| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Derive standard distribution PGF |
| Difficulty | Standard +0.3 This is a structured question guiding students through standard PGF results for Bernoulli and binomial distributions, with routine differentiation to find mean and variance. Part (iv) adds a Poisson distribution but requires only straightforward convolution of two standard distributions. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Choosing row \(C\) in column \(A\) | M1 | For choosing row \(C\) in column \(A\) |
| Choosing more than one entry from column \(C\) | M1 dep | For choosing more than one entry from column \(C\) |
| Correct entries chosen | A1 | For correct entries chosen |
| Order: \(A\ C\ E\ D\ B\ F\) | B1 | For correct order, listed or marked on arrows or table, or arcs listed \(AC\ CE\ ED\ CB\ DF\) |
| Minimum spanning tree diagram (B,D connected; A,C,E,F connected) | B1 | For tree (correct or follow through from table, provided solution forms a spanning tree) |
| Total weight: 23 miles | B1 | For 23 (or follow through from table or diagram, provided solution forms a spanning tree) [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| MST for reduced network \(= 18\) | M1 | For their 18 seen or implied |
| Two shortest arcs from \(B = 5 + 6 = 11\) | M1 | For 11 seen or implied |
| Lower bound \(= 29\) miles | A1 | For 29 (cao) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F-D-E-C-A-B-F\) | M1 | For \(F-D-E-C-A-B\) |
| Correct tour | A1 | For correct tour |
| \(8+3+4+3+6+14\) | M1 | For a substantially correct attempt at sum |
| \(= 38\) miles | A1 | For 38 (cao) [4] |
# Question 6 (Paper 4736):
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Choosing row $C$ in column $A$ | M1 | For choosing row $C$ in column $A$ |
| Choosing more than one entry from column $C$ | M1 dep | For choosing more than one entry from column $C$ |
| Correct entries chosen | A1 | For correct entries chosen |
| Order: $A\ C\ E\ D\ B\ F$ | B1 | For correct order, listed or marked on arrows or table, or arcs listed $AC\ CE\ ED\ CB\ DF$ |
| Minimum spanning tree diagram (B,D connected; A,C,E,F connected) | B1 | For tree (correct or follow through from table, provided solution forms a spanning tree) |
| Total weight: 23 miles | B1 | For 23 (or follow through from table or diagram, provided solution forms a spanning tree) **[6]** |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| MST for reduced network $= 18$ | M1 | For their 18 seen or implied |
| Two shortest arcs from $B = 5 + 6 = 11$ | M1 | For 11 seen or implied |
| Lower bound $= 29$ miles | A1 | For 29 (cao) **[3]** |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $F-D-E-C-A-B-F$ | M1 | For $F-D-E-C-A-B$ |
| Correct tour | A1 | For correct tour |
| $8+3+4+3+6+14$ | M1 | For a substantially correct attempt at sum |
| $= 38$ miles | A1 | For 38 (cao) **[4]** |
---6 The discrete random variable $X$ takes the values 0 and 1 with $\mathrm { P } ( X = 0 ) = q$ and $\mathrm { P } ( X = 1 ) = p$, where $p + q = 1$.\\
(i) Write down the probability generating function of $X$.
The sum of $n$ independent observations of $X$ is denoted by $S$.\\
(ii) Write down the probability generating function of $S$, and name the distribution of $S$.\\
(iii) Use the probability generating function of $S$ to find $\mathrm { E } ( S )$ and $\operatorname { Var } ( S )$.\\
(iv) The independent random variables $Y$ and $Z$ are such that $Y$ has the distribution $\mathrm { B } \left( 10 , \frac { 1 } { 2 } \right)$, and $Z$ has probability generating function $\mathrm { e } ^ { - ( 1 - t ) }$. Find the probability that the sum of one random observation of $Y$ and one random observation of $Z$ is equal to 2 .
\hfill \mbox{\textit{OCR S4 2007 Q6 [15]}}