| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Consecutive non-overlapping periods |
| Difficulty | Easy -1.2 This is a straightforward question testing basic understanding of Poisson distribution conditions and routine calculations. Part (a) requires recall of standard assumptions, parts (b)(i-iii) are textbook Poisson probability calculations, and part (c) is conceptual but accessible (both CDs have the same mean despite different variance structures). The question is well-scaffolded with clear guidance at each step, requiring no novel insight or complex multi-step reasoning. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | Each mistake has to be made independently of |
| Answer | Marks |
|---|---|
| Mistakes occur singly, oe | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.2 | Any one correct, B1 |
| Any two correct, full marks | Not ‘probability of a mistake is |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | (i) |
| [1] | 1.1 | BC. 0.1403738958... |
| 5 | (b) | (ii) |
| [1] | 1.1 | BC. 0.1246520195... |
| 5 | (b) | (iii) |
| awrt 0.07 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Allow M1 for 0.1301 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | E(M ) = E(Y + Y ) = E(Y ) + E(Y ) = 5 + 5 |
| Answer | Marks |
|---|---|
| both CDs. | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 2.4 | Realising that M = Y + Y and |
| Answer | Marks |
|---|---|
| needed | Can be implied by E(M ) = 5+5 and |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | Var(M ) = Var(Y + Y ) |
| Answer | Marks |
|---|---|
| correct CD | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 3.2a | Using correct rule for adding |
| Answer | Marks |
|---|---|
| provided that values are correct. | Condone use of n rather than n – 1: |
Question 5:
5 | (a) | Each mistake has to be made independently of
any other mistake
Mistakes have to occur at a constant average rate.
Mistakes occur singly, oe | B1
B1
[2] | 3.3
1.2 | Any one correct, B1
Any two correct, full marks | Not ‘probability of a mistake is
independent’
Not “average constant rate”.
Some context necessary
5 | (b) | (i) | awrt 0.14 | B1
[1] | 1.1 | BC. 0.1403738958...
5 | (b) | (ii) | awrt 0.12 | B1
[1] | 1.1 | BC. 0.1246520195...
5 | (b) | (iii) | Use of Po(10) soi
awrt 0.07 | M1
A1
[2] | 1.1
1.1 | Allow M1 for 0.1301
BC. 0.06708596288...
5 | (c) | E(M ) = E(Y + Y ) = E(Y ) + E(Y ) = 5 + 5
1 AM PM AM PM
= 10
E(M ) = E(2Y ) = 2E(Y ) = 25 = 10
2 AM AM
So the expected values would be the same for
both CDs. | M1
A1
[2] | 3.1b
2.4 | Realising that M = Y + Y and
1 AM PM
that is M = 2Y and using
B AM
expectation algebra accordingly
Both values = 10 and statement
needed | Can be implied by E(M ) = 5+5 and
1
E(M ) = 2*5
2
5 | (d) | Var(M ) = Var(Y + Y )
1 AM PM
= Var(Y ) + Var(Y ) = 5 + 5 = 10
AM PM
Var(M ) = Var(2Y ) = 4Var(Y ) = 45 = 20
2 AM AM
2
70
812−7
7 56
For A: s2 = = = awrt 19
n−1
6 3
2
7 2
8 0 0 − 7
7 2 0 8
For B: s 2n = = = awrt 10
− 1
6 2 1
19 is close to 20 and 9.9 is close to 10. So it is
likely that critic A has been given the wrongly
recorded CD.
Or 9.9 is close to Var(M )=E(M )=10 so B has the
1 1
correct CD | B1
B1
M1
A1
[4] | 3.1b
3.1b
1.1
3.2a | Using correct rule for adding
variances of independent RVs to
obtain 10
Using correct rule for variance of a
multiple to obtain 20
Writing correct calculation for at
least one of these.
Correct conclusion from (either set
of) cvo. It is sufficient to note that
A’s sample variance > B’s, or that
B’s variance is close to Var(M ),
1
provided that values are correct. | Condone use of n rather than n – 1:
awrt 16, awrt 85 Over a long period of time, it is found that the mean number of mistakes made by a certain player when playing a particular piece of music is 5 .
The number of mistakes that the player makes when playing the piece is denoted by the random variable $Y$.
\begin{enumerate}[label=(\alph*)]
\item State two assumptions necessary for $Y$ to be modelled by a Poisson distribution.
For the remainder of this question you may assume that $Y$ can be modelled by a Poisson distribution.
\item \begin{enumerate}[label=(\roman*)]
\item Find the probability that the player makes exactly 3 mistakes when playing the piece.
\item Find the probability that the player makes fewer than 3 mistakes when playing the piece.
\item Find the probability that the player makes fewer than 6 mistakes in total when playing the piece twice, assuming that the performances are independent.
In a recording studio, the player plays the piece once in the morning and once in the afternoon each day for one week (7 days). It can be assumed that all the performances are independent of each other.
The performances are recorded onto two CDs, one for each of two critics, A and B, to review. The critics are interested in the total number of mistakes made by the player per day. Unfortunately, there is a recording error in one of the CDs; on this CD, every piece that is supposed to be an afternoon recording is in fact just a repeat of that morning's recording.
The random variables $M _ { 1 }$ and $M _ { 2 }$ represent the total number of mistakes per day for the correctly recorded CD and for the wrongly recorded CD respectively.
\end{enumerate}\item By considering the values of $\mathrm { E } \left( M _ { 1 } \right)$ and $\mathrm { E } \left( M _ { 2 } \right)$ explain why it is not possible to use the mean number of mistakes per day on the CDs to determine which critic received the wrongly recorded CD.
Each critic counts the total number of mistakes made per day, for each of the 7 days of recordings on their CD. Summary data for this is given below.
Critic A: $\quad n = 7 , \quad \sum x _ { A } = 70 , \quad \sum x _ { A } ^ { 2 } = 812$\\
Critic B: $\quad \mathrm { n } = 7 , \sum \mathrm { x } _ { \mathrm { B } } = 72 , \sum \mathrm { x } _ { \mathrm { B } } ^ { 2 } = 800$
\item By considering the values of $\operatorname { Var } \left( M _ { 1 } \right)$ and $\operatorname { Var } \left( M _ { 2 } \right)$ determine which critic is likely to have received the wrongly recorded CD.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2024 Q5 [12]}}