| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric and negative binomial distributions with standard formulas. Part (a) is direct geometric probability, (b) uses complement rule, (c) applies negative binomial formula, and (d) requires simple inequality solving. All parts are textbook exercises requiring recall and routine calculation rather than problem-solving insight, making it slightly easier than average. |
| Spec | 2.04b 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 |
|---|---|---|
| 2 | (a) | (1 – 0.12)4 × 0.12 soi |
| = 0.0720 awrt isw | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.3 | |
| 1.1 | (0.0719634…) | NB 0.072 only is A0 |
| (b) | 0.889 = 0.3165 awrt | B1 |
| [1] | 1.1 | (0.3164783…) |
| (c) | 2 in first 19 then also 20th |
| Answer | Marks |
|---|---|
| (0.28026... × 0.12 =) 0.0336 awrt | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | For identifying required |
| Answer | Marks |
|---|---|
| distribution | (0.0336314…) |
| (d) | 1 – 0.88n ≥ 0.95 oe |
| Answer | Marks |
|---|---|
| (At least) 24 people | M1* |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 3.2a | 0.88n < 0.05 |
| Answer | Marks |
|---|---|
| and 0.8824=0.0465 must be seen | For both M marks |
Question 2:
2 | (a) | (1 – 0.12)4 × 0.12 soi
= 0.0720 awrt isw | M1
A1
[2] | 3.3
1.1 | (0.0719634…) | NB 0.072 only is A0
(b) | 0.889 = 0.3165 awrt | B1
[1] | 1.1 | (0.3164783…)
(c) | 2 in first 19 then also 20th
use B(19, 0.12) soi (to get 0.28026...)
(0.28026... × 0.12 =) 0.0336 awrt | B1
M1
A1
[3] | 3.1a
2.2a
1.1 | For identifying required
approach
Use of correct binomial
distribution | (0.0336314…)
(d) | 1 – 0.88n ≥ 0.95 oe
log0.05
n (23.43)oe soi
log0.88
(At least) 24 people | M1*
M1dep*
A1
[3] | 3.1b
1.1
3.2a | 0.88n < 0.05
log 0.05
0.88
T & I: both 0.8823=0.0529
and 0.8824=0.0465 must be seen | For both M marks
allow an equation or
strict inequality
Must see method2 A market researcher wants to interview people who watched a particular television programme. Audience research data used by the broadcaster indicates that $12 \%$ of the adult population watched this programme. This figure is used to model the situation.\\
The researcher asks people in a shopping centre, one at a time, if they watched the programme. You should assume that these people form a random sample of the adult population.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the fifth person the researcher asks is the first to have watched the programme.
\item Find the probability that the researcher has to ask at least 10 people in order to find one who watched the programme.
\item Find the probability that the twentieth person the researcher asks is the third to have watched the programme.
\item Find how many people the researcher would have to ask to ensure that there is a probability of at least 0.95 that at least one of them watched the programme.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2019 Q2 [9]}}