Question 6 - A-Level Maths

OCR MEI Further Statistics Minor 2020 November — Question 6 9 marks

Exam Board	OCR MEI
Module	Further Statistics Minor (Further Statistics Minor)
Year	2020
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Uniform Distribution
Type	Derive general variance formula
Difficulty	Challenging +1.2 This is a multi-part question requiring derivation of variance formulas using standard summation formulas (Σr and Σr²), then applying a linear transformation, and finally a straightforward calculation. While it requires algebraic manipulation and understanding of variance properties, the techniques are standard for Further Statistics and the question provides clear scaffolding through its parts. It's moderately above average difficulty due to the algebraic proof requirement and multi-step nature, but not exceptionally challenging for students at this level.
Spec	5.02b Expectation and variance: discrete random variables 5.02e Discrete uniform distribution

The random variable \(X\) has a uniform distribution over the values \(\{ 1,2 , \ldots , n \}\). Show that \(\operatorname { Var } ( X )\) is given by \(\frac { 1 } { 12 } \left( n ^ { 2 } - 1 \right)\).
The random variable \(Y\) has a uniform distribution over the values \(\{ 1,3,5 , \ldots , 2 n - 1 \}\). Using the result in part (a) or otherwise, show that \(\operatorname { Var } ( Y )\) is given by \(\frac { 1 } { 3 } \left( n ^ { 2 } - 1 \right)\).
Given that \(n = 100\), find the least value of \(k\) for which \(\mathrm { P } ( \mu - k \sigma \leqslant Y \leqslant \mu + k \sigma ) = 1\), where the mean and standard deviation of \(Y\) are represented by \(\mu\) and \(\sigma\) respectively.

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
6	(a)	2 1 2 2 2

E=(𝑋𝑋 ) = (1 +2 …𝑙𝑙 )

𝑙𝑙

1 𝑛𝑛(𝑛𝑛+1)(2𝑛𝑛+1)

= 𝑛𝑛 � 6 �

(𝑛𝑛+1)(2𝑛𝑛+1)

6

2 (𝑙𝑙+1)(2𝑙𝑙+1) 𝑙𝑙+1

Var(X)= −� �

2 6 2 2

(4𝑙𝑙 +6𝑙𝑙+2) (3𝑙𝑙 +6𝑙𝑙+3)

= 1 (n2 −1) −

Answer	Marks
12 12 12	B1

M1

A1

Answer	Marks
[3]	3.1a

2.1

Answer	Marks
1.1	For initial expression for E(X2)

using their E(X2) and correct E(X)

At least one constructive intermediate

step must be seen

Answer	Marks
AG	term may be implied

2 𝑙𝑙

Answer	Marks	Guidance
6	(b)	Uniform {1, 3, 5, …, 2n − 1} is a transformation of

Y = 2X – 1 of uniform {1, 2, …, n}

so Var(Y) = 22 Var(X) = 1(n2−1)

Answer	Marks
3	M1

A1

Answer	Marks	Guidance
[2]	2.2a
1.1	AG
6	(c)	E(Y) = µ = n (= 100)

Var(Y) = 3333 or σ = 57.73

100 – 57.73k = 1 or 100 + 57.73k = 199

k = 1.715 [1.71, 1,72]

Answer	Marks
⇒	B1

B1

M1

A1

Answer	Marks
[4]	3.1a

1.1 2.2a

Answer	Marks
1.1	100 substituted into 1(n2−1)

3 using their numerical E(Y) and

Answer	Marks
Var(Y)	Allow inequalities

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Question 6:
6 | (a) | 2 1 2 2 2
E=(𝑋𝑋 ) = (1 +2 …𝑙𝑙 )
𝑙𝑙
1 𝑛𝑛(𝑛𝑛+1)(2𝑛𝑛+1)
= 𝑛𝑛 � 6 �
(𝑛𝑛+1)(2𝑛𝑛+1)
6
2
(𝑙𝑙+1)(2𝑙𝑙+1) 𝑙𝑙+1
Var(X)= −� �
2 6 2 2
(4𝑙𝑙 +6𝑙𝑙+2) (3𝑙𝑙 +6𝑙𝑙+3)
= 1 (n2 −1) −
12 12 12 | B1
M1
A1
[3] | 3.1a
2.1
1.1 | For initial expression for E(X2)
using their E(X2) and correct E(X)
At least one constructive intermediate
step must be seen
AG | term may be implied
2
𝑙𝑙
6 | (b) | Uniform {1, 3, 5, …, 2n − 1} is a transformation of
Y = 2X – 1 of uniform {1, 2, …, n}
so Var(Y) = 22 Var(X) = 1(n2−1)
3 | M1
A1
[2] | 2.2a
1.1 | AG
6 | (c) | E(Y) = µ = n (= 100)
Var(Y) = 3333 or σ = 57.73
100 – 57.73k = 1 or 100 + 57.73k = 199
k = 1.715 [1.71, 1,72]
⇒ | B1
B1
M1
A1
[4] | 3.1a
1.1
2.2a
1.1 | 100 substituted into 1(n2−1)
3
using their numerical E(Y) and
Var(Y) | Allow inequalities
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

Show LaTeX source

6
\begin{enumerate}[label=(\alph*)]
\item The random variable $X$ has a uniform distribution over the values $\{ 1,2 , \ldots , n \}$. Show that $\operatorname { Var } ( X )$ is given by $\frac { 1 } { 12 } \left( n ^ { 2 } - 1 \right)$.
\item The random variable $Y$ has a uniform distribution over the values $\{ 1,3,5 , \ldots , 2 n - 1 \}$. Using the result in part (a) or otherwise, show that $\operatorname { Var } ( Y )$ is given by $\frac { 1 } { 3 } \left( n ^ { 2 } - 1 \right)$.
\item Given that $n = 100$, find the least value of $k$ for which $\mathrm { P } ( \mu - k \sigma \leqslant Y \leqslant \mu + k \sigma ) = 1$, where the mean and standard deviation of $Y$ are represented by $\mu$ and $\sigma$ respectively.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2020 Q6 [9]}}

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