| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Moderate -0.8 This is a straightforward application of geometric distribution formulas and basic probability. Part (i) uses the standard geometric probability formula P(X=5) = (0.6)^4(0.4), part (ii) recalls the mean formula 1/p = 2.5, part (iii) uses complement rule 1-(0.6)^5, and part (iv) solves 1-(0.6)^n ≥ 0.99 using logarithms. All parts are routine calculations with no problem-solving insight required, making this easier than average for A-level. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | 5 | 5 |
| 5 | (i) | P(5) = 0.64 × 0.4 |
| = 0.05184 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (ii) | Mean = 2.5 |
| [1] | 3.4 | |
| 5 | (iii) | [1 − ] 0.65 |
| 0.9222 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1a | |
| 1.1 | N | |
| 5 | (iv) | 1(cid:16)0.6n (cid:32)0.99 |
| Answer | Marks |
|---|---|
| 10 [people] | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 3.2a | E |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | E | |
| 1 | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(iii) | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(iv) | P | |
| 1 | 0 | 2 |
Question 5:
5 | 5 | 5 | 5 | 5 | 5 | 6
5 | (i) | P(5) = 0.64 × 0.4
= 0.05184 | M1
A1
[2] | 3.3
1.1
5 | (ii) | Mean = 2.5 | B1
[1] | 3.4
5 | (iii) | [1 − ] 0.65
0.9222 | M1
A1
[2] | 1.1a
1.1 | N
5 | (iv) | 1(cid:16)0.6n (cid:32)0.99
log0.01
n(cid:32) [(cid:32)9.015...]
log0.6
10 [people] | M1
M1
A1
[3] | 3.1b
1.1
3.2a | E
M
--- 5(i) ---
5(i) | E
1 | 0 | 0 | 1 | 2
--- 5(ii) ---
5(ii) | 0 | 0 | 0 | 1 | 1
--- 5(iii) ---
5(iii) | 2 | 0 | 0 | 0 | 2
--- 5(iv) ---
5(iv) | P
1 | 0 | 2 | 0 | 35 In a recent report, it was stated that $40 \%$ of working people have a degree. For the whole of this question, you should assume that this is true.
A researcher wishes to interview a working person who has a degree. He asks working people at random whether they have a degree and counts the number of people he has to ask until he finds one with a degree.\\
(i) Find the probability that he has to ask 5 people.\\
(ii) Find the mean number of people the researcher has to ask.
Subsequently, the researcher decides to take a random sample from the population of working people.\\
(iii) A random sample of 5 working people is chosen. What is the probability that at least one of them has a degree?\\
(iv) How large a random sample of working people would the researcher need to take to ensure that the probability that at least one person has a degree is 0.99 or more?
\hfill \mbox{\textit{OCR MEI Further Statistics A AS Q5 [8]}}