| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | State assumptions for geometric model |
| Difficulty | Standard +0.3 This question tests standard knowledge of geometric distribution assumptions and basic calculations. Parts (a) and (b) require recall of model assumptions with minimal analysis. Parts (c) and (d) involve routine probability calculations using standard formulas. The conceptual demand is low—students need only apply textbook knowledge about when geometric distributions are appropriate and perform straightforward computations. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | A geometric distribution should still be a good |
| Answer | Marks | Guidance |
|---|---|---|
| to be infinite/etc. | B1 | |
| [1] | 3.3 | OE. Allow “population is large” but not “n is large” or “sample is large”. |
| Answer | Marks |
|---|---|
| (b)(i) | Whether or not one book refers to the topic is |
| Answer | Marks | Guidance |
|---|---|---|
| or arrangement/selection of books is random | B1 | |
| [1] | 3.3 | One correct modelling assumption, stated in context (e.g. “each book is |
| Answer | Marks |
|---|---|
| (b)(ii) | E.g. arrangement/selection of the books not |
| Answer | Marks | Guidance |
|---|---|---|
| or lecturer chooses books by author, etc | B1 | |
| [1] | 3.3 | Reason why it might not be valid. Needn’t be very plausible but must |
| Answer | Marks |
|---|---|
| (c) | 1 – (1 – 0.05)n > 0.9 |
| Answer | Marks |
|---|---|
| n > 44.(8)9, so minimum 45 books | M1* |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | Correct expression soi, allow n – 1, allow any of >, , = |
| Answer | Marks |
|---|---|
| (d) | 3 in first 39, then 1 ⇒ 39C 0.9536 0.054 |
| Answer | Marks |
|---|---|
| = 0.009 01(215) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 1.1 | Allow if any indication of B(39, 0.05) used (but NB: 0.0901 from |
Question 1:
1 | (a) | A geometric distribution should still be a good
model as the number of books is large/can be taken
to be infinite/etc. | B1
[1] | 3.3 | OE. Allow “population is large” but not “n is large” or “sample is large”.
“Likely to finish long before lecturer gets to the end”: B1
“Only a good model if the books are selected with replacement”: B0
(b)(i) | Whether or not one book refers to the topic is
independent of whether another does
or the probability that one book refers to the topic
is the same for any book
or arrangement/selection of books is random | B1
[1] | 3.3 | One correct modelling assumption, stated in context (e.g. “each book is
independent”, “the probability of finding the book is constant”, etc)
Allow any number of correct statements but B0 if any definitely incorrect
or ‘scattergun’ statements seen, e.g. not “a book must either refer to
the topic or not”
(b)(ii) | E.g. arrangement/selection of the books not
independent of whether they refer to the topic,
or books containing the topic may be grouped
together,
or lecturer chooses books by author, etc | B1
[1] | 3.3 | Reason why it might not be valid. Needn’t be very plausible but must
show correct understanding of the concept.
Not “sampling without replacement means that the probabilities change”
(part (a) focusses on the fact that, although the [conditional]
probabilities change, they do so by only a very small amount and so
the model is not invalidated)
(c) | 1 – (1 – 0.05)n > 0.9
n > log (0.1) or log(0.1) log(0.95) etc
0.95
n > 44.(8)9, so minimum 45 books | M1*
depM1
A1
[3] | 3.3
2.1
3.4 | Correct expression soi, allow n – 1, allow any of >, , =
Correct solution method, e.g. use logarithms, ignore inequality
(T&I: 44 or 46 gets M1M1A0)
45 only, can be from T&I (command word is “find”). Not from 0.95n–1
(giving n – 1 > 44.9) but allow “n < 44.9 so 45” BOD. Not n > 45.
SC: N(20, 380) giving 45: B2
SC: 45 following wrong working and first M1 not gained: B1
(d) | 3 in first 39, then 1 ⇒ 39C 0.9536 0.054
3
= 0.009 01(215) | M1
A1
[2] | 3.1b
1.1 | Allow if any indication of B(39, 0.05) used (but NB: 0.0901 from
B(40, 0.05) is M0A0).
Correct to 3sf, www1 A certain section of a library contains several thousand books. A lecturer is looking for a book that refers to a particular topic. The lecturer believes that one-twentieth of the books in that section of the library contain a reference to that topic. However, the lecturer does not know which books they might be, so the lecturer looks in each book in turn for a reference to the topic. The first book the lecturer finds that refers to the topic is the $X$ th book in which the lecturer looks.
\begin{enumerate}[label=(\alph*)]
\item A student says, "There is a maximum value of $X$ as there is only a finite number of books. So a geometric distribution cannot be a good model for $X$."
Explain whether you agree with the student.
\item \begin{enumerate}[label=(\roman*)]
\item State one modelling assumption (not involving the total number of books) needed for $X$ to be modelled by a geometric distribution in this context.
\item Suggest a reason why this assumption may not be valid in this context.
Assume now that $X$ can be well modelled by the distribution $\operatorname { Geo } ( 0.05 )$.
\end{enumerate}\item The probability that the lecturer needs to look in no more than $n$ books is greater than 0.9 .
Find the smallest possible value of $n$.
\item The lecturer needs to find four different books that refer to the topic.
Find the probability that the lecturer wants to look in exactly 40 books.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2023 Q1 [8]}}