Question 3 - A-Level Maths

OCR MEI Further Statistics Minor 2023 June — Question 3 10 marks

Exam Board	OCR MEI
Module	Further Statistics Minor (Further Statistics Minor)
Year	2023
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Discrete Probability Distributions
Type	Verify probability from combinatorial selection
Difficulty	Standard +0.3 This is a straightforward Further Maths statistics question requiring systematic enumeration of outcomes for part (a), standard expectation and variance calculations from a given distribution in part (b), and basic knowledge of variance properties for independent random variables in part (c). While it involves multiple steps and careful counting, all techniques are routine applications of standard formulas with no novel insight required.
Spec	5.02a Discrete probability distributions: general 5.02b Expectation and variance: discrete random variables 5.02c Linear coding: effects on mean and variance 5.04a Linear combinations: E(aX+bY), Var(aX+bY)

3 A fair four-sided dice has its faces numbered \(0,1,2,3\). The dice is rolled three times. The discrete random variable \(X\) is the sum of the lowest and highest scores obtained.

Show that \(\mathrm { P } ( X = 1 ) = \frac { 3 } { 32 }\). The table below shows the probability distribution of \(X\).
\(r\) 0 1 2 3 4 5 6
\(\mathrm { P } ( X = r )\) \(\frac { 1 } { 64 }\) \(\frac { 3 } { 32 }\) \(\frac { 13 } { 64 }\) \(\frac { 3 } { 8 }\) \(\frac { 13 } { 64 }\) \(\frac { 3 } { 32 }\) \(\frac { 1 } { 64 }\)
In this question you must show detailed reasoning. Find each of the following.
- \(\mathrm { E } ( X )\)
- \(\operatorname { Var } ( X )\)
- The random variable \(Y\) represents the sum of 10 values of \(X\).
  1. State a property of the 10 values of \(X\) that would make it possible to deduce the standard deviation of \(Y\).
  2. Given that this property holds, determine the standard deviation of \(Y\).

Show mark scheme Show mark scheme source

Question 3:

Answer	Marks	Guidance
3	(a)	Both cases: 2 zeroes, 1 one and 1 zero 2 ones seen

1 3 1 3

3×( ) +3×( )

4 4

6 3

= =

Answer	Marks
64 32	B1

M1

A1

Answer	Marks
[3]	3.1a

1.1

Answer	Marks
1.1	Calculation needs to relate correctly to the two cases.

3

1 Following B0 M0, SC B1 for use of ( ) but without

4 reference to the two distinct cases

AG

Answer	Marks	Guidance
3	(b)	DR

E(𝑋) = 3 with justification

1 3 13 3

E(𝑋2) = (0× )+1× +4× +9×

64 32 64 8

13 3 1

+16× +25× +36×

64 32 64

167 Var(𝑋) = 𝑡ℎ𝑒𝑖𝑟 { −32}

16

23 =

Answer	Marks
16	B1

M1

A1

Answer	Marks
[4]	2.4

1.1a

1.1

Answer	Marks
1.1	1 3 13 3 13 3

(0× )+1× +2× +3× +4× +5× +

64 32 64 8 64 32

1 6× or ‘By symmetry’

64 E(𝑋2) = 167 = 10.4375 quoted without working is M0

16 Allow one slip in calculation.

Alt: M1 M1 for 1 {(0−3)2+6(1−3)2+13(2−3)2+

64 24(3−3)2+13(4−3)2+6(5−3)2+(6−3)2} oe

1.4375 allow 1.4 Must follow M1 M1

Answer	Marks	Guidance
3	(c)	(i)
[1]	2.2a
3	(c)	(ii)

Var(𝑌) = 10×𝑡ℎ𝑒𝑖𝑟

16

1 SD(𝑌) = 3.79 or √230

Answer	Marks
4	M1

A1 FT

Answer	Marks
[2]	1.1a
1.1	230

16 Allow 3.8

Question 3:
3 | (a) | Both cases: 2 zeroes, 1 one and 1 zero 2 ones seen
1 3 1 3
3×( ) +3×( )
4 4
6 3
= =
64 32 | B1
M1
A1
[3] | 3.1a
1.1
1.1 | Calculation needs to relate correctly to the two cases.
3
1
Following B0 M0, SC B1 for use of ( ) but without
4
reference to the two distinct cases
AG
3 | (b) | DR
E(𝑋) = 3 with justification
1 3 13 3
E(𝑋2) = (0× )+1× +4× +9×
64 32 64 8
13 3 1
+16× +25× +36×
64 32 64
167
Var(𝑋) = 𝑡ℎ𝑒𝑖𝑟 { −32}
16
23
=
16 | B1
M1
M1
A1
[4] | 2.4
1.1a
1.1
1.1 | 1 3 13 3 13 3
(0× )+1× +2× +3× +4× +5× +
64 32 64 8 64 32
1
6× or ‘By symmetry’
64
E(𝑋2) = 167 = 10.4375 quoted without working is M0
16
Allow one slip in calculation.
Alt: M1 M1 for 1 {(0−3)2+6(1−3)2+13(2−3)2+
64
24(3−3)2+13(4−3)2+6(5−3)2+(6−3)2} oe
1.4375 allow 1.4 Must follow M1 M1
3 | (c) | (i) | The values must be independent | B1
[1] | 2.2a
3 | (c) | (ii) | 23
Var(𝑌) = 10×𝑡ℎ𝑒𝑖𝑟
16
1
SD(𝑌) = 3.79 or √230
4 | M1
A1 FT
[2] | 1.1a
1.1 | 230
16
Allow 3.8

Show LaTeX source

3 A fair four-sided dice has its faces numbered $0,1,2,3$. The dice is rolled three times. The discrete random variable $X$ is the sum of the lowest and highest scores obtained.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 1 ) = \frac { 3 } { 32 }$.

The table below shows the probability distribution of $X$.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 1 } { 64 }$ & $\frac { 3 } { 32 }$ & $\frac { 13 } { 64 }$ & $\frac { 3 } { 8 }$ & $\frac { 13 } { 64 }$ & $\frac { 3 } { 32 }$ & $\frac { 1 } { 64 }$ \\
\hline
\end{tabular}
\end{center}
\item In this question you must show detailed reasoning.

Find each of the following.

\begin{itemize}
  \item $\mathrm { E } ( X )$
  \item $\operatorname { Var } ( X )$
\item The random variable $Y$ represents the sum of 10 values of $X$.
\begin{enumerate}[label=(\roman*)]
\item State a property of the 10 values of $X$ that would make it possible to deduce the standard deviation of $Y$.
\item Given that this property holds, determine the standard deviation of $Y$.
\end{itemize}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2023 Q3 [10]}}

This paper (7 questions)

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Q1 8 Q2 5 Q3 10 Q4 13 Q5 8 Q6 10 Q7 6

\(r\)	0	1	2	3	4	5	6
\(\mathrm { P } ( X = r )\)	\(\frac { 1 } { 64 }\)	\(\frac { 3 } { 32 }\)	\(\frac { 13 } { 64 }\)	\(\frac { 3 } { 8 }\)	\(\frac { 13 } { 64 }\)	\(\frac { 3 } { 32 }\)	\(\frac { 1 } { 64 }\)