OCR Further Statistics 2024 June — Question 8 10 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2024
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicThe Gamma Distribution
TypeMethod of moments estimation
DifficultyStandard +0.3 This is a straightforward method of moments estimation requiring integration of αt^α to find E(T), equating to the sample mean, and solving for α. The integration is routine for Further Maths students, and the equation α/(α+1) = 0.6188 is easily solved. Part (b) requires basic comparison of observed vs expected frequencies. This is easier than average even for Further Statistics as it involves standard techniques with no conceptual challenges.
Spec5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03d E(g(X)): general expectation formula

8 A random sample of 100 students were given a task and the time taken by each student to complete the task was recorded. The maximum time allowed to complete the task was one minute and all students completed the task within the maximum time. The times, \(T\) minutes, for the random sample of students are summarised as follows. \(n = 100 \quad \sum t = 61.88\) A researcher proposes that \(T\) can be modelled by the continuous random variable with probability density function \(f ( t ) = \begin{cases} \alpha t ^ { \alpha - 1 } & 0 \leqslant t \leqslant 1 , \\ 0 & \text { otherwise, } \end{cases}\) where \(\alpha\) is a positive constant. \section*{(a) In this question you must show detailed reasoning.} By finding \(\mathbf { E } ( T )\) according to the researcher's model, determine an approximation for the value of \(\alpha\). Give your answer correct to \(\mathbf { 3 }\) significant figures. Further information about the times taken for the sample of 100 students to complete the task is given in the table.
Time \(t\)\(0 \leqslant t < \frac { 1 } { 3 }\)\(\frac { 1 } { 3 } \leqslant t < \frac { 2 } { 3 }\)\(\frac { 2 } { 3 } \leqslant t \leqslant 1\)
Frequency183745
(b) Using the value of \(\alpha\) found in part (a), determine the extent to which the proposed model is a good model. (Do not carry out a goodness of fit test.)

Question 8:
AnswerMarks Guidance
8(a) DR: t = 0.6188
E(T) = 1 t . t 1 d t   −
0
=
+1
Choose α to make E(T) = t , so 0 .6 1 8 8 =
1  +
α = 0.6188α + 0.6188  0.3812α = 0.6188
 = 1.62(329) or 1547
AnswerMarks
953B1
M1
A1
B1
M1
A1
AnswerMarks
[6]1.1
3.3
1.1
3.1b
3.4
AnswerMarks
2.2aCorrect value of t
Correct integral stated or implied, correct limits somewhere.
1
Can be by parts, giving 1 −
1  +
Any indication that they know that t is not necessarily the same as
E(T), e.g. “put”, mention “unbiased estimate”, or use  (DR)
Solve their E(T) = 0.6188 to find α
Awrt 1.62 but allow 1.6 if only 2 SF given
AnswerMarks
(b)Probabilities are ( 13 )  , ( 23 ) ( 13 )   − , 1 ( 23 )  −
( ) ( )
100(1) , 100 ( 2 ) −( 1 ) , 1 0 0 1 ( 23 )   −
3 3 3
16.8(07), 35.0 (34.972), 48.2(21)
or Cumulative frequencies 16.8, 51.8 (100)
Reasonably close to 18, 37, 45 so a good model
AnswerMarks
or 18, 55 (100) if CFs usedM1
M1
A1
A1ft
AnswerMarks
[4]3.3
1.1
3.4
AnswerMarks
3.5aCorrect formula for one probability, e.g. attempt at definite
integral with their α, correct limits, or CDF F(t) = tα
Multiply at least one probability by 100 or divide observed
frequencies by 100, needs integral but allow limits confused
All correct. Allow [16.8, 16.9] and awrt 35.0 and 48.2, or no
100 factor if 0.18, 0.37, 0.45 used for comparison. If CFs
used, can ignore 100 (or 1)
Assessment, not “the model is correct” nor “the model is
wrong”, with reason (e.g. 16.8  18, 35.0  37, 48.2  45)
Needs all three values used (first 2 from CFD) (but they may
be wrong) and both M marks, but ft conclusion from wrong 
(which could lead to conclusion “not a good model”).
AnswerMarks Guidance
SCSample variance 0.0619, Var(T) = 0.0651, which M1
are close; so may be good modelA1 Conclusion as above based on these correct values. Max 2/4
1.4  α  1.8 gives frequencies that can be called “close”. E.g.: α = 2.62: E = 5.6, 28.9, 65.4 which is not close;
f
α = 1.44 (median): E = 20.6, 35.2, 44.2 which is close
f
PMT
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we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
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Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
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Question 8:
8 | (a) | DR: t = 0.6188
E(T) = 1 t . t 1 d t   −
0

=
+1

Choose α to make E(T) = t , so 0 .6 1 8 8 =
1  +
α = 0.6188α + 0.6188  0.3812α = 0.6188
 = 1.62(329) or 1547
953 | B1
M1
A1
B1
M1
A1
[6] | 1.1
3.3
1.1
3.1b
3.4
2.2a | Correct value of t
Correct integral stated or implied, correct limits somewhere.
1
Can be by parts, giving 1 −
1  +
Any indication that they know that t is not necessarily the same as
E(T), e.g. “put”, mention “unbiased estimate”, or use  (DR)
Solve their E(T) = 0.6188 to find α
Awrt 1.62 but allow 1.6 if only 2 SF given
(b) | Probabilities are ( 13 )  , ( 23 ) ( 13 )   − , 1 ( 23 )  −
( ) ( )
100(1) , 100 ( 2 ) −( 1 ) , 1 0 0 1 ( 23 )   −
3 3 3
16.8(07), 35.0 (34.972), 48.2(21)
or Cumulative frequencies 16.8, 51.8 (100)
Reasonably close to 18, 37, 45 so a good model
or 18, 55 (100) if CFs used | M1
M1
A1
A1ft
[4] | 3.3
1.1
3.4
3.5a | Correct formula for one probability, e.g. attempt at definite
integral with their α, correct limits, or CDF F(t) = tα
Multiply at least one probability by 100 or divide observed
frequencies by 100, needs integral but allow limits confused
All correct. Allow [16.8, 16.9] and awrt 35.0 and 48.2, or no
100 factor if 0.18, 0.37, 0.45 used for comparison. If CFs
used, can ignore 100 (or 1)
Assessment, not “the model is correct” nor “the model is
wrong”, with reason (e.g. 16.8  18, 35.0  37, 48.2  45)
Needs all three values used (first 2 from CFD) (but they may
be wrong) and both M marks, but ft conclusion from wrong 
(which could lead to conclusion “not a good model”).
SC | Sample variance 0.0619, Var(T) = 0.0651, which | M1 | Find sample variance and Var(T) & compare
are close; so may be good model | A1 | Conclusion as above based on these correct values. Max 2/4
1.4  α  1.8 gives frequencies that can be called “close”. E.g.: α = 2.62: E = 5.6, 28.9, 65.4 which is not close;
f
α = 1.44 (median): E = 20.6, 35.2, 44.2 which is close
f
PMT
Need to get in touch?
If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
Call us on
01223 553998
Alternatively, you can email us on
support@ocr.org.uk
For more information visit
ocr.org.uk/qualifications/resource-finder
ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2024 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
8 A random sample of 100 students were given a task and the time taken by each student to complete the task was recorded. The maximum time allowed to complete the task was one minute and all students completed the task within the maximum time. The times, $T$ minutes, for the random sample of students are summarised as follows.\\
$n = 100 \quad \sum t = 61.88$

A researcher proposes that $T$ can be modelled by the continuous random variable with probability density function\\
$f ( t ) = \begin{cases} \alpha t ^ { \alpha - 1 } & 0 \leqslant t \leqslant 1 , \\ 0 & \text { otherwise, } \end{cases}$\\
where $\alpha$ is a positive constant.

\section*{(a) In this question you must show detailed reasoning.}
By finding $\mathbf { E } ( T )$ according to the researcher's model, determine an approximation for the value of $\alpha$. Give your answer correct to $\mathbf { 3 }$ significant figures.

Further information about the times taken for the sample of 100 students to complete the task is given in the table.

\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
Time $t$ & $0 \leqslant t < \frac { 1 } { 3 }$ & $\frac { 1 } { 3 } \leqslant t < \frac { 2 } { 3 }$ & $\frac { 2 } { 3 } \leqslant t \leqslant 1$ \\
\hline
Frequency & 18 & 37 & 45 \\
\hline
\end{tabular}
\end{center}

(b) Using the value of $\alpha$ found in part (a), determine the extent to which the proposed model is a good model. (Do not carry out a goodness of fit test.)

\hfill \mbox{\textit{OCR Further Statistics 2024 Q8 [10]}}