| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | State assumptions for geometric model |
| Difficulty | Easy -1.2 This is a straightforward application of the geometric distribution requiring only recall of standard conditions, formulas, and basic calculations. Part (a) asks students to state textbook assumptions, while parts (b)-(d) involve direct substitution into standard formulas with no problem-solving or novel insight required. |
| Spec | 5.02f 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) | Because we can assume that successive trials are |
| Answer | Marks | Guidance |
|---|---|---|
| is being counted | B1 | |
| [1] | 2.4 | B1 for ‘the number of trials up to (and including) the first |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | 0.854×0.15 = 0.0783 |
| [1] | 1.1 | (0.078300…) |
| 2 | (c) | 0.8510 = 0.1969 |
| [1] | 1.1 | (0.196874….) |
| 2 | (d) | 20 |
| Answer | Marks |
|---|---|
| 9 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | 1−𝑝 |
Question 2:
2 | (a) | Because we can assume that successive trials are
independent, each trial has two possible outcomes with
an equal probability of success (for each trial), and the
number of trials up to (and including) the first success
is being counted | B1
[1] | 2.4 | B1 for ‘the number of trials up to (and including) the first
success’ and one of the other three of these comments seen
2 | (b) | 0.854×0.15 = 0.0783 | B1
[1] | 1.1 | (0.078300…)
2 | (c) | 0.8510 = 0.1969 | B1
[1] | 1.1 | (0.196874….)
2 | (d) | 20
Mean = [= 6.67 to 3 s.f.]
3
1−0.15
Variance =
0.152
340
= [=37.8 to 3 s.f.]
9 | B1
M1
A1
[3] | 3.1b
1.1
1.1 | 1−𝑝
M1 for used
𝑝2
SCB1 for 37.8 without justification2 A group of friends live by the sea. Each day they look out to sea in the hope of seeing a dolphin. The probability that they see a dolphin on any day is 0.15 . You should assume that this probability is not affected by whether or not they see a dolphin on any other day.
\begin{enumerate}[label=(\alph*)]
\item Explain why you can use a geometric distribution to model the number of days that it takes for them to first see a dolphin.
\item Find the probability that they see a dolphin for the first time on the fifth day.
\item Find the probability that they do not see a dolphin for at least 10 days.
\item Determine the mean and the variance of the number of days that it takes for them to see a dolphin.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2023 Q2 [6]}}