\
(ii) Find the cumulative distribution function of $H$, where $H = \frac { 1 } { T ^ { 4 } }$, and hence show that the probability density function of $H$ is given by $\mathrm { g } ( h ) = \frac { 1 } { h ^ { 2 } }$ over an interval to be stated.\\
(iii) Find $\mathrm { E } \left( 1 + 2 H ^ { - 1 } \right)$.
\hfill \mbox{\textit{OCR S3 2008 Q7 [12]}}