Question 2 - A-Level Maths

OCR MEI Further Statistics Major 2021 November — Question 2 13 marks

Exam Board	OCR MEI
Module	Further Statistics Major (Further Statistics Major)
Year	2021
Session	November
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Discrete Probability Distributions
Type	Verify probability from combinatorial selection
Difficulty	Moderate -0.3 This is a structured Further Maths statistics question with clear scaffolding. Part (a) requires basic combinatorial counting (all three dice showing same value: 6 cases out of 216). Parts (d)-(f) involve standard expectation and variance calculations using given probabilities. While it's from Further Maths, the techniques are routine applications of discrete probability formulas with no novel problem-solving required, making it slightly easier than average A-level difficulty.
Spec	5.02a Discrete probability distributions: general 5.02b Expectation and variance: discrete random variables 5.02c Linear coding: effects on mean and variance

2 In a game at a charity fair, a player rolls 3 unbiased six-sided dice. The random variable \(X\) represents the difference between the highest and lowest scores.

Show that \(\mathrm { P } ( X = 0 ) = \frac { 1 } { 36 }\). The table shows the probability distribution of \(X\).
\(r\) 0 1 2 3 4 5
\(\mathrm { P } ( \mathrm { X } = \mathrm { r } )\) \(\frac { 1 } { 36 }\) \(\frac { 5 } { 36 }\) \(\frac { 2 } { 9 }\) \(\frac { 1 } { 4 }\) \(\frac { 2 } { 9 }\) \(\frac { 5 } { 36 }\)
Draw a graph to illustrate the distribution.
Describe the shape of the distribution.
In this question you must show detailed reasoning. Find each of the following.
As a result of playing the game, the player receives \(30 X\) pence from the organiser of the game.
Find the variance of the amount that the player receives.
The player pays \(k\) pence to play the game. Given that the average profit made by the organiser is 12.5 pence per game, determine the value of \(k\).

Show mark scheme Show mark scheme source

Question 2:

Answer	Marks	Guidance
2	(a)	P(X =0)= 6×1×1

6 6 6

= 1

Answer	Marks
36	M1

A1

Answer	Marks	Guidance
[2]	3.1a
1.1	AG	Allow M1 for

1×1 = 1

6 6 36

Answer	Marks	Guidance
2	(b)	0.30

r

0.20 =

)X(P

0.10

0.00 0 1 2 3 4 5

Answer	Marks
r	B1

B1

Answer	Marks
[2]	1.1
1.1	For heights
For axes and labels	Roughly correct but

must have linear scale

Do not allow just P on

vertical axis

Answer	Marks	Guidance
2	(c)	The distribution has (slight) negative skew
[1]	1.1	Allow ‘roughly symmetrical’ or
‘unimodal’	Not ‘Normal

distribution’

Answer	Marks	Guidance
2	(d)	DR

E(X)=0× 1 +1× 5 +2×2+3×1 +4×2+5× 5

36 36 9 4 9 36

=105 = 35 =2.9166

36 12

E(X2)=02× 1 +12× 5 +22×2+32×1 +42×2+52× 5

36 36 9 4 9 36

= 371=10.3055

36 Var(X)=10.3055−(2.9166)2

= 259 =1.80 (1.7986…)

Answer	Marks
144	M1

A1

M1

A1

Answer	Marks
[5]	1.1a

1.1

1.2

Answer	Marks	Guidance
1.1	Allow fraction or decimal form
2	(e)	Variance=302×1.7986=1619 (pence2)
[1]	1.1
2	(f)	Average amount received = 30 × 2.916… = 87.5
k – 87.5 = 12.5 k = 100	B1

B1

Answer	Marks
[2]	3.1a

1.1

Question 2:
2 | (a) | P(X =0)= 6×1×1
6 6 6
= 1
36 | M1
A1
[2] | 3.1a
1.1 | AG | Allow M1 for
1×1 = 1
6 6 36
2 | (b) | 0.30
r
0.20
=
)X(P
0.10
0.00
0 1 2 3 4 5
r | B1
B1
[2] | 1.1
1.1 | For heights
For axes and labels | Roughly correct but
must have linear scale
Do not allow just P on
vertical axis
2 | (c) | The distribution has (slight) negative skew | B1
[1] | 1.1 | Allow ‘roughly symmetrical’ or
‘unimodal’ | Not ‘Normal
distribution’
2 | (d) | DR
E(X)=0× 1 +1× 5 +2×2+3×1 +4×2+5× 5
36 36 9 4 9 36
=105 = 35 =2.9166
36 12
E(X2)=02× 1 +12× 5 +22×2+32×1 +42×2+52× 5
36 36 9 4 9 36
= 371=10.3055
36
Var(X)=10.3055−(2.9166)2
= 259 =1.80 (1.7986…)
144 | M1
A1
M1
M1
A1
[5] | 1.1a
1.1
1.1
1.2
1.1 | Allow fraction or decimal form
2 | (e) | Variance=302×1.7986=1619 (pence2) | B1
[1] | 1.1
2 | (f) | Average amount received = 30 × 2.916… = 87.5
k – 87.5 = 12.5 k = 100 | B1
B1
[2] | 3.1a
1.1

Show LaTeX source

2 In a game at a charity fair, a player rolls 3 unbiased six-sided dice. The random variable $X$ represents the difference between the highest and lowest scores.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 0 ) = \frac { 1 } { 36 }$.

The table shows the probability distribution of $X$.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( \mathrm { X } = \mathrm { r } )$ & $\frac { 1 } { 36 }$ & $\frac { 5 } { 36 }$ & $\frac { 2 } { 9 }$ & $\frac { 1 } { 4 }$ & $\frac { 2 } { 9 }$ & $\frac { 5 } { 36 }$ \\
\hline
\end{tabular}
\end{center}
\item Draw a graph to illustrate the distribution.
\item Describe the shape of the distribution.
\item In this question you must show detailed reasoning.

Find each of the following.

\begin{itemize}
  \item $\mathrm { E } ( X )$
  \item $\operatorname { Var } ( X )$
\end{itemize}

As a result of playing the game, the player receives $30 X$ pence from the organiser of the game.
\item Find the variance of the amount that the player receives.
\item The player pays $k$ pence to play the game.

Given that the average profit made by the organiser is 12.5 pence per game, determine the value of $k$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2021 Q2 [13]}}

This paper (11 questions)

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Q1 6 Q2 13 Q3 10 Q4 8 Q5 17 Q6 14 Q7 10 Q8 16 Q9 6 Q10 9 Q11 11

\(r\)	0	1	2	3	4	5
\(\mathrm { P } ( \mathrm { X } = \mathrm { r } )\)	\(\frac { 1 } { 36 }\)	\(\frac { 5 } { 36 }\)	\(\frac { 2 } { 9 }\)	\(\frac { 1 } { 4 }\)	\(\frac { 2 } { 9 }\)	\(\frac { 5 } { 36 }\)