| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Gamma Distribution |
| Type | Verifying pdf normalization |
| Difficulty | Standard +0.3 This is a straightforward Further Maths Statistics question on the Gamma distribution requiring standard techniques: recognizing pdf normalization, performing a routine substitution to find MGF, and differentiating MGF to find moments. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(S - E - I - T\) | B1 | For this route (not in reverse), cao |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 6 litres per second | B1 | For 6 |
| From \(A\) to \(G\) | B1 | For direction \(AG\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6+2+4+0+8 = 20\) litres per second | M1, M1, A1 | For a substantially correct attempt with \(DF=0\); for dealing with \(EI\) (\(= 8\) or \(= 2+6\)); for 20, cao |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| e.g. flow 5 along \(S-A-G-T\) and 2 along \(S-C-F-H-G-T\) | M1, A1 | For describing a valid flow augmenting route; for correctly flowing 7 from \(S\) to \(T\) |
| Diagram correctly augmented | M1, M1, A1 | For a reasonable attempt at augmenting a flow; for correctly augmenting a flow; for a correct augmentation by a total of 7 |
| Cut \(\{S, A, B, C, D, E, F, G, H, I\}\), \(\{T\}\) | B1 | For identifying cut or arcs \(GT\) and \(IT\) |
| This cut has value 13 and flow already found is \(6+7=13\) litres per second. Or: arcs \(GT\) and \(IT\) are both saturated so no more can flow into \(T\) | B1 | For explaining how this shows the flow is a maximum, but NOT just stating max flow \(=\) min cut |
| 2/13 |
# Question 5:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $S - E - I - T$ | B1 | For this route (not in reverse), cao |
| | **1** | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| 6 litres per second | B1 | For 6 |
| From $A$ to $G$ | B1 | For direction $AG$ |
| | **2** | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $6+2+4+0+8 = 20$ litres per second | M1, M1, A1 | For a substantially correct attempt with $DF=0$; for dealing with $EI$ ($= 8$ or $= 2+6$); for 20, cao |
| | **3** | |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. flow 5 along $S-A-G-T$ and 2 along $S-C-F-H-G-T$ | M1, A1 | For describing a valid flow augmenting route; for correctly flowing 7 from $S$ to $T$ |
| Diagram correctly augmented | M1, M1, A1 | For a reasonable attempt at augmenting a flow; for correctly augmenting a flow; for a correct augmentation by a total of 7 |
| Cut $\{S, A, B, C, D, E, F, G, H, I\}$, $\{T\}$ | B1 | For identifying cut or arcs $GT$ and $IT$ |
| This cut has value 13 and flow already found is $6+7=13$ litres per second. Or: arcs $GT$ and $IT$ are both saturated so no more can flow into $T$ | B1 | For explaining how this shows the flow is a maximum, but NOT just stating max flow $=$ min cut |
| | **2/13** | |
The images provided appear to be blank (page 93) and a back cover/contact information page for OCR (Oxford Cambridge and RSA Examinations). Neither page contains any mark scheme content with questions, answers, or mark allocations to extract.
If you have additional pages from the mark scheme document that contain the actual question-by-question marking content, please share those and I'd be happy to extract and format that information for you.5 The continuous random variable $X$ has probability density function given by
$$\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { ( \alpha - 1 ) ! } x ^ { \alpha - 1 } \mathrm { e } ^ { - x } & x \geqslant 0 \\ 0 & x < 0 \end{cases}$$
where $\alpha$ is a positive integer.\\
(i) Explain how you can deduce that $\int _ { 0 } ^ { \infty } x ^ { \alpha - 1 } \mathrm { e } ^ { - x } \mathrm {~d} x = ( \alpha - 1 )$ !.\\
(ii) Write down an integral for the moment generating function $\mathrm { M } _ { X } ( t )$ of $X$ and show, by using the substitution $x = \frac { u } { 1 - t }$, that $\mathrm { M } _ { X } ( t ) = ( 1 - t ) ^ { - \alpha }$.\\
(iii) Use the moment generating function to find, in terms of $\alpha$,
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { E } ( X )$,
\item $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{OCR S4 2007 Q5 [12]}}