| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find PGF from probability distribution |
| Difficulty | Standard +0.8 This is a multi-part PGF question requiring careful enumeration of outcomes (0,1,2,3,4 points with different probabilities), construction of the PGF, differentiation for mean/variance, and convolution of PGFs. While methodical, it demands precision in probability calculations and fluency with PGF manipulation beyond routine exercises, placing it moderately above average difficulty. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): 0, 1, 2, 4; \(p\): \(\frac{1}{4}\), \(\frac{3}{16}\), \(\frac{9}{64}\), \(\frac{27}{64}\) | B2 | B1 any 2 correct. May be implied |
| \(G_X(t) = \frac{1}{4} + \frac{3}{16}t + \frac{9}{64}t^2 + \frac{27}{64}t^4\) | M1A1, [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G'_X(t) = \frac{3}{16} + \frac{9}{32}t + \frac{27}{16}t^3\) | B1ft | |
| \(G'_X(1) = \frac{69}{32} = 2.16\) | M1A1 | |
| \(G''_X(t) = \frac{9}{32} + \frac{81}{16}t^2\) | B1 | |
| \(\text{Var}(X) = \frac{171}{32} + \frac{69}{32} - \left(\frac{69}{32}\right)^2\) | M1 | Not if negative variance |
| \(\frac{2919}{1024} = 2.85\) | A1, [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{1}{4} + \frac{3}{16}t + \frac{9}{64}t^2 + \frac{27}{64}t^4\right)^2\) | B1ft | |
| coeff of \(t^4 = \frac{945}{4096} = 0.231\) | M1A1, [3] | coeff of \(t^4\) soi |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: 0, 1, 2, 4; $p$: $\frac{1}{4}$, $\frac{3}{16}$, $\frac{9}{64}$, $\frac{27}{64}$ | B2 | B1 any 2 correct. May be implied |
| $G_X(t) = \frac{1}{4} + \frac{3}{16}t + \frac{9}{64}t^2 + \frac{27}{64}t^4$ | M1A1, [4] | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G'_X(t) = \frac{3}{16} + \frac{9}{32}t + \frac{27}{16}t^3$ | B1ft | |
| $G'_X(1) = \frac{69}{32} = 2.16$ | M1A1 | |
| $G''_X(t) = \frac{9}{32} + \frac{81}{16}t^2$ | B1 | |
| $\text{Var}(X) = \frac{171}{32} + \frac{69}{32} - \left(\frac{69}{32}\right)^2$ | M1 | Not if negative variance |
| $\frac{2919}{1024} = 2.85$ | A1, [6] | |
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{1}{4} + \frac{3}{16}t + \frac{9}{64}t^2 + \frac{27}{64}t^4\right)^2$ | B1ft | |
| coeff of $t^4 = \frac{945}{4096} = 0.231$ | M1A1, [3] | coeff of $t^4$ soi |
---6 In each round of a quiz a contestant can answer up to three questions. Each correct answer scores 1 point and allows the contestant to go on to the next question. A wrong answer scores 0 points and the contestant is allowed no further question in that round. If all 3 questions are answered correctly 1 bonus point is scored, making a total score of 4 for the round. For a certain contestant, $A$, the probability of giving a correct answer is $\frac { 3 } { 4 }$, independently of any other question. The random variable $X _ { r }$ is the number of points scored by $A$ during the $r ^ { \text {th } }$ round.\\
(i) Find the probability generating function of $X _ { r }$.\\
(ii) Use the probability generating function found in part (i) to find the mean and variance of $X _ { r }$.\\
(iii) Write down an expression for the probability generating function of $X _ { 1 } + X _ { 2 }$ and find the probability that $A$ has a total score of 4 at the end of two rounds.
\hfill \mbox{\textit{OCR S4 2018 Q6 [13]}}