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Mark Scheme

Mock Set 5

Pearson Edexcel GCE in Mathematics (9MA0)

Paper 31 Statistics

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January 2024

Publications Code 9MA0_ 31 _MS 5 _MS

All the material in this publication is copyright

© Pearson Education Ltd 202 4

EDEXCEL GCE MATHEMATICS

General Instructions for Marking

1. The total number of marks for the paper is 5 0.
2. The Edexcel Mathematics mark schemes use the following types of marks:
   • M marks: method marks are awarded for ‘knowing a method and attempting to apply

it’, unless otherwise indicated.

• A marks: Accuracy marks can only be awarded if the relevant method (M) marks have

been earned.

• B marks are unconditional accuracy marks (independent of M marks)
• Marks should not be subdivided.

1. Abbreviations

These are some of the traditional marking abbreviations that will appear in the mark

schemes.

• bod – benefit of doubt
• ft – follow through
• the symbol will be used for correct ft
• cao – correct answer only
• cso - correct solution only. There must be no errors in this part of the question to

obtain this mark

• isw – ignore subsequent working
• awrt – answers which round to
• SC: special case
• oe – or equivalent (and appropriate)
• dep – dependent
• indep – independent
• dp decimal places
• sf significant figures
•  The answer is printed on the paper
• The second mark is dependent on gaining the first mark

1. For misreading which does not alter the character of a question or materially simplify it,

deduct two from any A or B marks gained, in that part of the question affected.

1. Where a candidate has made multiple responses and indicates which response they wish

to submit, examiners should mark this response.

If there are several attempts at a question which have not been crossed out, examiners

should mark the final answer which is the answer that is the most complete.

1. Ignore wrong working or incorrect statements following a correct answer.
2. Mark schemes will firstly show the solution judged to be the most common response

expected from candidates. Where appropriate, alternatives answers are provided in the

notes. If examiners are not sure if an answer is acceptable, they will check the mark

scheme to see if an alternative answer is given for the method used.

Question Scheme Marks AOs

1(a)(i) P ( X = 3 ) =0.

awrt 0.

B1 1.1b

(a)(ii)

P ( X 6 )

M1 3.

P ( X 6 ) =0.

awrt 0.

A1 1.1b

(b) H : 0 p = 0.36 H : 1 p 0.

B1 2.

X B 25, 0.36( ) seen or used M1 3.

P ( X 4 )=0.025487...

awrt 0.

A1 1.1b

[0.0255 < 0.05, reject H 0 o.e.]

Sufficient evidence to suggest the coach’s new system has reduced the A1 2.2b

proportion of matches in which the team commits serious disciplinary

offences

(4)

(c) One assumption from:

• Serious offences are committed in matches independently

➢ E.g. Unlikely to be valid as a player may be banned or withdrawn

by the coach for a serious offence in a previous game.

• Probability of a serious offence being committed in a match is the same for all matches

➢ E.g. Unlikely to be valid as may depend on tougher

opposition/different referees/different player selection

• There are only two states : the team committing a serious offence

in a match or not ➢ E.g. Likely to be valid as seriousness of offences can be

categorised by an objective measure such as any incidents

resulting in certain levels of sanctions from officials.

Accept other plausible reason in context to support assumptions

B

dB

(9 marks)

Notes:

(a)(i)

(ii)

B1:

M1:

A1:

awrt 0.

Writing or indicating that the required outcomes needed X 6

awrt 0.

(b) B1:

M1:

A1:

A1:

For both hypotheses correct and in p or

Writing the correct distribution, or evidence of using it

awrt 0.

Correct statement including new system and serious disciplinary offences

Do not allow contradictory statements

(c) B1:

dB1:

A correct contextual statement linking serious offences and either: independence,

constant probability, or two outcomes

A sensible corresponding contextual comment on the validity of their statement

Question Scheme Marks AOs

3 (a) Numbering the data 0-183 or 1-^184 B1^ 2.

Generate 20 (distinct) random numbers and select the data

corresponding to these numbers

B1 1.1b

(b)(i)

(ii)

Negative correlation B1 1. 2

The data loosely lies along a line with a negative gradient B1 2.

(c) H : 0 =^0 H : 1 ^0 B1^ 2.

CV = 0.3783 M1 1.1b

− 0.554  −0.378 3( ) or 0.554 0.378 3( )so in critical region

Evidence of a negative correlation

A1 2.2b

(d) No, as correlation coefficient is invariant to (linear) coding B1 2.

(8 marks)

Notes:

(a) B1:

B1:

Describes numbering the data in an appropriate manner that shows an

understanding of the LDS covering only six months (not including February)

Accept:

• a starting point of 0 and an end point of 179-185 inclusive

or

• a starting point of 1 and an end point of 180 - 186 inclusive

Describes finding 20 random numbers and selecting the corresponding data.

No reference to distinct or ignoring repetition is required for this mark.

(b)(i)

(ii)

B1:

B1:

For stating negative. ‘Negative skew’ is B0 though

For a suitable reason referencing linearity with a negative gradient.

Condone generally higher temperature implies lower pressure oe

(c) B1:

M1:

A1:

Both hypotheses in terms of , do not accept r

For the CV, sight of 0. 3783 or any CV such that0.25  r 0.

A comparison of 0.3783 and 0.554, must compare values of the same sign

Comment referencing ‘negative correlation’

Do not allow contradictory statements

(d) (^) B1: No with a reason referencing that the pmcc does not change with coding

Question Scheme Marks AOs

4(a) (^) Two of:

P ( 2 ) , P ( 3 ) , P ( 4 )

p p p (^) N = = N = = N = = M1 2.

Fully correct distribution in terms of 𝑝

n 1 2 3 4

P ( N = n ) p

p

9

p

27

p

A1 1.1b

(b) 1 3 9 27

p p p (^) p + + + = Or

p (^) = M1 1.1b

So,

p = * (^) A1cso* 1.1b

(c)

P P 3, 4

N N

 ^ =^ =

Or

3 n − (^21 4 7 )

P ( N = n )

And selecting P 3( N − 2 =7, 10)

M1 3.1a

0.1 o.e.

A1 1.1b

(6 marks)

Notes:

(a) M1:

A1:

Two correct probabilities associated with the correct outcome

(not including P ( N = 1 )= p )

Fully correct with correct notation and probabilities in terms of p

Allow fully defined probability function

1 1 1, 2, 3, 4 P (^3)

0 otherwise

n

p n N n

−  (^)    (^)   = = =   

 

(b) M1: A1cso:*

For clear use of sum of probabilities = 1 (all terms)

Fully correct with no incorrect working seen.

Must have clear evidence of the M1 scored before final answer.

(c) M1:

A1:

Simplifying the inequality to reach

P

N

, and selecting N = 3, 4

Or Forming a correct distribution for 3 n − 2 and selecting 3 N − 2 =7, 10

0.1 or equivalent

Question Scheme Marks AOs

6(a)(i)^ P^ (^ X^ ^15 )=0.1150697...

awrt 0.115 B^

1.1b

(a)(ii) ( ) ( )

(^2 ) 1 − P X  15  = 1 − 1 −'0.115'

or

2 2  P X  15  P X  15 + P X  15 

2 = 2  '0.115'  1 − '0.115' +'0.115'

M1 3.1b

awrt 0.217 A1ft 1.1b

(2)

(b)^ P^ (^ X^ ^ m^ |^ X ^15 )=0.

P

P 15

X m

X





M1 3.

P ( X  m ) = 0.25  P ( X  15 ) = 0.25  0.115... =0.028767...

awrt 0.028 8

A

1.1b

(c) H : 0 =^12 H : 1 ^12 B1^ 2.

2

N 12, N 12, 0. 25

X X

seen or used (^) M1 3.

P (^) ( X c 1 (^) )0.005 orP (^) ( X c 2 (^) )0.

12 2.57...

5

X − =  (^) M1 3.

c 1 (^) = 10.7, c 2 =13.

awrt 10.7 and 13.

A1 1.1b

( X^^ 10.7^ ) ( X 13.3)^ o.e.^ A1^ 1.1b

(10 marks)

Notes:

(a)(i) B1^ Finding^ P^ (^ X^ ^15 )=awrt 0.

(a)(ii) M1:

A1ft:

Correct method using their P ( X  15 )

awrt 0. 217

(b) M1: A1:

Correct use of conditional probability formula given as a fraction equal to 0.

awrt 0.028 8

(c) B1:^ For both hypotheses correct and in terms of

M1:

M1:

A1:

A1:

Correct distribution written or used. Condone X^ N 12, 0.5( )if used correctly

subsequently

Correct method or statement for finding two-tailed 1% CR

Both CVs given correctly to 3.s.f.

CR written correctly in terms of X (allow < and >)

Allow any letter in place of X providing it denotes a mean