| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Name geometric distribution and parameter |
| Difficulty | Moderate -0.8 This is a straightforward application of the geometric distribution with clearly stated parameters. Parts (a)-(c) are direct recall and formula application, while part (d) requires solving an inequality but uses standard geometric series summation. The question is easier than average as it requires no novel insight—just recognition of the distribution and routine calculations. |
| 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 |
|---|---|---|
| 7 | (a) | Geometric(0.2) |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | M1 for distribution. Allow Geo(…) |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (b) | 0 .8 3 0 .2 = 0 .1 0 2 4 |
| [1] | 1.1 | BC Allow 0.102 |
| 7 | (c) | Must find 1 voucher in first 9 so B(9, 0.2) then × 0.2 |
| Answer | Marks |
|---|---|
| = 0.060 (0.060397…) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 1.1 | BC Or 0.30198… × 0.2 | |
| 7 | (d) | 1 − 0 .8 n 0 .9 9 0 .8 n 0 .0 1 |
| Answer | Marks |
|---|---|
| ln 0 .8 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 3.2a | M1 for 0.8n < 0.01 |
Question 7:
7 | (a) | Geometric(0.2) | M1
A1
[2] | 3.3
1.1 | M1 for distribution. Allow Geo(…)
for parameter
7 | (b) | 0 .8 3 0 .2 = 0 .1 0 2 4 | B1
[1] | 1.1 | BC Allow 0.102
7 | (c) | Must find 1 voucher in first 9 so B(9, 0.2) then × 0.2
P(second on tenth) = 9 × 0.88 × 0.2 × 0.2
= 0.060 (0.060397…) | M1
A1
[2] | 1.1
1.1 | BC Or 0.30198… × 0.2
7 | (d) | 1 − 0 .8 n 0 .9 9 0 .8 n 0 .0 1
ln 0 .0 1
n so least n = 21
ln 0 .8 | M1
A1
[2] | 3.1b
3.2a | M1 for 0.8n < 0.01
A1 for n = 21 www.
SC B1 for n = 21 from 0.8n ≤ 0.01 or 0.8n = 0.01
If trial and improvement used award M1 if suitable
inequality shown and A1 for n = 21 with suitable working
shown.
SC B1 for n = 21 if inequality not shown but suitable
probabilities seen. e.g. P(X ≤ 20) = 0.9884.. and P(X ≤ 21) =
0.9907..
PMT
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Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.7 On average one in five packets of a breakfast cereal contains a voucher for a discount on the next packet bought. Whether or not a packet contains a voucher is independent of other packets, and can only be determined by opening the packet.
\begin{enumerate}[label=(\alph*)]
\item State the distribution of the number of packets that need to be opened in order to find one which contains a voucher.
\item Determine the probability that exactly 4 packets have to be opened in order to find one which contains a voucher.
\item Determine the probability that exactly 10 packets have to be opened in order to find two which contain a voucher.
\item I have $n$ packets, and I open them one by one until I find a voucher or until all the packets are open.
Given that the probability that I find a voucher is greater than 0.99 , determine the least possible value of $n$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2022 Q7 [7]}}