OCR MEI Further Statistics Major Specimen — Question 11 24 marks

Exam BoardOCR MEI
ModuleFurther Statistics Major (Further Statistics Major)
SessionSpecimen
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeThree or more independent values
DifficultyStandard +0.3 This is a straightforward application of discrete probability distributions requiring calculation of basic probabilities, expectation, and variance using standard formulas. Part (i) involves simple probability calculations for a discrete uniform distribution and recognizing that H=60 is extremely unlikely. Parts (iii)-(iv) are routine applications of E(aX) = aE(X) and Var(aX) = a²Var(X) for independent random variables. The conceptual reasoning in part (ii) is elementary. This is easier than average as it's mostly formula application with no novel problem-solving required.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p)5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions
11 Two girls, Lili and Hui, play a game with a fair six-sided dice. The dice is thrown 10 times. \(X _ { 1 } , X _ { 2 } , \ldots , X _ { 10 }\) represent the scores on the \(1 ^ { \text {st } } , 2 ^ { \text {nd } } , \ldots , 10 ^ { \text {th } }\) throws of the dice. \(L\) denotes Lili's score and \(L = 10 X _ { 1 }\). \(H\) denotes Hui's score and \(H = X _ { 1 } + X _ { 2 } + X _ { 3 } + \ldots + X _ { 10 }\).
  1. Calculate
    The spreadsheet below shows a simulation of 25 plays of the game. The cell E3, highlighted, shows the score when the dice is thrown the fourth time in the first game. \begin{table}[h]
    ABCDEFGHIJKLMN
    1Throw of diceLili'sHui's
    212345678910scorescore
    3Game 135211311143022
    4Game 263244353356038
    5Game 364265215236036
    6Game 415166314621035
    7Game 544316441624035
    8Game 621512515232027
    9Game 711344563421033
    10Game 811363445231032
    11Game 922243215562032
    12Game 1035335343113031
    13Game 1153655421155037
    14Game 1264324133536034
    15Game 1323212222212019
    16Game 1441331266134030
    17Game 1551263463645040
    18Game 1636115313333029
    19Game 1752524522345034
    20Game 1836355231123031
    21Game 1966315634166041
    22Game 2026456524332040
    23Game 2153545336615041
    24Game 2263556356116041
    25Game 2354556421365041
    26Game 2435232432333030
    27Game 2552424522525033
    28
    29mean37.6033.68
    30sd17.395.77
    \captionsetup{labelformat=empty} \caption{Fig. 11}
    \end{table}
  2. Use the simulation to estimate \(\mathrm { P } ( L > 40 )\) and \(\mathrm { P } ( H > 40 )\).
  3. (A) Calculate the exact value of \(\mathrm { P } ( L > 40 )\).
    (B) Comment on how the exact value compares with your estimate of \(\mathrm { P } ( L > 40 )\) in part (v). Hui wonders whether it is appropriate to use the Central Limit Theorem to approximate the distribution of \(X _ { 1 } + X _ { 2 } + X _ { 3 } + \ldots + X _ { 10 }\).
  4. (A) State what type of diagram Hui could draw, based on the output from the spreadsheet, to investigate this.
    (B) Explain how she should interpret the diagram.
  5. (A) Calculate an approximate value of \(\mathrm { P } \left( X _ { 1 } + X _ { 2 } + X _ { 3 } + \ldots + X _ { 10 } > 40 \right)\) using the Central Limit Theorem.
    (B) Comment on how this value compares with your estimate of \(\mathrm { P } ( H > 40 )\) in part (v). \section*{Copyright Information:} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.
    For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
    OCR is part of the
Question 11:
AnswerMarks Guidance
11(i) 1
P(L(cid:32)60)(cid:32)
6
(cid:167)1(cid:183) 10
P(H (cid:32)60)(cid:32)(cid:168) (cid:184)
(cid:169)6(cid:185)
AnswerMarks
(cid:32)1.65(cid:117)10(cid:16)8B1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks Guidance
1.1N
11(ii) Lili’s score because Hui’s score uses more
results so is likely to be closer to the meanE1
[1]2.2b E
Any reasonable explanation
e.g. Lili’s score has greater
M
standard deviation, as the
extreme scores have higher
probability
AnswerMarks Guidance
11(iii) Discrete uniform
E(X )(cid:32)3.5
1
35
Var(X )(cid:32) [(cid:32)2.917]
AnswerMarks
1 12B1
B1
B1
AnswerMarks
[3]I
C3.3
1.1
3.4
AnswerMarks Guidance
11(iv) (A)
(B)
(C)
AnswerMarks
(D)E(L)(cid:32)35 P
35
Var(L)(cid:32)102(cid:117)
12
S
875
(cid:32) (cid:32)291.7
3
E(H)(cid:32)35
35
Var(H)(cid:32)10(cid:117)
12
175
(cid:32)29.17
AnswerMarks
6E
M1
A1
B1
M1
A1
AnswerMarks
[5]1.2
1.1
1.1
2.4
AnswerMarks
1.1See (C)
Both expected values correct
AnswerMarks Guidance
11(v) 11
Estimate ofP(L(cid:33)40)(cid:32)
25
4
Estimate ofP(H(cid:33)40)(cid:32)
AnswerMarks
25B1
B1
AnswerMarks
[2]2.2b
1.1
AnswerMarks Guidance
11(vi) (A)
P(L(cid:33)40)(cid:32)
AnswerMarks Guidance
3B1
[1]1.1 N
11(vi) (B)
way off but not totally unreasonable
AnswerMarks Guidance
approximation with only 25 trials.E1
[1]3.2b E
MAny sensible relevant comment
Ft their (v)
AnswerMarks Guidance
11(vii) (A)
values of Hui’s scores (or of the scores (cid:121)10)B1
[1]I
1.2
AnswerMarks Guidance
11(vii) (B)
appear to be from Normal distribution…
… so Central Limit Theorem would seem to
AnswerMarks
apply.EE1
E1
AnswerMarks
[2]C
2.4
2.2b
AnswerMarks Guidance
11(viii) (A)
Mean ~N (cid:168) 35, (cid:184)
(cid:169) 12 (cid:185)
P(Mean > 40) = P(Normal >40.5)
AnswerMarks
So P(H(cid:33)40)(cid:124)0.154M1
B1
A1
AnswerMarks
[3]1.2
3.4
AnswerMarks
1.1Continuity correction – with
N
value of 40.5 as border (may
have 40.5 included)
BC
AnswerMarks Guidance
11(viii) (B)
Agrees well with (cid:32)0.16
AnswerMarks Guidance
25E1
[1]3.2b E
FT their (v)
AnswerMarks Guidance
QuestionAO1 AO2
1i1 0
1ii1 0
1iii5 0
2iA2 0
2iB1 2
2iiA2 0
2iiB2 0
2iii2 1
3i3 0
3iiA1 0
3iiB1 0
3iii0 2
3iv0 0
4i1 0
4ii2 0
2
AnswerMarks Guidance
4iii1 0
4iv1 0
02
4v1 2
5i1 0
5ii2 0
02 4
6i0 1
6ii0 1
00 1
7i0 1
7ii3 C
20 3
8i0 0
8iiAE
10 0
8iiB0 0
8iii2 0
8ivP
20 1
9iA0 1
S
AnswerMarks Guidance
9iB1 0
9iC0 1
9ii3 0
9iii1 1
9iv0 0
10i2 0
10ii2 0
10iii0 0
10iv0 2
10v1 0
QuestionAO1 AO2
11i3 0
11ii0 1
11iii1 0
11iv4 1
11v1 1
11viA1 0
11viB0 0
11viiA1 0
11viiB0 2
11viiiA2 0
11viiiB0 0
Totals61 22 
Question 11:
11 | (i) | 1
P(L(cid:32)60)(cid:32)
6
(cid:167)1(cid:183) 10
P(H (cid:32)60)(cid:32)(cid:168) (cid:184)
(cid:169)6(cid:185)
(cid:32)1.65(cid:117)10(cid:16)8 | B1
M1
A1
[3] | 1.1
1.1
1.1 | N
11 | (ii) | Lili’s score because Hui’s score uses more
results so is likely to be closer to the mean | E1
[1] | 2.2b | E
Any reasonable explanation
e.g. Lili’s score has greater
M
standard deviation, as the
extreme scores have higher
probability
11 | (iii) | Discrete uniform
E(X )(cid:32)3.5
1
35
Var(X )(cid:32) [(cid:32)2.917]
1 12 | B1
B1
B1
[3] | I
C3.3
1.1
3.4
11 | (iv) | (A)
(B)
(C)
(D) | E(L)(cid:32)35 P
35
Var(L)(cid:32)102(cid:117)
12
S
875
(cid:32) (cid:32)291.7
3
E(H)(cid:32)35
35
Var(H)(cid:32)10(cid:117)
12
175
(cid:32)29.17
6 | E
M1
A1
B1
M1
A1
[5] | 1.2
1.1
1.1
2.4
1.1 | See (C)
Both expected values correct
11 | (v) | 11
Estimate ofP(L(cid:33)40)(cid:32)
25
4
Estimate ofP(H(cid:33)40)(cid:32)
25 | B1
B1
[2] | 2.2b
1.1
11 | (vi) | (A) | 1
P(L(cid:33)40)(cid:32)
3 | B1
[1] | 1.1 | N
11 | (vi) | (B) | Estimate 0.44, calculated value 0.33. Some
way off but not totally unreasonable
approximation with only 25 trials. | E1
[1] | 3.2b | E
MAny sensible relevant comment
Ft their (v)
11 | (vii) | (A) | Produce a normal probability plot of the 25
values of Hui’s scores (or of the scores (cid:121)10) | B1
[1] | I
1.2
11 | (vii) | (B) | … if approximately a straight line then would
appear to be from Normal distribution…
… so Central Limit Theorem would seem to
apply. | EE1
E1
[2] | C
2.4
2.2b
11 | (viii) | (A) | (cid:167) 350(cid:183)
Mean ~N (cid:168) 35, (cid:184)
(cid:169) 12 (cid:185)
P(Mean > 40) = P(Normal >40.5)
So P(H(cid:33)40)(cid:124)0.154 | M1
B1
A1
[3] | 1.2
3.4
1.1 | Continuity correction – with
N
value of 40.5 as border (may
have 40.5 included)
BC
11 | (viii) | (B) | 4
Agrees well with (cid:32)0.16
25 | E1
[1] | 3.2b | E
FT their (v)
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Totals
1i | 1 | 0 | 0 | 0 | 1
1ii | 1 | 0 | 0 | 0 | 1
1iii | 5 | 0 | 0 | 0 | 5
2iA | 2 | 0 | 0 | 0 | 2
2iB | 1 | 2 | 0 | 0 | 3
2iiA | 2 | 0 | 0 | 0 | 2
2iiB | 2 | 0 | 0 | 0 | 2
2iii | 2 | 1 | 0 | 0 | 3
3i | 3 | 0 | 0 | 0 | 3
3iiA | 1 | 0 | 0 | 1 | 2
3iiB | 1 | 0 | 0 | 1 | 2
3iii | 0 | 2 | 0 | 0 | 2
3iv | 0 | 0 | 0 | 2 | 2
4i | 1 | 0 | 0 | 0 | 1
4ii | 2 | 0 | 0 | 0 | N
2
4iii | 1 | 0 | 1 | 0 | 2
4iv | 1 | 0 | 1 | E
0 | 2
4v | 1 | 2 | 0 | 0 | 3
5i | 1 | 0 | 0 | 2 | 3
5ii | 2 | 0 | M
0 | 2 | 4
6i | 0 | 1 | 0 | 1 | 2
6ii | 0 | 1 | I
0 | 0 | 1
7i | 0 | 1 | 1 | 1 | 3
7ii | 3 | C
2 | 0 | 3 | 8
8i | 0 | 0 | 0 | 2 | 2
8iiA | E
1 | 0 | 0 | 0 | 1
8iiB | 0 | 0 | 0 | 2 | 2
8iii | 2 | 0 | 0 | 1 | 3
8iv | P
2 | 0 | 1 | 1 | 4
9iA | 0 | 1 | 0 | 0 | 1
S
9iB | 1 | 0 | 0 | 0 | 1
9iC | 0 | 1 | 0 | 0 | 1
9ii | 3 | 0 | 0 | 1 | 4
9iii | 1 | 1 | 0 | 2 | 4
9iv | 0 | 0 | 0 | 3 | 3
10i | 2 | 0 | 0 | 0 | 2
10ii | 2 | 0 | 0 | 2 | 4
10iii | 0 | 0 | 0 | 1 | 1
10iv | 0 | 2 | 0 | 0 | 2
10v | 1 | 0 | 0 | 0 | 1
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Totals
11i | 3 | 0 | 0 | 0 | 3
11ii | 0 | 1 | 0 | 0 | 1
11iii | 1 | 0 | 0 | 2 | 3
11iv | 4 | 1 | 0 | 0 | 5
11v | 1 | 1 | 0 | 0 | 2
11viA | 1 | 0 | 0 | 0 | 1
11viB | 0 | 0 | 1 | 0 | 1
11viiA | 1 | 0 | 0 | 0 | 1
11viiB | 0 | 2 | 0 | 0 | 2
11viiiA | 2 | 0 | 1 | 0 | 3
11viiiB | 0 | 0 | 0 | 1 | 1
Totals | 61 | 22 | 6 | 31 | 120
11 Two girls, Lili and Hui, play a game with a fair six-sided dice. The dice is thrown 10 times.\\
$X _ { 1 } , X _ { 2 } , \ldots , X _ { 10 }$ represent the scores on the $1 ^ { \text {st } } , 2 ^ { \text {nd } } , \ldots , 10 ^ { \text {th } }$ throws of the dice.\\
$L$ denotes Lili's score and $L = 10 X _ { 1 }$.\\
$H$ denotes Hui's score and $H = X _ { 1 } + X _ { 2 } + X _ { 3 } + \ldots + X _ { 10 }$.
\begin{enumerate}[label=(\roman*)]
\item Calculate

\begin{itemize}
  \item $\mathrm { P } ( L = 60 )$ and
  \item $\mathrm { P } ( H = 60 )$.
\item Without doing any further calculations, explain which girl's score has the greater standard deviation.
\item Write down
  \item the name of the probability distribution of $X _ { 1 }$,
  \item the value of $\mathrm { E } \left( X _ { 1 } \right)$,
  \item the value of $\operatorname { Var } \left( X _ { 1 } \right)$.
\item Find\\
(A) $\mathrm { E } ( L )$,\\
(B) $\operatorname { Var } ( L )$,\\
(C) $\mathrm { E } ( H )$,\\
(D) $\operatorname { Var } ( H )$.
\end{itemize}

The spreadsheet below shows a simulation of 25 plays of the game. The cell E3, highlighted, shows the score when the dice is thrown the fourth time in the first game.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
 & A & B & C & D & E & F & G & H & I & J & K & L & M & N \\
\hline
1 & \multicolumn{12}{|c|}{Throw of dice} & Lili's & Hui's \\
\hline
2 &  & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 &  & score & score \\
\hline
3 & Game 1 & 3 & 5 & 2 & 1 & 1 & 3 & 1 & 1 & 1 & 4 &  & 30 & 22 \\
\hline
4 & Game 2 & 6 & 3 & 2 & 4 & 4 & 3 & 5 & 3 & 3 & 5 &  & 60 & 38 \\
\hline
5 & Game 3 & 6 & 4 & 2 & 6 & 5 & 2 & 1 & 5 & 2 & 3 &  & 60 & 36 \\
\hline
6 & Game 4 & 1 & 5 & 1 & 6 & 6 & 3 & 1 & 4 & 6 & 2 &  & 10 & 35 \\
\hline
7 & Game 5 & 4 & 4 & 3 & 1 & 6 & 4 & 4 & 1 & 6 & 2 &  & 40 & 35 \\
\hline
8 & Game 6 & 2 & 1 & 5 & 1 & 2 & 5 & 1 & 5 & 2 & 3 &  & 20 & 27 \\
\hline
9 & Game 7 & 1 & 1 & 3 & 4 & 4 & 5 & 6 & 3 & 4 & 2 &  & 10 & 33 \\
\hline
10 & Game 8 & 1 & 1 & 3 & 6 & 3 & 4 & 4 & 5 & 2 & 3 &  & 10 & 32 \\
\hline
11 & Game 9 & 2 & 2 & 2 & 4 & 3 & 2 & 1 & 5 & 5 & 6 &  & 20 & 32 \\
\hline
12 & Game 10 & 3 & 5 & 3 & 3 & 5 & 3 & 4 & 3 & 1 & 1 &  & 30 & 31 \\
\hline
13 & Game 11 & 5 & 3 & 6 & 5 & 5 & 4 & 2 & 1 & 1 & 5 &  & 50 & 37 \\
\hline
14 & Game 12 & 6 & 4 & 3 & 2 & 4 & 1 & 3 & 3 & 5 & 3 &  & 60 & 34 \\
\hline
15 & Game 13 & 2 & 3 & 2 & 1 & 2 & 2 & 2 & 2 & 2 & 1 &  & 20 & 19 \\
\hline
16 & Game 14 & 4 & 1 & 3 & 3 & 1 & 2 & 6 & 6 & 1 & 3 &  & 40 & 30 \\
\hline
17 & Game 15 & 5 & 1 & 2 & 6 & 3 & 4 & 6 & 3 & 6 & 4 &  & 50 & 40 \\
\hline
18 & Game 16 & 3 & 6 & 1 & 1 & 5 & 3 & 1 & 3 & 3 & 3 &  & 30 & 29 \\
\hline
19 & Game 17 & 5 & 2 & 5 & 2 & 4 & 5 & 2 & 2 & 3 & 4 &  & 50 & 34 \\
\hline
20 & Game 18 & 3 & 6 & 3 & 5 & 5 & 2 & 3 & 1 & 1 & 2 &  & 30 & 31 \\
\hline
21 & Game 19 & 6 & 6 & 3 & 1 & 5 & 6 & 3 & 4 & 1 & 6 &  & 60 & 41 \\
\hline
22 & Game 20 & 2 & 6 & 4 & 5 & 6 & 5 & 2 & 4 & 3 & 3 &  & 20 & 40 \\
\hline
23 & Game 21 & 5 & 3 & 5 & 4 & 5 & 3 & 3 & 6 & 6 & 1 &  & 50 & 41 \\
\hline
24 & Game 22 & 6 & 3 & 5 & 5 & 6 & 3 & 5 & 6 & 1 & 1 &  & 60 & 41 \\
\hline
25 & Game 23 & 5 & 4 & 5 & 5 & 6 & 4 & 2 & 1 & 3 & 6 &  & 50 & 41 \\
\hline
26 & Game 24 & 3 & 5 & 2 & 3 & 2 & 4 & 3 & 2 & 3 & 3 &  & 30 & 30 \\
\hline
27 & Game 25 & 5 & 2 & 4 & 2 & 4 & 5 & 2 & 2 & 5 & 2 &  & 50 & 33 \\
\hline
28 &  &  &  &  &  &  &  &  &  &  &  &  &  &  \\
\hline
29 &  &  &  &  &  &  &  &  &  &  &  & mean & 37.60 & 33.68 \\
\hline
30 &  &  &  &  &  &  &  &  &  &  &  & sd & 17.39 & 5.77 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{table}
\item Use the simulation to estimate $\mathrm { P } ( L > 40 )$ and $\mathrm { P } ( H > 40 )$.
\item (A) Calculate the exact value of $\mathrm { P } ( L > 40 )$.\\
(B) Comment on how the exact value compares with your estimate of $\mathrm { P } ( L > 40 )$ in part (v).

Hui wonders whether it is appropriate to use the Central Limit Theorem to approximate the distribution of $X _ { 1 } + X _ { 2 } + X _ { 3 } + \ldots + X _ { 10 }$.
\item (A) State what type of diagram Hui could draw, based on the output from the spreadsheet, to investigate this.\\
(B) Explain how she should interpret the diagram.
\item (A) Calculate an approximate value of $\mathrm { P } \left( X _ { 1 } + X _ { 2 } + X _ { 3 } + \ldots + X _ { 10 } > 40 \right)$ using the Central Limit Theorem.\\
(B) Comment on how this value compares with your estimate of $\mathrm { P } ( H > 40 )$ in part (v).

\section*{Copyright Information:}
}{www.ocr.org.uk}) after the live examination series.

If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.\\
For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.\\
OCR is part of the 
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major  Q11 [24]}}
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