Question 1 - A-Level Maths

OCR MEI Further Statistics A AS 2023 June — Question 1 12 marks

Exam Board	OCR MEI
Module	Further Statistics A AS (Further Statistics A AS)
Year	2023
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Discrete Probability Distributions
Type	Verify probability from combinatorial selection
Difficulty	Moderate -0.3 This is a straightforward Further Statistics question involving basic combinatorial probability and standard expectation/variance calculations. Part (a) requires simple counting of combinations (choosing 3 coins from 6 one-pound coins), parts (c) and (d) are routine applications of E(X) and Var(X) formulas with the linear transformation Y = 2X + 3. While it's a multi-part question worth several marks, each component is a standard textbook exercise requiring only direct application of learned techniques without novel insight or complex problem-solving.
Spec	5.02a Discrete probability distributions: general 5.02b Expectation and variance: discrete random variables 5.02c Linear coding: effects on mean and variance

1 Ryan has 6 one-pound coins and 4 two-pound coins. Ryan decides to select 3 of these coins at random to donate to a charity. The total value, in pounds, of these 3 coins is denoted by the random variable \(X\).

Show that \(\mathrm { P } ( X = 3 ) = \frac { 1 } { 6 }\). The table below shows the probability distribution of \(X\).
\(r\) 3 4 5 6
\(\mathrm { P } ( \mathrm { X } = \mathrm { r } )\) \(\frac { 1 } { 6 }\) \(\frac { 1 } { 2 }\) \(\frac { 3 } { 10 }\) \(\frac { 1 } { 30 }\)
Draw a graph to illustrate the distribution.
In this question you must show detailed reasoning. Find each of the following.
Ryan's friend Sasha decides to give the same amount as Ryan does to the charity plus an extra three pounds. The random variable \(Y\) represents the total amount of money, in pounds, given by Ryan and Sasha.
Determine each of the following.

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
1	(a)	6 5 4 6 10

× × or for ( )÷( )

10 9 8 3 3

1 =

Answer	Marks
6	M1

A1

Answer	Marks
[2]	1.1a
1.1	M1 for correct denominator

𝟏𝟎 𝟏 𝟏 𝟏

AG correct working only. A0 for ( )× × ×

𝟑 𝟏𝟎 𝟗 𝟖

Answer	Marks	Guidance
1	(b)	1/2 ytilibaborP

1/3

1/6

0 3 4 5 6

Answer	Marks
Value (£X)	B1

B1

Answer	Marks
[2]	1.1
1.1	For heights. 3, 4, 5 & 6 labelled. B0 if tops joined.

For labels to identify both axes and appropriate scale on the

probability axis.

Answer	Marks	Guidance
1	(c)	DR

1 1 3 1

E(𝑋) = 3× +4× +5× +6×

6 2 10 30

21 = (= 4.2)

5 E(𝑋2) = 32× 1 +42× 1 +52× 3 +62× 1 [= 91 ]

6 2 10 30 5

2 91 21

Var(𝑋) = −( )

5 5

14 = (=0.56)

Answer	Marks
25	M1

A1

M1

A1

Answer	Marks
[5]	1.1a

1.1

1.2

Answer	Marks
1.1	Allow 0.5 + 2 + 1.5 + 0.2. Condone 1 error for M1.

Allow 1.5 + 8 + 7.5 + 1.2. Condone 1 error for M1.

FT their E(X) and E(𝑋2). M0 if this leads to a non-positive

answer.

cao SCB1 for 0.56 if M0M1 or M1M0 awarded

Answer	Marks	Guidance
1	(d)	21 57

E(Y) = 2× +3 = (= 11.4)

5 5

Var(𝑌) = 𝟐𝟐× 𝟏𝟒

𝟐𝟓

56 = (=2.24)

Answer	Marks
25	B1

M1

A1

Answer	Marks
[3]	2.2a

3.5a

Answer	Marks
1.1	FT their E(X)

For 4 × their Var(X) seen provided this is positive.

cao SCB1 for correct answer without working seen.

Question 1:
1 | (a) | 6 5 4 6 10
× × or for ( )÷( )
10 9 8 3 3
1
=
6 | M1
A1
[2] | 1.1a
1.1 | M1 for correct denominator
𝟏𝟎 𝟏 𝟏 𝟏
AG correct working only. A0 for ( )× × ×
𝟑 𝟏𝟎 𝟗 𝟖
1 | (b) | 1/2 ytilibaborP
1/3
1/6
0
3 4 5 6
Value (£X) | B1
B1
[2] | 1.1
1.1 | For heights. 3, 4, 5 & 6 labelled. B0 if tops joined.
For labels to identify both axes and appropriate scale on the
probability axis.
1 | (c) | DR
1 1 3 1
E(𝑋) = 3× +4× +5× +6×
6 2 10 30
21
= (= 4.2)
5
E(𝑋2) = 32× 1 +42× 1 +52× 3 +62× 1 [= 91 ]
6 2 10 30 5
2
91 21
Var(𝑋) = −( )
5 5
14
= (=0.56)
25 | M1
A1
M1
M1
A1
[5] | 1.1a
1.1
1.1
1.2
1.1 | Allow 0.5 + 2 + 1.5 + 0.2. Condone 1 error for M1.
Allow 1.5 + 8 + 7.5 + 1.2. Condone 1 error for M1.
FT their E(X) and E(𝑋2). M0 if this leads to a non-positive
answer.
cao SCB1 for 0.56 if M0M1 or M1M0 awarded
1 | (d) | 21 57
E(Y) = 2× +3 = (= 11.4)
5 5
Var(𝑌) = 𝟐𝟐× 𝟏𝟒
𝟐𝟓
56
= (=2.24)
25 | B1
M1
A1
[3] | 2.2a
3.5a
1.1 | FT their E(X)
For 4 × their Var(X) seen provided this is positive.
cao SCB1 for correct answer without working seen.

Show LaTeX source

1 Ryan has 6 one-pound coins and 4 two-pound coins. Ryan decides to select 3 of these coins at random to donate to a charity. The total value, in pounds, of these 3 coins is denoted by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 3 ) = \frac { 1 } { 6 }$.

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

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$r$ & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( \mathrm { X } = \mathrm { r } )$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 2 }$ & $\frac { 3 } { 10 }$ & $\frac { 1 } { 30 }$ \\
\hline
\end{tabular}
\end{center}
\item Draw a graph to illustrate 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}

Ryan's friend Sasha decides to give the same amount as Ryan does to the charity plus an extra three pounds. The random variable $Y$ represents the total amount of money, in pounds, given by Ryan and Sasha.
\item Determine each of the following.

\begin{itemize}
  \item E (Y)
  \item $\operatorname { Var } ( Y )$
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2023 Q1 [12]}}

This paper (6 questions)

View full paper

Q1 12 Q2 6 Q3 7 Q4 10 Q5 10 Q6 15

\(r\)	3	4	5	6
\(\mathrm { P } ( \mathrm { X } = \mathrm { r } )\)	\(\frac { 1 } { 6 }\)	\(\frac { 1 } { 2 }\)	\(\frac { 3 } { 10 }\)	\(\frac { 1 } { 30 }\)