OCR MEI Further Statistics Major 2019 June — Question 10 14 marks

Exam BoardOCR MEI
ModuleFurther Statistics Major (Further Statistics Major)
Year2019
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeFind parameter from median
DifficultyStandard +0.8 This is a multi-part Further Maths statistics question requiring integration to find k, deriving the CDF, then solving a non-trivial equation involving substitution p=2^m. Part (c) requires algebraic manipulation and insight to reach the given form, followed by solving for m from a quadratic. While systematic, it demands careful algebra across multiple steps and is above standard A-level difficulty due to the Further Maths content and the non-obvious substitution.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration
10 The probability density function of the continuous random variable \(X\) is given by \(f ( x ) = \begin{cases} k x ^ { m } & 0 \leqslant x \leqslant a , \\ 0 & \text { otherwise, } \end{cases}\) where \(a , k\) and \(m\) are positive constants.
  1. Show that \(k = \frac { m + 1 } { a ^ { m + 1 } }\).
  2. Find the cumulative distribution function of \(X\) in terms of \(x , a\) and \(m\).
  3. Given that \(\mathrm { P } \left( \frac { 1 } { 4 } a < X < \frac { 1 } { 2 } a \right) = \frac { 1 } { 10 }\),
    1. show that \(2 p ^ { 2 } - 10 p + 5 = 0\), where \(p = 2 ^ { m }\),
    2. find the value of \(m\). \section*{END OF QUESTION PAPER}
Question 10:
AnswerMarks Guidance
10(a) a
 kxm dx1
0
kxm1 a
  1
m1
0
kam1 m1
1k 
AnswerMarks
m1 am1M1
M1
A1
AnswerMarks
[3]For integral equated to 1
AG
AnswerMarks Guidance
10(b) xm1
 um du
 am1
0
m1 um1  x
= 
am1 m1
0
xm1
=
am1
0 x0
xm1
F(x) 0 xa
am1
1 xa
AnswerMarks
B1
M1
A1
A1
AnswerMarks
[4]Limits needed.
Can get first two marks if in terms of k
oe use of constant of integration
oe but not with k
Fully simplified
For fully correct answer with no
incorrect working allow all 4 marks.
AnswerMarks Guidance
10(c) (i)
P 1a X  1a  2  4
4 2 am1 am1
1 1
 
2m1 4m1
1 1 1
  
2p 4p2 10
AnswerMarks
10p52p2 2p210p50M1
M1
M1
A1
AnswerMarks
[4] F  
For use of F 1a 1a , oe
2 4
Forming equation in p
AnswerMarks Guidance
AGAllow in terms of k
10(c) (ii)
log0.5635 log4.4365
so m or
log2 log2
giving m = 2.149 (and reject negative value
AnswerMarks
0.8275)B1
M1
A1
AnswerMarks
[3]BC
For attempt to find m5±√15
Or 𝑝 =
2
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
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Cambridge
CB2 8EA
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OCR is an exempt Charity
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Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
Question 10:
10 | (a) | a
 kxm dx1
0
kxm1 a
  1
m1
0
kam1 m1
1k 
m1 am1 | M1
M1
A1
[3] | For integral equated to 1
AG
10 | (b) | xm1

 um du
 am1
0
m1 um1  x
= 
am1 m1
0
xm1
=
am1
0 x0

xm1
F(x) 0 xa
am1
1 xa
 | B1
M1
A1
A1
[4] | Limits needed.
Can get first two marks if in terms of k
oe use of constant of integration
oe but not with k
Fully simplified
For fully correct answer with no
incorrect working allow all 4 marks.
10 | (c) | (i) | 1a m1 1a m1
P 1a X  1a  2  4
4 2 am1 am1
1 1
 
2m1 4m1
1 1 1
  
2p 4p2 10
10p52p2 2p210p50 | M1
M1
M1
A1
[4] |  F  
For use of F 1a 1a , oe
2 4
Forming equation in p
AG | Allow in terms of k
10 | (c) | (ii) | p = 0.5635 or p = 4.4365
log0.5635 log4.4365
so m or
log2 log2
giving m = 2.149 (and reject negative value
0.8275) | B1
M1
A1
[3] | BC
For attempt to find m | 5±√15
Or 𝑝 =
2
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
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 (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
10 The probability density function of the continuous random variable $X$ is given by\\
$f ( x ) = \begin{cases} k x ^ { m } & 0 \leqslant x \leqslant a , \\ 0 & \text { otherwise, } \end{cases}$\\
where $a , k$ and $m$ are positive constants.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { m + 1 } { a ^ { m + 1 } }$.
\item Find the cumulative distribution function of $X$ in terms of $x , a$ and $m$.
\item Given that $\mathrm { P } \left( \frac { 1 } { 4 } a < X < \frac { 1 } { 2 } a \right) = \frac { 1 } { 10 }$,
\begin{enumerate}[label=(\roman*)]
\item show that $2 p ^ { 2 } - 10 p + 5 = 0$, where $p = 2 ^ { m }$,
\item find the value of $m$.

\section*{END OF QUESTION PAPER}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2019 Q10 [14]}}
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Q1 11 Q2 9 Q3 9 Q4 7 Q5 13 Q6 18 Q7 11 Q8 13 Q9 15 Q10 14