Cambridge International A Level

MATHEMATICS 9709/61
Paper 6 Probability & Statistics 2 May/June 2021
MARK SCHEME
Maximum Mark: 50

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes for the May/June 2021 series for most Cambridge
IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

This document consists of 13 printed pages.

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9709/61 Cambridge International A Level – Mark Scheme May/June 2021
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Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the
mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:
• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:
• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the
syllabus and mark scheme, referring to your Team Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as
indicated by the mark scheme. The meaning, however, should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited
according to the quality of the candidate responses seen).

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in
mind.

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Mathematics Specific Marking Principles

1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required
then no marks will be awarded for a scale drawing.

2 Unless specified in the question, answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of
accuracy is not affected.

3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points.

4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).

5 Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.

6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.

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Mark Scheme Notes

The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons
outside the scope of these notes.

Types of mark

M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units.
However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea
must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula
without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method
mark is earned (or implied).

B Mark for a correct result or statement independent of method marks.

DM or DB When a part of a question has two or more ‘method’ steps, the M marks are generally independent unless the scheme specifically says otherwise;
and similarly, when there are several B marks allocated. The notation DM or DB is used to indicate that a particular M or B mark is dependent on
an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full
credit is given.

FT Implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are
given for correct work only.

• A or B marks are given for correct work only (not for results obtained from incorrect working) unless follow through is allowed (see abbreviation FT above).
• For a numerical answer, allow the A or B mark if the answer is correct to 3 significant figures or would be correct to 3 significant figures if rounded (1
decimal place for angles in degrees).
• The total number of marks available for each question is shown at the bottom of the Marks column.
• Wrong or missing units in an answer should not result in loss of marks unless the guidance indicates otherwise.
• Square brackets [ ] around text or numbers show extra information not needed for the mark to be awarded.

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Abbreviations

AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent

AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)

CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)

CWO Correct Working Only

ISW Ignore Subsequent Working

SOI Seen Or Implied

SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)

WWW Without Wrong Working

AWRT Answer Which Rounds To

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Question Answer Marks Guidance

1 λ = (3.1 + 1.7) × 2 M1 Attempt combined mean. Allow 3.1 + 1.7 for M1

= 9.6 A1 Correct mean

1 – e–9.6 (1 + 9.6 + 9.62 3

+ 9.63! ) M1 Allow incorrect mean. Allow one end error.
2

= 0.986 (3 sf) A1 SC If 9.6 seen and unsupported 0.986 M1A1B1.

SC Unsupported correct answer of 0.986 only if 9.6
also not seen scores B2 only.

Question Answer Marks Guidance

2(a) ± 123 − 125 [= –2.108...] M1 Must have √40

6 No standard deviation/variance mix. Ignore any
40 continuity correction attempts for this mark.

P(z < ‘–2.108’) = 1 – Φ(‘2.108’) M1 For correct probability area consistent with their
working.

= 0.0175 or 0.0176 (3 sf) A1

2(b) No, population is normal B1 Need both.

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Question Answer Marks Guidance

3 20 20 1 M1 For either expression seen.

1– or –
27 27 2
20  20   20 1 
– 1 −  or  − 
27  27   27 2 

13 A1 OE. Accept 0.481 or 0.482.

27

Question Answer Marks Guidance

4 3820 B1
[= 38.2]
100

100  182200 1  38202  M1 Use of biased (362.76) scores M0

2
 − '38.2 '  or  182200 − 
99  100  99  100 

12092 A1 Accept SD=19.1422 or 19.1(3sf)

= or 366.424 or 366 (3 sf)
33

'366.424' M1 Expression of the correct form must be a z-value.

‘38.2’ ± z×
100

z = 1.881 or 1.882 B1 Seen.

34.6 to 41.8 (3 sf) A1 Allow use of biased giving (34.6,41.8)

Must be an interval.

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Question Answer Marks Guidance

5(a)  2 M1 SOI. Allow Po(0.133).

Po  
 15 

−
2 M1 Allow incorrect λ allow one end error
P(X ⩾ 1) = 1 – e 15

= 0.125 (3 sf) A1 SC Partially unsupported final answer:

 2
Po   stated B1 then unsupported 0.125 B1
 15 
SC Use of Binomial (0.1248) B1 only
Use of Normal scores M0

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Question Answer Marks Guidance

5(b) n B1
λ=
75000

−
n M1 Allow ‘=’
e 75000 > 0.9 Allow incorrect λ

n M1 Attempt ln both sides

– > ln 0.9 [n < 7902.04]
75000

Largest value of n is 7902 A1 CWO. Must be an integer.

Alternative method for Question 5(b)

e- μ > 0.9 M1 Allow ‘=’

–μ > ln 0.9 [μ < 0.10536] M1 Attempt ln both sides

n = μ × 75000 B1

Largest value of n is 7902 A1 CWO. Must be an integer.

Alternative method for Question 5(b)

74999 B1
75000

 74999 
n M1
  > 0.9
 75000 

74999 M1 Attempt ln or log both sides

nln > ln 0.9
75000

Largest value of n is 7901 A1 CWO Must be an integer

4
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Question Answer Marks Guidance

6 E(X) = 3 B1 N.B. E(X)=108k is B0 until correct k substituted in.

6 M1 Attempt integration of f(x) and =1.

k (6 x − x 2 )dx = 1
0
Ignore limits at this stage.

 x3  6
k 3x 2 −  [= 1]
 3 0

 216  A1
k 108 −  =1
 3 
3 1
k= or
108 36

3
6 *M1 Attempt integration of their k × x2f(x).
'
108 0
' (6 x3 − x 4 )dx Ignore limits at this stage. Accept in terms of k.

3  3 x 4 x5  6
=  −  = 10.8
108  2 5 0

'10.8' – '3'2 DM1 Their 10.8 (from use of limits 0 and 6) minus their
(E(X))2.
Accept in terms of k: 388.8k–(108k)2

9 A1 CWO. Must be convincingly obtained as AG.

or 1.8
5

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Question Answer Marks Guidance

7(a) E(T)= 3 × 55 + 6 × 27 [ = 327 ] B1 OE. Accept unsimplified.

Var(T) = 3×32 + 6×2.52 [= 64.5] B1 Accept unsimplified.

340 − '327' M1 Must have √

[= 1.619]
'64.5'

P(z < '1.619') = Φ('1.619') M1 Correct probability area consistent with their
working.

0.947 (3 sf) A1

7(b) E(L–S1–S2) = 55 – 2 × 27 [=1] B1 OE e.g. E(S1+S2 – L)= –1. Accept unsimplified.

Var(L– S1–S2) = 32 + 2 × 2.52 [= 21.5] B1 Accept unsimplified.

0 − '1' M1 Standardising with their values. Must come from a

[= –0.216] combination attempt.
'21.5'

P(L–S1–S2 > 0) = Φ(‘0.216’) M1 Correct probability area consistent with their

working.

0.586 or 0.585 (3 sf) A1

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Question Answer Marks Guidance

8(a) Not representative (of all students in the school) B1 OE idea of ‘not being representative’ e.g. different
grades in the school have different
characteristics/proportions....
Don’t accept ‘not random’ or ‘biased’ without
further explanation.

8(b) H0: P(not correct uniform) = 0.15 B1 Allow "p"

H1: P(not correct uniform) < 0.15

8(c) Any two probs attempted using B(50,0.15) M1

P(X ⩽ 3) = 0.8550 + 50 × 0.8549 × 0.15 + 50C2 × 0.8548 × 0.152 + 50C3 × 0.8547 × M1 Attempt the tail probability P(0,1,2,3) with
0.153 B(50,0.15) must be added.

P(X ⩽ 4) = 0.04605 + 50C4×0.8546×0.154 M1 OE. Their P(X ⩽ 3) + P(X = 4) or P(0,1,2,3,4) with

B(50,0.15) must be added.

P(X ⩽ 3) = 0.0460 or 0.0461 [<0.05] A1 Both correct.

P(X ⩽ 4) = 0.112 or [>0.05] OR if P(X ⩽ 4) not seen; P(4)=0.06606 and
0.06606>0.05 and P(X ⩽ 3)=0.0460 scores M1 A1

P(Type I) = 0.0460 or 0.0461 (3 sf) A1 Dependent on second M1.

SC If M1M1M1A0 scored allow A1FT for
incorrect P(X ⩽ 3) as long as <0.05

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Question Answer Marks Guidance

8(d) 4 is outside critical region (⩽3) OE or P(X ⩽ 4) = 0.112 which is > 0.05 M1 FT working from (c).

No evidence that proportion not wearing the correct uniform has decreased A1 In context not definite, e.g. not ‘Proportion has not
(Accept Ho) decreased’. No contradiction.

8(e) Not rejected H0 *B1 FT FT If Reject H0 in (d)

Type II DB1 FT FT Type I

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