OCR MEI Further Statistics Minor 2024 June — Question 6 9 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
Year2024
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeMultiple unknowns from expectation and variance
DifficultyStandard +0.8 This is a multi-part question requiring systematic algebraic manipulation of expectation and variance formulas, optimization using calculus (finding maximum variance), and conceptual understanding that minimum variance occurs when the distribution is degenerate. The optimization in (b)(ii) and the conceptual insight needed for (c) elevate this above routine exercises, though the individual techniques are standard for Further Statistics.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance

6 The probability distribution of a discrete random variable, \(X\), is shown in the table below.
\(x\)012
\(\mathrm { P } ( X = x )\)\(1 - a - b\)\(a\)\(b\)
  1. Find \(\mathrm { E } ( X )\) in terms of \(a\) and \(b\).
    1. In the case where \(\mathrm { E } ( \mathrm { X } ) = \mathrm { a } + 0.4\), find an expression for \(\operatorname { Var } ( X )\) in terms of \(a\).
    2. In this case, show that the greatest possible value of \(\operatorname { Var } ( X )\) is 0.65 . You must state the associated value of \(a\).
  2. You are now given instead that \(\mathrm { E } ( X )\) is not known.
    1. State the least possible value of \(\operatorname { Var } ( X )\).
    2. Give all possible pairs of values of \(a\) and \(b\) which give the least possible value of \(\operatorname { Var } ( X )\) stated in part (c)(i).

Question 6:
AnswerMarks Guidance
6(a) E(X) = a + 2b
[1]1.1 Correct expression for E(X)
6(b) (i)
Var(X) = 12a + 220.2 – (a + 0.4)2
= a + 0.8 – a2 – 0.8a – 0.16
AnswerMarks
= –a2 + 0.2a + 0.64 oeB1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1Correct expression for Var(X) =
12*a+22*b – (E(X))2 using their b
(or the symbol b) and their E(X)
Like terms must be collected
AnswerMarks Guidance
6(b) (ii)
= –((a – 0.1)2 – 0.12) + 0.64
= 0.65 – (a – 0.1)2 whose maximum value is 0.65
AnswerMarks
when a = 0.1M1
A1
AnswerMarks
[2]3.1a
2.2aValid method for finding TP, either
by completing the square,
differentiating their quadratic in a
and equating to zero, or by clear
use of -b/2a from quadratic
AnswerMarks
formula to find ad V
= − 2 a + 0 .2 = 0
d a
Solving, a = 0.1 and substituting, max
V = 0.65
AnswerMarks Guidance
6(c) (i)
[1]2.2a
6(c) (ii)
soi
AnswerMarks
So (a, b) pairs are (0, 0), (1, 0), (0, 1)M1
A1
AnswerMarks
[2]3.1a
3.2aFor clear understanding that the
minimum variance comes when the
distribution is as concentrated as
possible, implied by at least one
correct pair of values
All 3 pairs correct and no others.
Allow any unambiguous format
(eg a = 0 and b = 0 or a = 1 and...
etc)
PMT
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Question 6:
6 | (a) | E(X) = a + 2b | B1
[1] | 1.1 | Correct expression for E(X)
6 | (b) | (i) | a + 2b = a + 0.4 => b = 0.2
Var(X) = 12a + 220.2 – (a + 0.4)2
= a + 0.8 – a2 – 0.8a – 0.16
= –a2 + 0.2a + 0.64 oe | B1
M1
A1
[3] | 3.1a
1.1
1.1 | Correct expression for Var(X) =
12*a+22*b – (E(X))2 using their b
(or the symbol b) and their E(X)
Like terms must be collected
6 | (b) | (ii) | Var(X) = –(a2 – 0.2a) + 0.64
= –((a – 0.1)2 – 0.12) + 0.64
= 0.65 – (a – 0.1)2 whose maximum value is 0.65
when a = 0.1 | M1
A1
[2] | 3.1a
2.2a | Valid method for finding TP, either
by completing the square,
differentiating their quadratic in a
and equating to zero, or by clear
use of -b/2a from quadratic
formula to find a | d V
= − 2 a + 0 .2 = 0
d a
Solving, a = 0.1 and substituting, max
V = 0.65
6 | (c) | (i) | 0 | B1
[1] | 2.2a
6 | (c) | (ii) | Probabilities must be 1, 0, 0 or 0, 1, 0 or 0, 0, 1
soi
So (a, b) pairs are (0, 0), (1, 0), (0, 1) | M1
A1
[2] | 3.1a
3.2a | For clear understanding that the
minimum variance comes when the
distribution is as concentrated as
possible, implied by at least one
correct pair of values
All 3 pairs correct and no others.
Allow any unambiguous format
(eg a = 0 and b = 0 or a = 1 and...
etc)
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.
6 The probability distribution of a discrete random variable, $X$, is shown in the table below.

\begin{center}
\begin{tabular}{ | l | c | l | l | }
\hline
$x$ & 0 & 1 & 2 \\
\hline
$\mathrm { P } ( X = x )$ & $1 - a - b$ & $a$ & $b$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( X )$ in terms of $a$ and $b$.
\item \begin{enumerate}[label=(\roman*)]
\item In the case where $\mathrm { E } ( \mathrm { X } ) = \mathrm { a } + 0.4$, find an expression for $\operatorname { Var } ( X )$ in terms of $a$.
\item In this case, show that the greatest possible value of $\operatorname { Var } ( X )$ is 0.65 . You must state the associated value of $a$.
\end{enumerate}\item You are now given instead that $\mathrm { E } ( X )$ is not known.
\begin{enumerate}[label=(\roman*)]
\item State the least possible value of $\operatorname { Var } ( X )$.
\item Give all possible pairs of values of $a$ and $b$ which give the least possible value of $\operatorname { Var } ( X )$ stated in part (c)(i).
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2024 Q6 [9]}}