4040 w15 ms 12 - Max Papersmaxpapers.com/wp-content/uploads/2012/11/4040_w15_ms_complete.pdf4040 STATISTICS 4040/12 Paper 1, maximum raw mark 100 This mark scheme is published as an

® IGCSE is the registered trademark of Cambridge International Examinations.

CAMBRIDGE INTERNATIONAL EXAMINATIONS

Cambridge Ordinary Level

MARK SCHEME for the October/November 2015 series

4040 STATISTICS

4040/12 Paper 1, 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 2015 series for most Cambridge IGCSE

®, Cambridge International A and AS Level components and some

Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper

Cambridge O Level – October/November 2015 4040 12

© Cambridge International Examinations 2015

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 marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For

accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier, asterisked, mark in the scheme. The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only. Abbreviations AG answer given on question paper awrt answer which rounds to cao correct answer only dep dependent ft follow through after error oe or equivalent SC special case soi seen or implied www without wrong working




1 (i) systematic B1 (ii) quota B1 (iii) stratified B1 2 (i) mean or mode B1 (ii) median B1 (iii) standard deviation OR variance OR range B1 (iv) interquartile range B1 correct method for Q1 and Q3 (cf = 8, Q1 = 2; cf = 24, Q3 = 5) M1 3 A1 3 (i) one two-way table M1 with rows/columns headed M, F and columns/rows headed T, C, X A1 cell values 2, 5, 1 3, 7, 2 in correct places, totals not required A2 allow A1 for four or five correct (ii) from these data, for males no, for females yes B1 but sample too small for general conclusion B1 4 (i) (a) B B1 (b) C B1 (c) C B1 (ii) 12 × mean for any shop M1 sum of three such products (41.04, 56.04, 45) M1 142 A1 5 (i) any four from 1, 5, 7, 11, 13, 14, 16, 17 for first four numbers written down allow B1 for three correct B2 (ii) for 4, 4, 4 (1/6) × (1/6) × (1/6) (=1/216) B1 for 6, 6, not6 (1/6) × (1/6) × (5/6) B1 × 3 (= 15/216) B1 addition of all cases for 4, 4, 4 and 6, 6, not6 M1 16/216 oe (2/27, 0.0741) A1




6 (i) cumulative frequency polygon B1 (ii) 2.8 (hours) B1 (iii) attempt to read length of stay corresponding to cf = 102 M1 6.5 – 6.6 (hours) A1 (iv) correct method for numbers in paying categories (88 – 48 or 112–88 or 120–112, 40 or 24 or 8) M1 correct payment in a paying category (88–48) × 6 or (112–88) × 9 or (120–112) × 12 (40 × 6 or 24 × 9 or 8 × 12) and total of numbers in two adjacent categories is 64 or 32 A1 correct method for total payment (at least one correct product) (240 + 216 + 96) M1 $552 A1 7 (i) correct method for overall mean M1 overall mean (12.5, 60.3) A1 correct method for LSA or USA M1 LSA (5, 77.7) A1 USA (20, 43) A1 (ii) correct method for gradient M1 correct method for c M1 m = –2.306 to –2.320 and c = 89.1 to 89.4 A1 (iii) use of x = 30 in their equation M1 20 °C A1 ft only if gradient negative and answer is less than 38° (iv) correctly plotted points B2 allow B1 for five correct (v) straight line with negative gradient, for t = 0 to t = 30 M1 correct line joining (0, 89) and (30, 20) OR line joining (0, their c) and (30, their 20) A1 (vi) relationship between the variables is not linear B1 (vii) will be higher than that calculated B1




8 (i) 748 B1 (ii) (525/1614) × 100 AG B1 (iii) ((53 + 60 + 51 + 39)/1614) × 100 M1 awrt 12.6(%) A1 (iv) indication of area being proportional to class frequency M1 rectangles width 2 height 33 A1 width 3 height 24 A1 width 5 height 12 A1 (v) (90/360) × 144 M1 36 A1 (vi) finds (80/360) × 144 (= 32) M1* (their 32/525) × 100 M1dep awrt 6.10(%) or awrt 6.1(%) A1 (vii) finds (70/360) × 144 × 0.5 (= 14) M1* (their 14/their 203) × 100 M1dep 6.90(%) or 6.9(%) A1 9 (i) (a) 11 B1 (b) 6 B1 (c) 4 B1 (d) 2 B1 (e) 8/30 oe B1 (f) 6/14 oe B1 (ii) (a) 4 B1 (b) 9 B1 (c) 16 B1 (d) 0 B1 (iii) for the swimmer 17/30 B1 for the track athlete 8/20 B1 multiplication of their swimmer and track athlete probabilities not multiplied by 2 M1 (provided at least one B1 earned)

17/75 oe A1 (iv) any Venn diagram with a triple intersection of 1 and double intersections of 5, 2, 0 M1 fully correct and annotated diagram A1




10 (i) 6 – under 9 B1 (ii) attempted use of class mid-points (1.5 4.5 7.5 10.5 13.5 17.5) M1* correct method for mean (Σfx = 386) M1dep 7.72 A1 finding values of f × variable squared M1 correct method for SD or variance (Σfx2 = 3798.5, Σfx2 / Σf = 75.97) M1dep 4.04 to 4.05 A1 4.05 A1 (iii) km2 B1

(iv) 12 B1 20 B1 (v) (1) × (1/5) × p × q M1 (1) × (1/5) × (1/4) × (1/3) A1 1/60 oe (0.0167) A1 (vi) (1 – their 1/60) × their 1/60 M1 59/3600 oe (0.0164) A1 11 (i) 6 + 35 + 680 + 961 (= 1682) M1 4000 + 5600 + 8500 + 6200 (= 24300) M1 (their 1682/their 24300) × 1000 M1 69.22 A1 (ii) correct method for any medical condition M1 1.5 6.25 80 155 A1 (iii) any one medical condition rate multiplied by standard population figure M1 sum of four such products M1 (1.5 × 0.15) + (6.25 × 0.25) + (80 × 0.40) + (155 × 0.20) oe A1 64.79 or 64.7875 A1 (iv) (1.4 × 0.15) + (6.25 × 0.25) + (85 × 0.40) + (162 × 0.20) oe M1 68.17 or 68.1725 A1 (v) correct method for deaths at Southshore (1.4 × 5) + (6.25 × 6.4) + (85 × 7.8) + (162 × 5.5) (=1601) M1 81 (ft only on their 1682) A1 (vi) because it has the higher standardised mortality rate OR because mortality rates are higher for groups most at risk M1 Southshore A1





4040 STATISTICS








MARK SCHEME NOTES






1 (i) 5 B1 (ii) 4 + 5+ 6 + 6 + 7 + 3 M1 31 A1 (iii) 4 B1 (iv) identifies their 16th student M1 3 A1 2 (i) 100° ± 2° B1 (ii) “100”/360 × 212 (= 58.88888888 … ) M1 59 A1 (iii) √(137/212) × 3 M1 2.41 cm A1 (iv) larger as a proportion B1 3 (i) ordered list/register B1 175/25 = 7 → every 7th B1 random start in range 1–7 B1 (ii) 52/175 × 25 ( = 7.42857 … ) M1 7 A1 (iii) depends on where they start OR 60 not a multiple of 7 OR differently ordered lists B1 4 (i) 49 B1 (ii) 36/41 × 100 M1 87.8 (%) A1 (iii) (57 + 45)/492 M1 20.7 (%) A1 (iv) boys increasing OR girls constant OR overall increase B1 5 (i) 0.3 B1 (ii) 7/10 B1 × 3/9 B1 (iii) 7/10 × 6/9 × 3/8 M1 7/40 A1 7/24 OR 1 – (7/30 + their P(1) + their P(3)) calculated B1




6 (i) (a) colour/shape/material etc. B1 (b) e.g. number of sides etc. B1 (c) mass/length of side/area etc. B1 (ii) (a) 11 B1 (b) 22 B1 (c) 30 B1 7 (a) product of correct pair of probabilities for SS, SG, GS or GG (stop stop, stop go etc.) M1 1 – GG or SS + SG + GS M1 0.791 A1 (b) (i) 0.5 × 0.78 M1 0.39 A1 (ii) 0.5 × (1 – 0.64) M1 0.18 A1 (iii) 1 – (0.39 + 0.18) M1 0.43 A1 (iv) (1 – “0.43”) × (0.78)2 M1 0.347 A1 (v) 0.36 × 0.78 (= 0.2808) M1 0.36 × 0.22 × 0.36 (= 0.028512) M1 0.64 × 0.36 × 0.78 (= 0.179712) M1 addition M1 0.489 A1




8 (i) 8 points correct B2 (B1 for 6 or 7 correct) (ii) any correct method for either semi-average M1 (37.5, 31.7) A1 (57.5, 45.7) A1 (iii) all 3 correctly plotted (ft their semi-averages) B1 straight line through at least two of their plotted averages B1 (iv) correct ratio using two of the averages or two points on their line M1 m = 0.7 A1 substitution M1 c = 5.45 A1 (v) original length of spring B1 (vi) (a) 35 cm B1 (b) 58 cm B1 (ft their line or equation, ± 0.5 if using line) (vii) Because 75 g is outside the range of the data M1 the reading at 42 g is likely to be more reliable A1 9 (i) 10 B1 (ii) (a) 47 s B1 (b) reading from cf of 35 (= 44 s) B1 reading from cf of 105 (= 51 s) B1 “51” – “44” (provided at least one B1 earned) M1 7 s A1 (c) 33 s–33.5 s B1 (iii) (a) reading from time of 40 s M1 18 A1 (b) reading from time of 49 s (= 88) – “18” M1 70 A1 (iv) 89/140 × 100 M1 64th A1 (v) 60/100 × 130 M1 78 A1 So max. is 70 (ft their (iii) (b) and “78”) B1




10 (i) correct use of cross-over M1 3 A1 (ii) (a) 2.5 × 2 M1 5 A1 (b) 1.5 × 6 + “5” M1 14 A1 (iii) 8/27 × 1.5 (= 0.444 …) M1 + 2.5 M1 2.94 cm A1 (iv) 443/60 M1 7.38 A1 3489/60 – (443/60)2 (= 3.6363888888…) M1* √ M1dep 1.91 A1 (v) 9.38 (ft their mean) B1 1.91 (ft their sd) B1 11 (i) 21/3500 × 1000 = 6 B1 (ii) 12/b × 1000 = 2.5 M1 4800 A1 (iii) 5 B1 7.5 B1 (iv) 75 (21 + 12 + 27 + 15) M1 15700 (3500 + 4800 + 5400 + 2000) M1 “75”/“15 700” × 1000 M1 4.78 A1 (v) 15 B1 (vi) Any correct product of death rate and standard pop M1 Sum of 4 such products M1 (6 × 0.21) + (2.5 × 0.29) + (5 × 0.35) + (7.5 × 0.15) ft A1 4.86 A1 (vii) Because it has a lower SDR M1 Birchville A1





4040 STATISTICS








MARK SCHEME NOTES






1 (i) awrt 16.7 B1 awrt 3.7 B1 (ii) Hotter in 2010 oe B1 Less varied in 2010 oe B1 2 (a) Insufficient information to decide Insufficient information to decide Definitely not mutually exclusive All 3 correct B2 B1 for two correct (b) (i) Use of P(C∩D) = P(C) × P(D) M1 0.4 × 0.3 = 0.12 A1 (ii) Use of P(C∪D) = P(C) + P(D) – P(C∩D) M1 0.4 + 0.3 – 0.12 = 0.58 A1 3 (i) (151.9 – 148.5)/148.5 × 100 OR (151.9/148.5 × 100 – 100) OR 3.4/148.5 × 100 B1 (ii) 4.3[28…] B1 [–] 1.5[21…] B1 (iii) Attempt at change chart illustrating positive and negative change B1* Suitable scale, labelled as percentage change and all bars labelled B1dep Correct bars (within ± ½ small square) B1 dep 4 (i) (a) (x – 50)/10 = (48 – 58.1)/8.1 OR x = 50 + 10/8.1(48 – 58.1) M1 awrt 37.5 A1 (b) (x – 50)/10 = (x – 58.1)/8.1 M1 awrt 92.6 or 93 A1 (ii) (30 × 58.1 – 23 × 56)/7 One correct product seen, 30 × 58.1 OR 23 × 56 [1743 OR 1288] M1* (30 × 58.1 – 23 × 56) [455] M1dep /7 = 65 A1 5 (i) Attempt at reading from graph – 27 or attempt at reading from graph + 41 M1 588 – 589 A1 650 – 651 A1 (ii) [Original data] below the trend line [on average]/on average $38 below trend line B1 (iii) [Daily/quarterly] sales reducing (but not each quarter) oe B1 (iv) 24 B1




6 (i) 135 (allow 135.75 or 136) B1 139.5 + or 149.5 – M1 (‘135’ – 104)/43 × 10 (147 – ‘135’)/43 × 10 oe M1 awrt 146.7 A1 SC B1 for 123.0 (ii) (116 – 109.5)/20 × 31 M1 Some fraction of 31 + 24 M1 34 www A1 7 (a) (i) Advantage: quicker, cheaper, easier to handle (oe) B1 Disadvantage: less accurate, may not be representative (oe) B1 (ii) 100, 200, 300, 400, 500, 600 Any systematic sample B1 Starting value 100 B1 Gaps of 100 and 6 values in range B1 (iii) One that gives each member of the population an equal chance of being selected B1 (b) (i) Attempt at job type totals [20, 30, 10] (can be implied) M1 Evidence of 2, 3, 1 of each (only implied by a fully correct answer) A1 24(T), 19(C), 50(E), 43(T), 38(T), 13(C) B3 –1 each independent error (ii) M, F, M, F, M, F, so 3 of each (identifying the genders in their sample) B1 Should have 4 males and 2 females/twice as many males as females B1* So not representative B1 dep (iii) Because it is likely to be most relevant to enjoyment (or any related reason) B1* Sample stratified by job type more appropriate B1dep (gender could score here if reason clearly connected to enjoyment of work)




8 (a) (i) Non numerical B1* so qualitative B1dep (ii) 22% represents 33 students (can be implied) B1 Using 33/“22” M1 Correct method for any one subject (can be implied) M1 Plumbing = 54, Carpentry = 129, Building = 117 A2 (A1 for 2 correct) (iii) (a) Plumbing = 46%, Carpentry = 80% /greater percentage studying Carpentry M1 so definitely false A1 (b) Numbers of students in 2013 not known B1* so insufficient information to decide B1dep (b) (i) Can take any value [in a range] OR can be measured B1* so continuous B1dep (ii) 23 AND 26 B1 3 B1 (iii) Speedy Wheelers cycled further oe B1 9 (i) (a) 8/18 or 4/9 or 0.44 B1 (b) 4/18 or 2/9 or 0.22 B1 (c) 3/8 or 0.375 or 0.38 B1 (ii) 7/18 × 11/17 × 2 7 × 11 seen in numerator (oe 4 × 5 + 3 × 6 + 3 × 5 + 4 × 6) B1* Product of 2 probabilities × 2 oe M1dep n × n – 1 in denominator M1 77/153 o. e. 0.50[3…] A1 (iii) 15/18 × 14/17 × 13/16 × 3/15 “18” – 3 seen in numerator M1 n × (n – 1) × (n – 2) × (n – 3) in denominator M1 91/816 oe A1 (iv) (4/10 × 6/9 × 5/8 × 4/7) × 2 + (6/10 × 5/9 × 3/8 × 5/7) × 2 10 and 8 seen multiplied in a denominator M1* One ( ) correct M1dep At least 2 products of 4 probabilities with 4 × 6 × 5 × 4 in one numerator

and 6 × 5 × 3 × 5 in the other M1 31/84 oe A1 (v) 1/3 × 6 or 2/5 × 5 seen M1 4/11 (or 0.36..) A1




10 (i) 8.52/7.96 [× 100] OR (8.52 – 7.96)/7.96 × 100 OR 7.96 × 107 / 100 = M1 Fully correct method, 8.52/7.96 × 100 = 107 OR (8.52 – 7.96)/7.96 × 100 + 100 AG A1 (ii) 7.96 × 103 / [100] = [8.1988] oe M1 8.20 A1 (iii) Price/cost fell by 3% B1 Between 2011/base year and 2012 B1 (iv) Any one correct method (can be implied) M1 awrt 106, 96, 97, 107 A2 A1 for any 2 or 3 correct (v) (106 × 12 + 96 × 9 + 97 × 4 + 107 × 2)/(12 + 9 + 4 + 2) Σ any price rels × weight M1 Σ their (iv) × weights / Σw (27) M1 101.4–101.7 www A1 (vi) 319 000 × their (v) /100 M1 323 000 A1 (vii) As price changes [in A OR D] have been accounted for in the price relatives B1* A AND D B1dep 11 (i) 0.05 B1 (ii) 0.4 × 1 + 0.2 × 2 + 0.2 × 3 + 0.15 × 4 + ‘0.05’ × 5 [= 2.25] M1 “2.25” – 2.40 M1 Loss of 0.15 (must state ‘loss’ somewhere or –0.15) A1 (iii) (a) P(3 or less) = 0.4 × 0.4 + 0.4 × 0.2 × 2 (condone × 2 missing) M1 = 0.32 A1 (b) “0.32” × y = 2.40 y = 7.50 (c) “0.32” × 100 = 32 B1 (“7.50” – 9) × “32” or 2.40 × 100 – “32” × 9 (±), M1 Loss of $48 A1 (iv) P(1) = 150/360 = 5/12 oe P(2) = 120/360 = 1/3 oe P(3) = 90/360 = 1/4 oe All 3 correct B2 “5/12” × x + “1/3” × 2x + “1/4” × 3x = 11 M1 x = 6 A1 Prizes = 6, 12, 18 A1





4040 STATISTICS








MARK SCHEME NOTES






1 (i) 9, 12 B1 (ii) Pair of polygons M1

Labelled or key B1 Correct plots vertically A1 Correct plots horizontally A1 (ft their boundaries, provided difference of 3. All consistent with possible exception of end points)

(iii) Children at aqua splash are older oe B1 (general comment required, it is not enough to comment on one age group only) 2 (i) (45 – 50)/10 = (x – 62.7)/7.4 M1

59 A1 (ii) (82 – 45.1)/8.2 = (x – 62.7)/7.4 M1

96 A1 (iii) (37.5 – 50)/10 = (39 – 48.5)/a M1

7.6 A1 3 (a) (i) P(A) = 0.3 or (1 – 0.7) seen B1

Use of P(A∩B) = P(A) + P(B) – P(A∪B) or “0.3” + 0.6 – 0.7 M1 0.2 A1

(ii) P(A) × P(B) = “0.3” × 0.6 ≠ “0.2” M1

So not independent A1 (b) C and D, D and F B1 4 (i) Suitable scale and axis labelling B1

Key/bars labelled B1 Correct bars for country A: 9, 35, 56 B1 Correct bars for country B: 14, 47, 39 B1

(ii) Country A has greater urban area oe B1 Country B had greater proportion of its area that is urban oe B1

5 (a) Advantage: Quicker, cheaper, easier to handle (less data) (oe) B1

Disadvantage: May not be representative, less accurate (oe) B1 (b) (i) True if the original population contains equal numbers of males and females/gender

is relevant, otherwise not true B1* So sometimes true B1dep

(ii) A random sample could produce these numbers/true if there is some order to the

list, but not true if the list is random B1* So sometimes true B1dep




6 (i) 7 × 39 – 6 × 38 or 38 + 1 × 7 M1 45 A1

(ii) Σx²/6 – 38² = 71 use of formula for var/sd M1

Σx² = 9090 for 6 days A1 Σx² for 7 days = “9090” + “45”² (11 115) M1 Var = 11 115/7 – 39² = 66.9 (awrt) A1

7 (i) 2/5 or 3/5 seen B1

(2/5 × 3/5) M1 White and black or black and white Product of 2 probs × 2 (oe) M1 12/25 A1

(ii) (3/5 × 3/5 × 2/5) M1

Product of 3 probs × 3 (oe) M1 (3/5)³ M1 “P(2 black)” + “P(3 black)” (dep on at least one previous M) M1dep 81/125 A1

OR

(2/5 × 2/5 × 3/5) (M1 Product of 3 probs × 3 (oe) M1 (2/5)3 M1 1 – “P(0 black)” – “P(1 black)” (dep on at least one previous M) M1dep 81/125 A1)

(iii) Without replacement understood, i.e. n × (n – 1) in denominator M1

(2/5 × 1/4 × [3/3]) M1 “(2/5 × 1/4 × [3/3])” × 3 M1 3/10 A1

(iv) Evidence of bwbwb M1

3/5 × 2/4 × 2/3 × 1/2 [× 1] or 3!2!/5! M1 1/10 A1




8 (i) 16, 37, 70, 92, 108, 116, 120 B1 (ii) 60th value (allow 60.5th) B1

20 + M1 (“60” – 37)/33 × 10 (26.9696…) M1 27.0 (condone 27) A1

(iii) 80/100 × 120 or 120 – 20/100 × 120 [96] M1

40 + M1 (“96” – 92)/16 × 20 M1 45 A1 (SC B1 for 13.8)

(iv) 8/10 × 21 + 16 = 32.8 (33 people less than 18)

7/20 × 8 + 108 = 110.8 (111 people less than 67) 8/10 × 21 or 2/10 × 21 or 7/20 × 8 or 13/20 × 8 M1 Full attempt at total less than 18 or ⩾ 18 or < 67 or ⩾ 67 OR 2/10 × 21 (= 4.2) AND 7/20 × 8 (= 2.8) M1* “111” – “33” OR “4.2” + 33 + 22 + 16 + “2.8” M1dep 78 A1 78/120 × 100 = 65% A1

(v) Data is grouped/actual ages not known B1

and assumed to be evenly distributed within each class B1 9 (i) 7.50 × 98/100 oe or 7.50 × 106/100 oe M1

7.35 and 7.95 A1 (ii) 100s in first column B1

8.52/8.10 [× 100] or 8.36/8.10 [× 100] or 7.01/7.20 [× 100] M1 105, 103, 97 and 97 (awrt) (A1 for two or three correct) A2 (–1 if all correct but not to nearest whole number)

(iii) 10 × 8.10, 6 × 7.50, 5 × 7.20 [81:45:36] M1

÷ 9 gives 9, 5 and 4 M1 Each worker does same number of hours B1

(iv) Any one weight × price relative M1

9 × “103” + 5 × 106 + 4 × “97” + 2 × 108 M1 ÷ (9 + 5 + 4 + 2) M1 awrt 103 A1

(v) There has been an increase of 3% B1

in the total wage bill between 2011 and 2013 B1 assuming that number of workers/hours worked at each grade has remained the same B1




10 (i) To remove variation, in order to find the trend/to make predictions B1 B1 (OR To find the trend, in order to make predictions B1 B1)

(ii) So that moving average values coincide with original data items B2

(B1 for mention of 4 being even) (iii) a = 95.9 B1

b = 226.2 B1 c = 58.2 B1

(iv) 64.1 – 57.5 = [6.6]

63.2 – 56.4 = [6.8] attempt at a suitable difference (may be negative) M1 Sum of two such differences ÷ 2 (may be negative) M1 6.7 (thousand) A1

(v) Correct plots (B1 for six or seven correct plots) ft their c B2 Suitable trend line B1

(vi) Number of marriages is decreasing (not each quarter) B1 (vii) Reading from graph + their (iv) (e.g. 55.5 + 6.7) M1

61.9 to 62.3 thousand or 61 900 to 62 300 (ft their (iv) and their trend line) A1 11 (i) (a) 1/2, 1/3, 1/6 oe seen B2

(B1 for 1 or 2 correct)

“1/2” × 1 + “1/3” × 2 + “1/6” × 3 = [5/3 or 1.67] M1 “5/3” – 2 [= –1/3] allow (±) award earlier if “1/2” × –1 + “1/3” × 0 + “1/6” × 1 M1 = loss of 0.33 (must state ‘loss’) A1

(b) (i) At least one of 1/2 × 1/2 or 1/3 × 1/3 or 1/6 × 1/6 M1

1/2 × 1/2 + 1/3 × 1/3 + 1/6 × 1/6 [= 14/36 = 7/18] M1 “7/18” × x = 2 M1 x = 36/7 = 5.1… A1 $5 A1

(ii) “7/18” × 90 [= 35] (“7/18” must be a probability) M1

90 × 2 – “35” × “5” or (“36/7” – 5) × “35” M1 $5 profit (condone ‘profit’ missing) A1

(ii) “2/3” × y + “1/3” × 2y = 2 where “2/3” and “1/3” are probabilities M1

1.50 A1 3 A1

Documents

4040 w15 ms 12 - Max Papersmaxpapers.com/wp-content/uploads/2012/11/4040_w15_ms_complete.pdf4040 STATISTICS 4040/12 Paper 1, maximum raw mark 100 This mark scheme is published as an