OCR MEI Further Statistics Minor 2019 June — Question 4 17 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
Year2019
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeGoodness-of-fit test for Poisson
DifficultyStandard +0.3 This is a standard goodness-of-fit test for a Poisson distribution with straightforward calculations: estimating the parameter from the mean, computing expected frequencies using Poisson probabilities, calculating chi-squared contributions, and comparing to critical values. Part (a) requires recognizing that mean ≈ variance suggests Poisson. The calculations are routine for Further Statistics, though combining categories and determining degrees of freedom requires care. Overall, this is a textbook application slightly easier than average for Further Maths statistics.
Spec5.02i 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 = lambda5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous
4 Zara uses a metal detector to search for coins on a beach.
She wonders if the numbers of coins that she finds in an area of \(10 \mathrm {~m} ^ { 2 }\) can be modelled by a Poisson distribution. The table below shows the numbers of coins that she finds in randomly chosen areas of \(10 \mathrm {~m} ^ { 2 }\) over a period of months.
Number of coins found0123456\(> 6\)
Frequency1328301410230
  1. Software gives the sample mean as 1.98 and the sample standard deviation as 1.4212. Explain how these values suggest that a Poisson distribution may be an appropriate model for the numbers of coins found. Zara decides to carry out a chi-squared test to investigate whether a Poisson distribution is an appropriate model.
    Fig. 4 is a screenshot showing part of the spreadsheet used to analyse the data. Some values in the spreadsheet have been deliberately omitted. \begin{table}[h]
    ABCD
    1Number of coins foundObserved frequencyExpected frequencyChi-squared contribution
    201313.80690.0472
    3128
    423027.06430.3184
    531417.86250.8352
    64108.84190.1517
    7\(\geqslant 5\)50.0015
    \captionsetup{labelformat=empty} \caption{Fig. 4}
    \end{table}
  2. Showing your calculations, find the missing values in each of the following cells.
    For the rest of this question, you should assume that the number of coins that Zara finds in an area of \(10 \mathrm {~m} ^ { 2 }\) can be modelled by a Poisson distribution with mean 1.98.
    Zara also finds pieces of jewellery independently of the coins she finds. The number of pieces of jewellery that she finds per \(10 \mathrm {~m} ^ { 2 }\) area is modelled by a Poisson distribution with mean 0.42 .
  3. Find the probability that Zara finds a total of exactly 3 items (coins and/or jewellery) in an area of \(10 \mathrm {~m} ^ { 2 }\).
  4. Find the probability that Zara finds a total of at least 30 items (coins and/or jewellery) in an area of \(100 \mathrm {~m} ^ { 2 }\).
Question 4:
AnswerMarks Guidance
4(a) Variance = 1.42122 = 2.0(198094…)
Because the variance is close to the mean oeB1
E1
AnswerMarks
[2]1.1
2.2b1.98 ≈ 2.0 scores 2 marks
NB E1 only awarded if a
relevant numerical comparison
AnswerMarks
is seenN.B. comparing 1.98
with 1.4212 is B0 E0
For E1 accept
“Both are close to 2”
AnswerMarks
(b)For C3 probability = 0.273377 so C3 = 27.3377
For C7 probability = 0.050867 so C7 = 5.0867
(28their27.3377)2
D3 =
their27.3377
AnswerMarks
= 0.0160M1
A1
M1
A1
AnswerMarks
[4]3.4
2.2a
1.1a
AnswerMarks
1.1For either C3 or C7
Allow second value to be found
by subtraction
(27.337728)2
Allow
AnswerMarks
27.3377Values should all be to
4dp to score full marks
Deduct at most one A
mark if any values are
not given to 4dp
AnswerMarks
(c)Because otherwise some of the expected
frequencies would be too low
AnswerMarks
(<5) (and so the test would not be valid)E1
[1]2.4
(d)H : Poisson model is a good fit
0
H : Poisson model is not a good fit
1
(Test statistic =) 1.37
Refer to 2
4
Critical value at 5% level = 9.488
1.37 < 9.488
Insufficient evidence to reject H
0
There is insufficient evidence to suggest that the
Poisson model is not a good fit for the number of
AnswerMarks
coins foundB1
B1FT
M1
A1
M1
A1
AnswerMarks
[6]1.2
1.1
3.4
1.1
1.1
AnswerMarks
2.2bAccept similar wording e.g.:
FT (1.354 + their D3)
4 degrees of freedom
their TS compared correctly to
their CV* and conclusion
For non-assertive conclusion in
AnswerMarks
context‘a suitable model’
‘an appropriate model’
*their CV must be
from upper tail of χ2
table
Allow “Accept H ”
0
For A1 TS & CV must
be correct
AnswerMarks
(e)λ = 1.98 + 0.42 (= 2.4) used soi
P(3) = 0.2090 awrtM1
A1
AnswerMarks Guidance
[2]3.1b
1.1(0.209014…) N.B. 0.209 only is A0
(e)
AnswerMarks
altp(C=0)×p(J=3) + p(C=1)×p(J=2) +
p(C=2)×p(J=1) + p(C=3)×p(J=0)
0.138069×0.008113 + 0.273377×0.057952 +
0.270643×0.275960 + 0.178625×0.657047
AnswerMarks
0.2090M1
A1
AnswerMarks Guidance
[2]3.1b 0.00112 + 0.01584 + 0.07469 +
0.11737N.B. 0.209 only is A0
(f)Po(10 × their λ from (e) ) used soi (their λ ≠ 1.98)
P(≥ 30) = 1 – 0.8679 = 0.1321 awrtM1
A1
AnswerMarks
[2]3.3
3.4Following M0
allow SC1 for Normal
approximation leading to answer
rounding to 0.13
Question 4:
4 | (a) | Variance = 1.42122 = 2.0(198094…)
Because the variance is close to the mean oe | B1
E1
[2] | 1.1
2.2b | 1.98 ≈ 2.0 scores 2 marks
NB E1 only awarded if a
relevant numerical comparison
is seen | N.B. comparing 1.98
with 1.4212 is B0 E0
For E1 accept
“Both are close to 2”
(b) | For C3 probability = 0.273377 so C3 = 27.3377
For C7 probability = 0.050867 so C7 = 5.0867
(28their27.3377)2
D3 =
their27.3377
= 0.0160 | M1
A1
M1
A1
[4] | 3.4
2.2a
1.1a
1.1 | For either C3 or C7
Allow second value to be found
by subtraction
(27.337728)2
Allow
27.3377 | Values should all be to
4dp to score full marks
Deduct at most one A
mark if any values are
not given to 4dp
(c) | Because otherwise some of the expected
frequencies would be too low
(<5) (and so the test would not be valid) | E1
[1] | 2.4
(d) | H : Poisson model is a good fit
0
H : Poisson model is not a good fit
1
(Test statistic =) 1.37
Refer to 2
4
Critical value at 5% level = 9.488
1.37 < 9.488
Insufficient evidence to reject H
0
There is insufficient evidence to suggest that the
Poisson model is not a good fit for the number of
coins found | B1
B1FT
M1
A1
M1
A1
[6] | 1.2
1.1
3.4
1.1
1.1
2.2b | Accept similar wording e.g.:
FT (1.354 + their D3)
4 degrees of freedom
their TS compared correctly to
their CV* and conclusion
For non-assertive conclusion in
context | ‘a suitable model’
‘an appropriate model’
*their CV must be
from upper tail of χ2
table
Allow “Accept H ”
0
For A1 TS & CV must
be correct
(e) | λ = 1.98 + 0.42 (= 2.4) used soi
P(3) = 0.2090 awrt | M1
A1
[2] | 3.1b
1.1 | (0.209014…) | N.B. 0.209 only is A0
(e)
alt | p(C=0)×p(J=3) + p(C=1)×p(J=2) +
p(C=2)×p(J=1) + p(C=3)×p(J=0)
0.138069×0.008113 + 0.273377×0.057952 +
0.270643×0.275960 + 0.178625×0.657047
0.2090 | M1
A1
[2] | 3.1b | 0.00112 + 0.01584 + 0.07469 +
0.11737 | N.B. 0.209 only is A0
(f) | Po(10 × their λ from (e) ) used soi (their λ ≠ 1.98)
P(≥ 30) = 1 – 0.8679 = 0.1321 awrt | M1
A1
[2] | 3.3
3.4 | Following M0
allow SC1 for Normal
approximation leading to answer
rounding to 0.13
4 Zara uses a metal detector to search for coins on a beach.\\
She wonders if the numbers of coins that she finds in an area of $10 \mathrm {~m} ^ { 2 }$ can be modelled by a Poisson distribution. The table below shows the numbers of coins that she finds in randomly chosen areas of $10 \mathrm {~m} ^ { 2 }$ over a period of months.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Number of coins found & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $> 6$ \\
\hline
Frequency & 13 & 28 & 30 & 14 & 10 & 2 & 3 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Software gives the sample mean as 1.98 and the sample standard deviation as 1.4212.

Explain how these values suggest that a Poisson distribution may be an appropriate model for the numbers of coins found.

Zara decides to carry out a chi-squared test to investigate whether a Poisson distribution is an appropriate model.\\
Fig. 4 is a screenshot showing part of the spreadsheet used to analyse the data. Some values in the spreadsheet have been deliberately omitted.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
 & A & B & C & D \\
\hline
1 & Number of coins found & Observed frequency & Expected frequency & Chi-squared contribution \\
\hline
2 & 0 & 13 & 13.8069 & 0.0472 \\
\hline
3 & 1 & 28 &  &  \\
\hline
4 & 2 & 30 & 27.0643 & 0.3184 \\
\hline
5 & 3 & 14 & 17.8625 & 0.8352 \\
\hline
6 & 4 & 10 & 8.8419 & 0.1517 \\
\hline
7 & $\geqslant 5$ & 5 &  & 0.0015 \\
\hline
 &  &  &  &  \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{table}
\item Showing your calculations, find the missing values in each of the following cells.

\begin{itemize}
  \item C3
  \item C7
  \item D3
\item Explain why the numbers for 5, 6 and more than 6 coins found have been combined into the single category of at least 5 coins found, as shown in the spreadsheet.
\item Complete the hypothesis test at the $5 \%$ level of significance.
\end{itemize}

For the rest of this question, you should assume that the number of coins that Zara finds in an area of $10 \mathrm {~m} ^ { 2 }$ can be modelled by a Poisson distribution with mean 1.98.\\
Zara also finds pieces of jewellery independently of the coins she finds. The number of pieces of jewellery that she finds per $10 \mathrm {~m} ^ { 2 }$ area is modelled by a Poisson distribution with mean 0.42 .
\item Find the probability that Zara finds a total of exactly 3 items (coins and/or jewellery) in an area of $10 \mathrm {~m} ^ { 2 }$.
\item Find the probability that Zara finds a total of at least 30 items (coins and/or jewellery) in an area of $100 \mathrm {~m} ^ { 2 }$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2019 Q4 [17]}}
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