OCR MEI Further Statistics Minor 2024 June — Question 2 7 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
Year2024
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicUniform Distribution
TypeProbability within standard deviations
DifficultyStandard +0.3 This is a straightforward application of discrete uniform distribution formulas with standard bookwork for parts (a)-(b). Part (c) requires translating an inequality involving standardization and then counting outcomes, which is slightly beyond pure recall but still routine for Further Statistics students.
Spec5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution
2 The sides of a fair 12 -sided spinner are labelled \(1,2 , \ldots , 12\). The spinner is spun and \(X\) is the random variable denoting the number on the side of the spinner that it lands on.
  1. Suggest a suitable distribution to model \(X\). You should state the value(s) of any parameter(s).
  2. Find each of the following.
    You are given that \(\mathrm { E } ( X )\) is denoted by \(\mu\) and \(\operatorname { Var } ( X )\) is denoted by \(\sigma ^ { 2 }\).
  3. Determine \(\mathrm { P } \left( \left| \frac { 2 ( X - \mu ) } { \sigma } \right| > 1 \right)\).
Question 2:
AnswerMarks Guidance
2(a) (Discrete) Uniform or U...
U(12), n =12, or p=1/12 or (1,2,….,12)M1
A1
AnswerMarks
[2]1.2
2.5Specification of either the name or the
values must make it clear that this is a
AnswerMarks
discrete distribution.If Uniform or U… absent,
List (1, 2,….,12) gets M1
with p=1/12 for A1
AnswerMarks Guidance
2(b) E(X) = (12 + 1) / 2 = 6.5
1 4 3
Var(X) = (122 – 1) / 12 =
AnswerMarks
1 2B1
B1
AnswerMarks Guidance
[2]1.1
1.11 1 11 12 o r 1 1 .9 1 6 or awrt 11.9
2(c) 2(X −) 1 1
1 X +  or X − 
 2 2
1 1 4 3 1 1 4 3
X  6 .5 + o X r  6 .5 −
2 1 2 2 1 2
X  8 .2 2 ... o r X  4 .7 7 ...

Total of 8 possible numbers

AnswerMarks
(9, 10, 11, 12 or 1, 2, 3, 4) so prob = 8/12 = 2/3M1
M1 FT
AnswerMarks
A13.1a
1.1
AnswerMarks
2.2aUntangling the inequality correctly. This
mark can be awarded if wrong value(s) used
consistently. Condone “and” if recovered
(eg by correct answer)
Substituting their values into at least one of
the inequalities to produce a ‘decimal’
inequality in X, or using one inequality to
AnswerMarks
identify either (1, 2, 3, 4) or (9, 10, 11, 12)Could use complementary event,
2 ( X )   − 
considering P 1 
(or <) leading to
 
−  X + M1
2 2
or 4 .7 7 ..  X  8 .2 2 ... M1 FT
SC (1, 2, 3, 4 and 9, 10, 11, 12)
identified without working,
B1 only

Total of 4 possible numbers (5,

6, 7, 8) so required prob = 1 –
4/12 = 2/3 A1
AnswerMarks
Alternative method 1M1 FT
A1
AnswerMarks
A1n 1 2 3 4 5 6
value 3.2 2.6 2.0 1.4 0.9 0.3
n 7 8 9 10 11 12
value 0.3 0.9 1.4 2.0 2.6 3.2
For n =1-6, condone negative values with the correct modulus provided
AnswerMarks Guidance
final answer is correct.n 1
value3.2 2.6
Draw table of values for n= 1, 2, 3,……,12 and
2(𝑋−𝜇)
AnswerMarks Guidance
attempt to calculate for at least four
𝜎
AnswerMarks Guidance
n7 8
values of nvalue 0.3
For n =1-6, condone negative values with the correct modulus provided
Calculate at least four correct values (to 1 d.p)
final answer is correct.
Identify the 8 possible values of n so prob = 8/12 =
2/3
[3]
M1 FT
A1
A1
Question 2:
2 | (a) | (Discrete) Uniform or U...
U(12), n =12, or p=1/12 or (1,2,….,12) | M1
A1
[2] | 1.2
2.5 | Specification of either the name or the
values must make it clear that this is a
discrete distribution. | If Uniform or U… absent,
List (1, 2,….,12) gets M1
with p=1/12 for A1
2 | (b) | E(X) = (12 + 1) / 2 = 6.5
1 4 3
Var(X) = (122 – 1) / 12 =
1 2 | B1
B1
[2] | 1.1
1.1 | 1 1 11 12 o r 1 1 .9 1 6 or awrt 11.9
2 | (c) | 2(X −) 1 1
1 X +  or X − 
 2 2
1 1 4 3 1 1 4 3
X  6 .5 + o X r  6 .5 −
2 1 2 2 1 2
X  8 .2 2 ... o r X  4 .7 7 ...
Total of 8 possible numbers
(9, 10, 11, 12 or 1, 2, 3, 4) so prob = 8/12 = 2/3 | M1
M1 FT
A1 | 3.1a
1.1
2.2a | Untangling the inequality correctly. This
mark can be awarded if wrong value(s) used
consistently. Condone “and” if recovered
(eg by correct answer)
Substituting their values into at least one of
the inequalities to produce a ‘decimal’
inequality in X, or using one inequality to
identify either (1, 2, 3, 4) or (9, 10, 11, 12) | Could use complementary event,
2 ( X )   − 
considering P 1 

(or <) leading to
 
−  X + M1
2 2
or 4 .7 7 ..  X  8 .2 2 ... M1 FT
SC (1, 2, 3, 4 and 9, 10, 11, 12)
identified without working,
B1 only
Total of 4 possible numbers (5,
6, 7, 8) so required prob = 1 –
4/12 = 2/3 A1
Alternative method 1 | M1 FT
A1
A1 | n 1 2 3 4 5 6
value 3.2 2.6 2.0 1.4 0.9 0.3
n 7 8 9 10 11 12
value 0.3 0.9 1.4 2.0 2.6 3.2
For n =1-6, condone negative values with the correct modulus provided
final answer is correct. | n | 1 | 2 | 3 | 4 | 5 | 6
value | 3.2 | 2.6 | 2.0 | 1.4 | 0.9 | 0.3
Draw table of values for n= 1, 2, 3,……,12 and
2(𝑋−𝜇)
attempt to calculate | | for at least four
𝜎
n | 7 | 8 | 9 | 10 | 11 | 12
values of n | value | 0.3 | 0.9 | 1.4 | 2.0 | 2.6 | 3.2
For n =1-6, condone negative values with the correct modulus provided
Calculate at least four correct values (to 1 d.p)
final answer is correct.
Identify the 8 possible values of n so prob = 8/12 =
2/3
[3]
M1 FT
A1
A1
2 The sides of a fair 12 -sided spinner are labelled $1,2 , \ldots , 12$.

The spinner is spun and $X$ is the random variable denoting the number on the side of the spinner that it lands on.
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable distribution to model $X$. You should state the value(s) of any parameter(s).
\item Find each of the following.

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

You are given that $\mathrm { E } ( X )$ is denoted by $\mu$ and $\operatorname { Var } ( X )$ is denoted by $\sigma ^ { 2 }$.
\item Determine $\mathrm { P } \left( \left| \frac { 2 ( X - \mu ) } { \sigma } \right| > 1 \right)$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2024 Q2 [7]}}
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