Cambridge International AS & A Level

MATHEMATICS 9709/52
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 18 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(a) 0.2  x + 0.1 2x + 0.7  0.25 = 0.235 M1 0.2  x + 0.1 2x + 0.7  0.25 or 0.2x + 0.2x + 0.175 seen.

M1 Equating their 3 term expression (2 terms involving x) to 0.235

x = 0.15 A1

1(b)  P ( car and not late )  M1 0.1  (1 − 2  their x ) or 0.1  0.7 as numerator
 P(car |not late ) = 
 P ( not late )  and
0.2  (1 – their x) + 0.1  (1 – 2 × their x ) + 0.7  0.75 with values
0.1  (1 − 0.3)
substituted or 1 – 0.235 or 0.765 as denominator of fraction.
1 − 0.235 Condone 0.2  (1 – their x) + 0.1  (1 –  their x) + 0.7  0.75 as
denominator consistent with 1(a).

 0.07  70 14 A1 0.091503267 to at least 3SF.

 0.765 =  0.0915, 765 , 153 If M0 scored SC B1 for 0.091503267 to at least 3SF.
 

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Question Answer Marks Guidance

2(a) 54.8 − 55.6 M1 Use of ± standardisation formula, with 54.8, 55.6 and 1.2
[P(X<54.8)] = P( Z  )
1.2 substituted. condone 1.22 , 1.2 or continuity correction of 54.75 or
54.85

[= P( Z  −0.6667)] = 1 − 0.7477 M1 Appropriate area Φ, from final process, must be probability.

= 0.2523 A1 0.252 ⩽ p ⩽ 0.2525

If A0 scored S CB1 for 0.252 ⩽ p ⩽ 0.2525

[Expected number =] 400  0.2523 = 100.92 B1 FT FT their 4SF (or better) probability from a normal calculation.
100 or 101 Must be a single integer answer.

2(b) 1 1 M1 {Both ½ and –½ seen as z-values

[P( −  Z  ) = Φ(½ ) – Φ(−½) =] or appropriate use of +½ or −½}
2 2
and {no other z-values in part}.
1
2Φ   − 1  56.2 − 55.6 55.0 − 55.6
2 Condone and seen as z-values.
1.2 1.2
= 2  their 0.6915 −1
or their 0.6915 − (1 − their 0.6915 ) M1 Calculating the appropriate area from stated phis of z-values which
or 2  ( 0.6915 − 0.5 ) must be ± the same number.

0.383 A1 0.3829 ⩽ z ⩽ 0.383

If A0 scored SC B1 for 0.3829 ⩽ z ⩽ 0.383

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Question Answer Marks Guidance

3(a) 4 1 B1 May be seen used in calculation.

[P(17 or 18) =] = ,0.0185(185…)
216 54
5
 53  1 M1 p(1 – p)5, 0 < p < 1
P(X = 6) =   .
 54  54

0.0169 A1 0.01686 < p ⩽ 0.0169

If A0 scored SC B1 for 0.01686 < p ⩽ 0.0169

3(b)  53 
7 M1  
r
[P(X < 8) =] 1 −    53 
1 −  their  or 0.98148  or correct  ,
 54    54  
r = 7,8 0 < their p < 1

0.123 A1 0.1225 ⩽ p ⩽ 0.123

Alternative method for Question 3(b)

[P(X < 8) =] M1 q + pq + p 2 q + p3q + p 4 q + p5 q  + p 6 q  , p + q = 1, 0  p, q  1, q

2 3 4
 1   53  1   53   1   53   1   53   1 
                 +
+ + + +
= their
53
 54   54  54   54   54   54   54   54   54  54
5 6
 53   1   53   1 
   +   
 54   54   54   54 

0.123 A1 0.1225 ⩽ p ⩽ 0.123

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Question Answer Marks Guidance

4(a) Cw 20 20 10 10 40 M1 At least 4 frequency densities calculated

Fd 1.6 2.3 9.6 5.2 0.6 f 32  f 
eg  condone if unsimplified  , accept unsimplified,
cw 20  cw  0.5 
may be read from graph using their scale no lower than 1 cm = fd 1

A1 All bar heights correct on graph, using their suitable linear scale
with at least 3 values indicated, no lower than 1 cm = fd 2.

B1 Bar ends at [0,] 20, 40, 50, 60, 100 (at axis), 5 bars drawn 0 ⩽ time
axis ⩽ 100, linear scale with at least 3 values indicated.

B1 Axes labelled frequency density (fd), time (t) and minutes

(mins, m) or appropriate title.
(Axes may be reversed).

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Question Answer Marks Guidance

4(b) Midpoints 10 30 45 55 80 B1 At least 4 correct midpoints seen (check data table).

[Mean = 43.2 given] M1 Appropriate variance formula with their 5 midpoints (not upper
32  102 + 46  302 + 96  452 + 52  552 + 24  802 bound, lower bound, class width, frequency density, frequency or
[Var =] − 43.22 cumulative frequency).
250
Or Condone 1 frequency error.
If correct midpoints seen accept
32 (10 − 43.2 ) + 46 ( 30 − 43.2 ) + 96 ( 45 − 43.2 )
2 2 2
 3200 + 41400 + 194400 + 157300 + 153600 549900 
 or 
+52 ( 55 − 43.2 ) + 24 (80 − 43.2 )
2 2
 250 250 
250 −{43.22 or 1866.24} .

 549900  A1 www, final answer 18.25814887 to at least 3SF.

= − 43.22 = 333.36 If M0 earned SC B1 for final answer 18.25814887 to at least 3SF.
 250 
Sd = 18.3

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Question Answer Marks Guidance

5(a) Method 1: Scenarios identified ignoring unbiased coin

1 3 3 M1 All 3 different calculations seen unsimplified.

P(BH1 BT2) =  =
4 4 16
3 1 3
P(BT1 BH2) =  =
4 4 16
1 1 1
P(BH1 BH2) =  =
4 4 16

3 3 1 7 A1 Clear identification of all scenarios, linked probabilities and sum.

+ + = AG
16 16 16 16

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Question Answer Marks Guidance

5(a) Method 2: Scenarios identified with all 3 coins

1 1 3 3 M1 All 6 different calculations seen unsimplified.

P(H BH1 BT2) =   =
2 4 4 32
1 1 3 3
P(T BH1 BT2) =   =
2 4 4 32
1 3 1 3
P(H BT1 BH2) =   =
2 4 4 32
1 3 1 3
P(T BT1 BH2) =   =
2 4 4 32
1 1 1 1
P(H BH1 BH2) =   =
2 4 4 32
1 1 1 1
P(T BH1 BH2) =   =
2 4 4 32

1 + 3 + 3 + 1 + 3 + 3 14 7 A1 Clear identification of all scenarios, linked probabilities and sum.

P(B) = = = AG
32 32 16

Method 3: 1- P(BT1 BT2) ignoring unbiased coin

3
2 M1 Calculation seen unsimplified
1 – P(BT1 BT2) = 1 −   and 1 – probability seen.
4

7 A1 Clear identification of scenario used, linked probability and

= calculation. AG
16

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Question Answer Marks Guidance

5(a) Method 4: 1- P(BT1 BT2) with all 3 coins

1 3 3 1 3 3 M1 Both calculations seen unsimplified

1 – P(H BT1 BT2) – P(T BT1 BT2) = 1 −     −     and 1 – 2 probabilities seen.
2 4 4 2 4 4

9 9 7 A1 Clear identification of all scenarios used, linked probabilities and

= 1− − = calculation. AG
32 32 16

5(b) 1 1 1 1 M1 Their identified P(HHH) or correct as numerator

 P ( A  B )  2  4  4 32 and their identified P(B) or correct as denominator.
 P ( A|B ) = = =
 P ( B) 
7 7 Either numerical expression acceptable.
16 16

1 A1 Accept 0.071428… rounded to at least 3SF.

= , 0.0714
14

5(c) 1 3 1 1 1 3 1 3 3 15 B1 Table with correct X values and at least one probability.

P(1H) =   +   +   = Condone any additional X values if probability stated as 0.
2 4 4 2 4 4 2 4 4 32
1 1 1 1 3 1 1 1 3 7
P(2H) =   +   +   = B1 P(1) or P(2) correct, need not be in table, accept unsimplified.
2 4 4 2 4 4 2 4 4 32
B1 4 correct probabilities linked with correct outcomes, may not be in
X 0 1 2 3 table.
p(X) 9 15 7 1 Decimals correct to at least 3 SF.
32 32 32 32
0.28125 0.46875 0.21875 0.03125 SC B1 for 4 probabilities (0 < p < 1) sum to 1 ± 0.005 with P(1)
and P(2) incorrect.

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Question Answer Marks Guidance

6(a) [1 – P(10, 11, 12) =] M1 One term 12Cx p x (1 − p )12− x , for 0 < x < 12, 0 < p < 1
1 − ( 12C10 0.910 0.12 + 12C11 0.911 0.11 + 12C12 0.912 0.10 )
= 1 – (0.230128 + 0.376573 + 0.282430) A1 Correct expression, accept unsimplified, no terms omitted, leading
to final answer.

0.111 B1 Mark the final answer at the most accurate value,

0.1108 < p ⩽ 0.111 WWW.

Alternative method for Question 6(a)

[P(0,1,2,3,4,5,6,7,8,9) =] M1 One term 12Cx p x (1 − p )12− x , for 0 < x < 12, 0 < p < 1
12
C0 0.90 0.112 +12C1 0.91 0.111 +12C2 0.92 0.110 +12C3 0.93 0.19
+12C4 0.94 0.18 +12C5 0.95 0.17 +12C6 0.96 0.16 +12C7 0.97 0.15 +12C8 A1 Correct expression, accept unsimplified, no terms omitted, leading
0.98 0.14 +12C9 0.99 0.13 ) to final answer. If answer correct condone omission of any 7 of the
8 middle terms.

0.111 B1 Final answer 0.1108 < p ⩽ 0.111 WWW.

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Question Answer Marks Guidance

6(b) [Mean = 80  0.9 =] 72, B1 72 and 7.2 seen, allow unsimplified.

 Variance = 80  0.9  0.1 = 7.2 May be seen in standardisation formula.
(2.683 ⩽ σ < 2.684 imply correct variance).

69.5 − 72 M1 Substituting their mean and their variance into ±standardisation

P(X > 69) = P( Z  )
7.2 formula (any number for 69∙5), not their 7.2, not √their 2.683

M1 Using continuity correction 69∙5 or 68∙5 in their standardisation

formula.

[= P( Z  −0.9317) =] M1 Appropriate area Φ, from final process, must be probability.

Φ ( 0.9317 )

0.824 A1 0.8239 ⩽ p ⩽ 0.8243 WWW.

6(c) np = 72, nq = 8 Both greater than 5, [so approximation is valid] B1 np, nq evaluated accurately.
both np & nq referenced correctly.
> 5 or greater than 5 seen.

Question Answer Marks Guidance

7(a) 7! M1 7!
b,c = 1,2
b ! c !
2! 2!
7!   oe, no further terms present.
2! 2!

5040 A1

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Question Answer Marks Guidance

7(b) Method 1 for first 3 marks: Arrangements of 6 letters including Ls between As

5!  5  2 M1 5!  d, d integer > 1

M1 e!  f  g, e = 5, 6, 7; f = 1, 5; g = 1, 2; f ≠ g,
1 can be implicit.

1200 A1

Method 2 for first 3 marks: Number of arrangements of LL^^^^^ – number of arrangements with the Ls split by an A

6!  2 – 5!  2 M1 6!  2 – h h an integer 1 < h < 1440

M1 k – 5!  2 k an integer k > 240

1200 A1

Method 3 for first 3 marks: Alternative approaches to Method 1

^A ^ ^ ^ ^ ^ A 5
P1  1P1  5P5  1P1 = 600 M1 LL treated as a single unit.

M1

1200 A1

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Question Answer Marks Guidance

7(b) Final 2 marks of Question 7(b)

 9!  B1 Accept unsimplified.
[Total number of arrangements =]  =  90720 May be seen as denominator of probability.
 2!2! 

1200 5 B1 FT their 1200

Probability = , , 0.0132 unsimplified B1 FT if their 1200 and their 90 720
90720 378 their 90720
supported by work in this part.

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Question Answer Marks Guidance

7(c) Method 1: Scenarios identified Both As and Ls removed

5
A____ C4 = 5 B1 1 correct, identified outcome/value
5
AA _ _ _ C3 = 10 for A, AL or AAL scenario, accept unsimplified
5 5
AL _ _ _ C3 = 10 C5–x cannot be used in place of 5Cx
5
AAL _ _ C2 = 10
M1 Add 4 values of appropriate scenarios,
no incorrect scenarios, no repeated scenarios, accept unsimplified,
condone use of permutations.

[Total =] 35 A1 Value stated WWW.

Method 2: 1 A fixed, 1 L removed

No other scenarios can be present anywhere in solution
7 7
A^^^^ C4 M1 Ch, 3 ⩽ h ⩽ 5
7
B1 C4 oe, no other terms, scenario identified.

[Total =] 35 A1 Value stated.

Method 3: 1 A fixed, both Ls removed

A ^ ^ ^ ^ = 6C4 = 15 B1 Correct outcome/value for 1 identified scenario, accept

A L ^ ^ ^ = 6C3 = 20 unsimplified. WWW

M1 Add 2 values of appropriate scenarios,

no incorrect scenarios, no repeated scenarios, accept unsimplified,
condone use of permutations.

[Total =] 35 A1 Value stated.

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