| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Probability distribution from context |
| Difficulty | Standard +0.3 This is a straightforward probability distribution question requiring basic setup from a verbal description, calculation of expectation and variance using standard formulas, and scaling for multiple trials. The context is simple (fair dice with clear rules), and all steps follow routine procedures taught in Further Statistics with no conceptual surprises or proof elements. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) | x (£) 1 2 3 6 10 |
| Answer | Marks |
|---|---|
| 6 6 6 6 6 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | x-values correct |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (ii) | (cid:166)xP(cid:11)x(cid:12)(cid:32) 1 (cid:14) 4(cid:14) 3(cid:14) 6(cid:14)10 (cid:32)4 |
| Answer | Marks |
|---|---|
| 3 | B1 |
| Answer | Marks |
|---|---|
| [5] | 2.2a |
| Answer | Marks |
|---|---|
| 1.1 | n |
| Answer | Marks | Guidance |
|---|---|---|
| x (£) | 1 | 2 |
| P(cid:11)X (cid:32)x(cid:12) | 1 | |
| 6 | 2 | |
| 6 | 1 | |
| 6 | 1 | |
| 6 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| y (£) | –4 | –3 |
| Answer | Marks |
|---|---|
| P(cid:11)Y (cid:32) y(cid:12) | 1 |
| 6 | 2 |
| 6 | 1 |
| 6 | 1 |
| 6 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(i) | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(ii) | 3 | 20 |
Question 3:
3 | (i) | x (£) 1 2 3 6 10
1 2 1 1 1
P(cid:11)X (cid:32)x(cid:12)
6 6 6 6 6 | M1
A1
A1
[3] | 3.1b
1.1
1.1 | x-values correct
At least 2 probabilities correct
All correct
3 | (ii) | (cid:166)xP(cid:11)x(cid:12)(cid:32) 1 (cid:14) 4(cid:14) 3(cid:14) 6(cid:14)10 (cid:32)4
6 6 6 6 6
p
(cid:166)x2P(cid:11)x(cid:12)(cid:16)(cid:80)2 (cid:32) 1 (cid:14)8(cid:14)9(cid:14)36(cid:14)100(cid:16)(cid:80)2
6 6 6 6 6
S
77
(cid:32) (cid:16)(cid:80)2
3
2
(cid:32)9
3
Therefore for 120 games the standard deviation
is
2
120(cid:117)9 (cid:32)34.1
3 | B1
c
e
M1
A1
M1
A1FT
[5] | 2.2a
i
1.1
1.1
2.2a
1.1 | n
For dismissing the £5 loss, or using
e
profit y:
y (£) –4 –3 –2 1 5
m
P(cid:11)Y (cid:32) y(cid:12) 1 2 1 1 1
6 6 6 6 6
giving (cid:166)yP(cid:11)y(cid:12)(cid:32)(cid:16)1 and
32
(cid:166)y2P(cid:11)y(cid:12)(cid:32)
3
Allow their value of μ
Multiply by 120 and take
In range [34(.0), 35.1]
x (£) | 1 | 2 | 3 | 6 | 10
P(cid:11)X (cid:32)x(cid:12) | 1
6 | 2
6 | 1
6 | 1
6 | 1
6
y (£) | –4 | –3 | –2 | 1 | 5
e
P(cid:11)Y (cid:32) y(cid:12) | 1
6 | 2
6 | 1
6 | 1
6 | 1
6
--- 3(i) ---
3(i) | 2 | 0 | 1 | 0 | 3
--- 3(ii) ---
3(ii) | 3 | 20 | 053 A game is played as follows. A fair six-sided dice is thrown once. If the score obtained is even, the amount of money, in $\pounds$, that the contestant wins is half the score on the dice, otherwise it is twice the score on the dice.\\
(i) Find the probability distribution of the amount of money won by the contestant.\\
(ii) The contestant pays $\pounds 5$ for every time the dice is thrown. Find the standard deviation of the loss made by the contestant in 120 throws of the dice.
\hfill \mbox{\textit{OCR Further Statistics Q3 [8]}}