OCR MEI Further Statistics Minor 2019 June — Question 6 7 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBivariate data
TypeCalculate r from summary statistics
DifficultyStandard +0.8 This Further Maths statistics question requires understanding of discrete uniform distributions, careful handling of floor functions for odd/even cases, and variance properties. The algebraic manipulation needed for parts (a)-(b) and applying variance rules for sums in part (c) goes beyond standard A-level, though it's methodical rather than requiring deep insight.
Spec5.02e Discrete uniform distribution5.04b Linear combinations: of normal distributions
6 The discrete random variable \(X\) has a uniform distribution over \(\{ n , n + 1 , \ldots , 2 n \}\).
  1. Given that \(n\) is odd, find \(\mathrm { P } \left( X < \frac { 3 } { 2 } n \right)\).
  2. Given instead that \(n\) is even, find \(\mathrm { P } \left( X < \frac { 3 } { 2 } n \right)\), giving your answer as a single algebraic fraction.
  3. The sum of 6 independent values of \(X\) is denoted by \(Y\). Find \(\operatorname { Var } ( Y )\).
Question 6:
AnswerMarks Guidance
6(a) P(X  3n) 1
2 2B1
[1]1.1
(b)X can take n1 values soi
of which 1n are below 3n
2 2
n n
P(X  3n) 
AnswerMarks
2 2(n1) 2n2M1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1convincing reasoning required if
using algebraic approach e.g.
(3n1) n1 
2
1(n11)
2
1(n1)1
AnswerMarks
2 22n n1 
accept correct
conclusion reached
from list(s) of
numbers
SC3 for correct
expression with no
working
AnswerMarks
(c)(n + 1) values so Var(X) 1 [(n1)2 1]
12
Var(Y)6theirVar(X)
Var(Y) 1(n2 2n) 1n(n2) isw
AnswerMarks
2 2M1
M1
A1
AnswerMarks
[3]2.2a
1.1
AnswerMarks
1.1Use of discrete uniform variance
over n1 values
AnswerMarks
Simplified form1 (n2 2n)
12
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Question 6:
6 | (a) | P(X  3n) 1
2 2 | B1
[1] | 1.1
(b) | X can take n1 values soi
of which 1n are below 3n
2 2
n n
P(X  3n) 
2 2(n1) 2n2 | M1
M1
A1
[3] | 3.1a
1.1
1.1 | convincing reasoning required if
using algebraic approach e.g.
(3n1) n1 
2
1(n11)
2
1(n1)1
2 2 | 2n n1 
accept correct
conclusion reached
from list(s) of
numbers
SC3 for correct
expression with no
working
(c) | (n + 1) values so Var(X) 1 [(n1)2 1]
12
Var(Y)6theirVar(X)
Var(Y) 1(n2 2n) 1n(n2) isw
2 2 | M1
M1
A1
[3] | 2.2a
1.1
1.1 | Use of discrete uniform variance
over n1 values
Simplified form | 1 (n2 2n)
12
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
6 The discrete random variable $X$ has a uniform distribution over $\{ n , n + 1 , \ldots , 2 n \}$.
\begin{enumerate}[label=(\alph*)]
\item Given that $n$ is odd, find $\mathrm { P } \left( X < \frac { 3 } { 2 } n \right)$.
\item Given instead that $n$ is even, find $\mathrm { P } \left( X < \frac { 3 } { 2 } n \right)$, giving your answer as a single algebraic fraction.
\item The sum of 6 independent values of $X$ is denoted by $Y$.

Find $\operatorname { Var } ( Y )$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2019 Q6 [7]}}
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Q1 7 Q2 9 Q3 4 Q4 17 Q5 16 Q6 7