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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Cambridge Ordinary Level
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MARK SCHEME for the October/November 2014 series
4040 STATISTICS
4040/23 Paper 2, maximum raw mark 100
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 will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the October/November 2014 series for
most Cambridge IGCSE®, Cambridge International A and AS Level components and some
Cambridge O Level components.
® IGCSE is the registered trademark of Cambridge International Examinations.
Page 2 Mark Scheme Syllabus Paper
Cambridge O Level – October/November 2014 4040 23
1 (i) Mode = 17 B1
(ii) Attempt at valid method to find median M1
Median = 16 A1
(iii) (a) Any attempt to work with a cumulative frequency of 29 M1
k = 11 A1
(b) k = 9 B1
2 Sight of 60% or 0.6 being used B1
Any attempt to multiply a ‘1st class probability’ by 0.4 AND a ‘2nd class probability’ by 0.6 M1
Any attempt to multiply at least two of these products by the appropriate value of
the variable M1
Attempt to sum five such ‘expectations’ M1
(0.4 × 0.8 × 1) + (0.4 × 0.2 × 2) + (0.6 × 0.5 × 2) + (0.6 × 0.3 × 3) + (0.6 × 0.2 × 4)
0.32 + 0.16 + 0.6 + 0.54 + 0.48
Five correct terms summed, either evaluated or unevaluated A1
2.1 A1
3 (i) 430 B1
(ii) 17.2 B1
(iii) 8131 B1
(iv) Variance = (8131/25) – (17.2)²
Use of a correct formula for variance M1*
Attempt to take square root of ‘their variance’ M1dep
5.42 cm A1
4 (i) (x – 27)/12 = (x – 30)/6
An appropriate equation in any form in which the two 'unknowns' are the same. M1
A correct such equation. A1
x = 33 A1
(ii) An attempt at a standardised term with the unknown s.d. in the denominator M1
(51 – 27)/12 = (100 – 50)/s.d.
Correct equation in any equivalent form A1
25 A1
5 (i) Bar chart of correct structure B1
Bar of correct heights and chart fully annotated B1
(ii) Two bars of equal height and full annotation B1
Percentage components correct (27-33-40) and (31-33-36) B1
(iii) Because it directly compares the share which each item has of overall expenditure
(or similar valid reason) M1
the percentage sectional chart is more useful. A1
© Cambridge International Examinations 2014
Page 3 Mark Scheme Syllabus Paper
Cambridge O Level – October/November 2014 4040 23
6 (a) Any reference to frequency being proportional to area in a histogram B1
A qualitative variable has no ‘class widths’ which can be used to form/evaluate
such areas. B1
(b) Any valid comparison, e.g.
A discrete variable can only take certain values within its range, whereas a
continuous variable can take all values within its range. B2
(Or, a discrete variable is counted, a continuous variable is measured.)
(c) (i) 15 B1
(ii) 14.5 B1
© Cambridge International Examinations 2014
Page 4 Mark Scheme Syllabus Paper
Cambridge O Level – October/November 2014 4040 23
7 (i) Number of boxes of balls purchased = 75/3 = 25 M1
Therefore cost of balls = 25 × 50 = $1250 A1
Total wages = 12.50 × 600 = $7500 B1
Required ratio = 10 000 : 1250 : 2500 : 7500 = 8 : 1 : 2 : 6 AG B1
(ii) Balls 90 B1
Maintenance 102, Services 105, Wages 103 (B1 for two correct) B2
(iii) [(102 × 8) + (90 × 1) + (105 × 2) + (103 × 6)] / 17
For any one product (weight × price ratio) (except for weight of 1) M1
For attempt to sum four such products M1*
Division by 17 M1dep
1734/17 = 102 A1
(iv) Total 2012 expenditure = $21 250 B1
Estimate of 2013 expenditure = $(21 250 × 102)/100 (with or without /100) M1
$21 675 (or 21 700 as 3sf value) A1
(v) Any valid reasons not accounted for by information included in the calculations
(i.e. not ‘inflation’)
e.g. Varying membership or number of matches played may affect the number of balls
purchased. B2
8 (i) 2 – under 3 B1
(ii) 8 cm B1
(iii) 12 209 242 255 379 401 412 500 (–1 each independent error) B2
(iv) 4 + M1
(8 or 8.5)/13 M1
4.62 or 4.65 A1
(v) (Use of formulae must be consistent throughout)
UQ = 5 + (120 or 120.75)/124 M1
= 5.97 (using either formula) A1
LQ = 2 + (113 or 113.25)/197 M1
= 2.57 or 2.58 A1
IQR = UQ – LQ = awrt 3.40 A1
(IQR A1 dep on at least one of the M1s)
(vi) (a) (1.35 or 1.32) and (2.04 or 2.05 or 2.07 or 2.08) B1ft
(b) Any valid comment relating to skewness or lack of symmetry B1ft
(vii) The gradient will be steepest where the class frequency is highest, M1
around the 2 – under 3 class. A1
© Cambridge International Examinations 2014
Page 5 Mark Scheme Syllabus Paper
Cambridge O Level – October/November 2014 4040 23
9 (a) (i) Any comment meaning the events cannot occur simultaneously B1
(ii) Any valid examples, but the two events must both be possible outcomes of the
same ‘experiment’ B1
(iii) (a) Any reference to the probabilities of possible outcomes not summing to more
than 1 B1
(b) Use of P(A) × P(B) M1
0.3 A1
(b) (i) Valid probability with a denominator of 60 M1
24/60 = 2/5 = 0.4 A1
(ii) Valid probability with a denominator of 35 or a numerator of 23 M1
23/35 = 0.657 A1
(iii) Valid probability with a denominator of 25 or a numerator of 11 M1
11/25 = 0.44 A1
(iv) Product of two valid probabilities with denominators of 60 and 59 M1
(5/60) × (4/59) = 1/177 = 0.00565 A1
(v) (35/60) × [(7/35 × 12/59) + (28/35 × 13/59)]
(35/60) × an attempt at the second probability, seen M1
Product of two probabilities with denominators 35 and 59 seen M1
112/885 = 0.127 (correct result) A1
OR
(7/60 × 12/59) + (28/60 × 13/59)
Correct numerators in an expression of this form M1
Correct denominators in an expression of this form M1
112/885 = 0.127 A1
© Cambridge International Examinations 2014
Page 6 Mark Scheme Syllabus Paper
Cambridge O Level – October/November 2014 4040 23
10 (i) 12 00 07 09 01 (–1 each independent error) B2
(ii) (a) 00 02 B1
(b) 00 B1
(c) 03 06 09 12 B1
(iii) (a) 3 friends, 2 relatives B1
(b) 06 09 08 04 02 (–1 each error) B3
(iv) (a) Group I 2, Group II 2, Group III 1 B1
(b) 11 13 10 02 09 (–1 each error) B2
(v) Sample in (iii) obviously representative for F/R and also for age, so totally representative.
Clear indication of valid method M1
Correct conclusion A1
Sample in (iv) obviously representative for age but over-represents friends. (Equivalent
comment regarding under-representation equally acceptable.)
Clear indication of valid method M1
Correct conclusion A1
11 (i) Because each ‘cycle’ is of length 5 days (or equivalent comment) B1
(ii) Because the MA values are at the same point in time as the original values
or some comment relating to each cycle containing an odd number of
observations. B1
(iii) Plots correct vertically B1
Plots correct horizontally B1
Either a clear cyclical pattern, or no clear upward or downward long-term trend B1
(iv) x = 127 B1
y = 24.8 B1
(v) Plots correct vertically B1
Plots correct horizontally B1
(vi) To eliminate seasonal variation, M1
achieved well in this case. A1
(vii) Suitable straight line through plotted MA points. B1
(viii) Use of seasonal components summing to 0. M1
q = –3 A1
(ix) Correct use of reading from their graph and Tuesday component. M1
17 (result must be an integer as discrete variable). A1ft
© Cambridge International Examinations 2014
