Cambridge International AS & A Level

MATHEMATICS 9709/53
Paper 5 Probability & Statistics 1 October/November 2022
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 October/November 2022 series for most
Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level
components.

This document consists of 16 printed pages.

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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 ∑ x − 50  20 = 35 ; ∑ x = 1035 B1 Correct value for ∑ x or x .

35 1035
or x = + 20 = [= 20.7]
50 50

25036   x 
2
25036  1035 
2 M1
25036   x 
2
−  = −  −  their 
 50   50    
50 50 50   50  

72.23 A1 Exact answer only

SC B1 for 72.23 with no substitution in formula.

Question Answer Marks Guidance

2 Mean = 80  0.32 = 25.6, B1 25.6 and 17.4[08] seen, allow unsimplified.

var = 80  0.32  0.68 = 17.408 4.172… implies correct variance.

19.5 − 25.6 M1 Substituting their 25.6 and 17.408 into ±standardisation

P(X < 20) = P( Z  ) = P( Z  −1.462) formula (any number for 19∙5), not σ2, √ σ.
17.408
M1 Using continuity correction 19∙5 or 20∙5 in their
standardisation formula.

= [1 − Φ (1.462)] = 1 – 0.9282 M1 Appropriate area Φ, from final process, must be probability.

(Expect final ans < 0∙5 ).
Note: the correct final answer may imply M1 from use of
calculator.

0.0718 A1 0.0718 ⩽ p ⩽0.0719

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Question Answer Marks Guidance

3(a) Cumulative frequency graph M1 At least 3 points plotted accurately at class upper end points:
(20,32), (30, 66), (35, 112), (40, 178), (50, 228),
(60, 250).
Linear cf scale 0 ⩽ cf ⩽ 250 and linear time scale 0 ⩽ time
⩽ 60 with at least 3 values identified on each.

A1 All points plotted correct, curve drawn (within tolerance) and

joined to (0,0).
Axes labelled cumulative frequency (cf), time (t) and
minutes (min or m) – or a suitable title.
Axes can be the other way round.

3(b) Line drawn from 150 on cf axis to meet graph at about B1 FT Must be an increasing cf graph with correct upper bounds.
t =38 minutes Use of graph must be seen.
Expect an answer in range 37 ⩽ t ⩽ 39 for a correct graph

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Question Answer Marks Guidance

3(c) [Frequencies] [32] 34 46 66 50 22 B1 May be unsimplified and/or in variance calculation.

[Midpoints] 10 25 32.5 37.5 45 55 M1 At least 5 correct midpoints seen , may be unsimplified.

[Variance] = M1 Correct unsimplified Variance formula with their midpoints

32  102 + 34  252 + 46  32.52 + 66  37.52 + 50  452 + 22  552 and their frequencies for var or sd.
− 34.42
250
( − mean2 included)
333650
[= − 34.42 = 151.24]
250

[Sd =] 12.3 A1 Awrt WWW

SC B1 for 12.3 if second M1 not awarded.

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Question Answer Marks Guidance

4(a) Method 1: Scenarios identified

[no of ways for score of 2 are] 222, 211, 212, 221, 122, 112, 121 B1 7 correct scenarios identified, no incorrect.
[Total options = 64]

7 7 M1 a
[So P(X = 2) =] = , a = their number of correct identified
4 4 4 64 4 4 4
scenarios > 4

A1 Approach identified, WWW.

Method 2: P(2 on all spinners) + P(2 on two spinners and 1 on one spinner) + P(2 on one spinner and 1 on two spinners)
3
1 3 1 1 1 3 1 1 1 B1 3
1 3 1 1 1
  + C2     + C1    
 4 4 4 4 4 4 4  4
( )
  + C2 or C1     + d , 0 < d< 1
3

 4 4 4

M1 1
3
1
3
1
3

  + e   + f   1 < e <5 and 1 < f < 5

 
4  
4  4

7 A1 Approach identified, WWW.

[So P(X = 2) =] =
64

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Question Answer Marks Guidance

Method 3: P(1 or 2 on each spinner) – P(1 on all spinners)

3
1 1
3 B1 1
3

  −    − b seen, 0 < b < 1

2 4 2

M1 1
3

  −c , 0 < c < ½
3

 2

7 A1 Approach identified, WWW.

[So P(X = 2) =]
64

4(b) 1 B1 P(X = 1) or P(X = 4) correct. Condone answers not in

P( X = 1) = probability distribution table if clearly identified.
64

 1 7 19  37 B1 FT All 4 probabilities summing to 1.

P(X = 4) = 1 − − − =
 64 64 64  64

4(c)  3 5 1  243 B1 Accept 0.059326… to 4 or more SF.

P(Y = 6) =    =  0.0593,
 4  4  4096

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Question Answer Marks Guidance

4(d)  3
4 M1 3
g
4
    , g = 4, 5 or p where 0 < p < 1
 4 4

81 A1 Accept 0.316406…to 4 or more SF.

= ,0.316
256

Alternative method for Question 4(d)

 1 3 1  3  2 1  3 3 1  M1 Correct or
P(Y >4) = 1 – P( Y⩽4) = 1 −  +  +    +      1 3 1  3  2 1  3 3 1  3  4 
4 4 4 4 4 4 4
  1−  +  +    +    +   
4 4 4 4 4 4 4 4 
 175   
 = 1 − 256  ( )
or 1 − p + qp + q 2 p + q3 p where 0 < p < 1 and q = 1–p
 

81 A1 Accept 0.316406…to 4 or more SF.

= ,0.316
256

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Question Answer Marks Guidance

5(a) [P(0, 1, 2) =] 10C0 0.10 0.910 + 10C1 0.11 0.99 + 10C2 0.12 0.98 M1 One term 10Cx p x (1 − p )10− x , 0  p  1, x  0

= 0.348678+0.38742+0.19371 A1 Correct expression, accept unsimplified.

0.930 B1 0.9298 ⩽ p ⩽ 0.9303

Alternative method for Question 5(a)

[1 – P(3, 4, 5, 6, 7, 8, 9, 10) = 1 – (10C3 0.97 0.13 +10C4 0.96 0.14 +10C5 M1 One term 10Cx p x (1 − p )10− x ,
0.95 0.15 +10C6 0.94 0.16 +10C7 0.93 0.17 +10C8 0.92 0.18 +10C9 0.91 0.19 0  p  1, x  0
+10C10 0.90 0.110 )

A1 Correct expression, accept unsimplified.

0.930 B1 0.9298 ⩽ p ⩽ 0.9303

5(b) 1.11 − 1.04 M1 1.11, 1.04 and 0.06 substituted into ±Standardisation
[P(X > 1.11) = ]P( Z  ) = P( Z  1.167) formula, no continuity correction not 0.062 or √0.06
0.06

= 1 – 0.8784 M1 1 – their 0.8784 as final answer, must be probability. (Expect

final ans < 0∙5 ).

0.122 A1 0.1216 ⩽ p ⩽ 0.122

SC M0 M1 B1 for 0.122 with no standardisation formula.

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Question Answer Marks Guidance

5(c) w − 1.04 B1 0.8775 < z ⩽ 0.878 or −0.878 ⩽ z < −0.8775 seen.

[P(X < w) = P( Z  ) = 0.81]
0.06
w − 1.04 M1 1.04 and 0.06 substituted in ±standardisation formula, no
= 0.878 continuity correction, not σ2, √ σ, equated to a z-value.
0.06

w = 1.09 A1 1.09 ⩽ w ⩽ 1.093

Question Answer Marks Guidance

6(a) 9! M1 h!
, h = 7, 8, 9; j = 1, 2
2!2! 2! j !

90720 A1

6(b) Arrangements with 5 letters between As + Arrangements with 6 letters between As + Arrangements with 7 letters between As

7! M1 7!
With gap of 5:  3 [= 7560]  k , k positive integer 1< k < 7
2! 2!
7!
With gap of 6:  2 [= 5040] M1 Add their no of ways for 3 identified correct scenarios, no
2!
additional incorrect scenarios, accept unsimplified.
7!
With gap of 7: 1 [= 2520]
2!

7! A1
[Total no =  6 =] 15120
2!

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Question Answer Marks Guidance

6(c) Method 1: Summing number of ways

AT _ _ _ 2×2× 5C3 40 B1 Correct no of ways for 4 correctly identified scenarios,

A____ 2×5C4 10 accept unsimplified.
5
AATT _ C1 5
AAT _ _ 2×5C2 20 M1 Add no of ways for 5 or 6 identified correct scenarios, no
5
AA _ _ _ C3 10 additional incorrect scenarios, no repeated scenarios, accept
5
_____ C5 1 unsimplified.

[Total no of ways not containing more Ts than As = ] A1 All correct and added
= 40+10+5+20+10+1 [=86]

86 M1 their 86
Probability = 9
accept numerator unevaluated
C5 9C 5 or their identified total

86 43 A1
, , 0.683
126 63

Method 2: Subtracting no of ways with more Ts from total

T____ 2×5C4 10 B1 Correct no of ways for 2 correctly identified scenarios, no

TTA _ _ 2×5C2 20 additional incorrect scenarios, no repeated scenarios, accept
5
TT _ _ _ C3 10 unsimplified, condone use of permutations

M1 Add no of ways for 2 or 3 correct scenarios and subtract from

their total no of ways
All correct and subtracted

Total no of ways with more Ts than As =40 A1

9
C5 − 40 = 86

86 M1 their 86
Probability = 9
accept numerator unevaluated
C5 9C 5 or their identified total

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Question Answer Marks Guidance

6(c) 43 A1
, 0.683
63

Question Answer Marks Guidance

7(a) 3 2 5 3 M1 3 2 5 3
[P(SR TR) + P(SW TR) =]  +   + k or l +  0 < k,l < 1
8 7 8 7 8 7 8 7

21 3 A1 3
= , , 0.375 SC B1 for with no explanation.
56 8 8

7(b) [RRWR, WRRR, WRWR] M1 m n o q

   1 ⩽ m,n,o,q ⩽ 5, m ≠ n ≠ o ≠ q
3 2 5 1 5 3 2 1 5 3 4 2 8 7 6 5
   +    +   
8 7 6 5 8 7 6 5 8 7 6 5
1 1 1 A1 Probability for one scenario correct, accept unsimplified.
[= + + ]
56 56 14
M1 Adding probabilities for 3 correct scenarios and no incorrect.

180 3 A1 Or 0.1071428… to 4SF or better.

= , , 0.107 SC B1 for 3/28 with inadequate explanation.
1680 28

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Question Answer Marks Guidance

7(c) 30 1 M1 3 2 5 1
their P ( RRWR ) or   
[P(S first disc R |T2 ) =] 1680 = 56 8 7 6 5
3 3 3
their 7 ( b ) − must be a prob or
28 28 28

1 A1
, 0.167
6

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