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® IGCSE is the registered trademark of Cambridge International Examinations.

CAMBRIDGE INTERNATIONAL EXAMINATIONS

Cambridge Ordinary Level

MARK SCHEME for the October/November 2014 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 2014 series for most Cambridge IGCSE

®, Cambridge International A and AS Level components and some

Cambridge O Level components.

Mark Scheme Syllabus Paper

Cambridge O Level – October/November 2014 4040 12

© Cambridge International Examinations 2014

1 (i) correct method for mean M1 46.5 A1 correct method for SD or variance M1 4.46 or better A1

(ii) mean smaller B1

SD larger B1 2 (i) SD/variance = 6, 36 or 36, 6 M1

SD = 6 and variance = 36 A1 med = 48 B1 LQ = 43 B1 UQ = 53 B1

if zero scored allow SC1 for their LQ, their med, their UQ in ascending order

(ii) their UQ B1 3 (i) (a) citizens not in the telephone directory excluded B1 (b) better response rate/questions can be clarified by interviewer B1 (c) can reach a wide range of people/efficient distribution/

responses obtained very quickly B1

excludes those without internet access/ responses may be from non citizens B1

(ii) (a) limited number of answers to questions possible/

respondent may feel none of allowed answers appropriate B1 (b) any relevant open question B1 4 (i) (a) 19 in correct place B1 (b) 20 in correct place B1 (c) 17 in correct place B1 (d) 34 in correct place B1 (ii) attempt to find frequencies for variable values 1, 2 (63, 81) M1

2 A1




5 (i) 15/40 or 3/8 or 0.375 B1 (ii) 5/40 or 1/8 or 0.125 B1 (iii) 6/15 or 2/5 or 0.4 B1

(iv) (17/40) × M1 (their 17 – 1)/(their 40 – 1) M1

272/1560 or 136/780 or 68/390 or 34/195 or 0.174 or 0.17 A1 6 (i) addition of scale readings of 10 km/h wide columns (26 + 43 + 47) M1

116 A1 (ii) appreciation of area being proportional to frequency

(may be earned here or in (iii) or (iv)) M1 62 A1

(iii) 40 A1 (iv) 6 A1




7 (i) 1 + 8 + 3 + 37 (=49) M1 25 + 167 + 40 + 228 (=460) M1

(their 49/their 460) × 1000 M1 106.5 A1

(ii) correct method for any job group M1

40 47.9 75 162.3 A1 (iii) any one job group rate multiplied by standard population figure M1

sum of four such products M1

(40 × 0.08) + (47.9 × 0.35) + (75 × 0.12) + (162.3 × 0.45) A1 102 or 102.0 A1

(iv) because its standardised accident rate is lower M1

Fastbuild A1 (v) Kwikbuild 30.7 (or Fastbuild 32.5) B1

Fastbuild 32.5 (or Kwikbuild 30.7) and Kwikbuild B1 (vi) crude B1*

standardised rate is to eliminate differences in population structures so is meaningless for one category B1dep

8 (i) 280 B1

(ii) (35/100) × 120 AG B1

(iii) (45/100) × 160 M1 72 A1

(iv) number completing = 65 + 39 + 10 + 71 + 46 + 37 (=268) M1

their (i) – their 268 M1 12 A1

(v) attempted use of class mid points (75, 105, 135, 165) M1*

correct method for mean (Σfx = 5205) M1dep 141 A1

(vi) finds 18 + 2 + 15 + 8 + 3 + 6 (=52) M1*

((3 + 6)/their 52) × 100 M1dep 17.3% or better or 17% A1

(vii) finds 30% of 160 (=48) M1*

((3 + 15)/their 48) × 100 M1dep 37.5% or 38% A1




9 (i) (a) 23.5–23.8 B1 (b) 26.2–26.5 B1 (c) 21.2–21.5 B1 (d) 29.5–29.8 B1 (ii) (a) attempt to read cf% values for BMI = 18.5 and 25 and subtract

on either graph (65 – 7) (40 – 4) M1 57(%)–59(%) A1

(b) 36(%) A1 (iii) 60% overweight B1

attempt to read BMI for cf% = 40% + ½ × 60% (=70%) on 2010 graph M1 28.8–29.1 A1

(iv) attempt to read BMI for cf% = 93% on 1980 graph M1

30 A1

Attempt to read cf% for BMI = 30 on 2010 graph M1 22(%) A1

(v) population has become more unhealthy, with specific support

median BMI increased or percentage healthy decreased or percentage obese increased B1

support strengthened by reference to more than one of these changes or citation of specific values for any change B1




10 (i) correctly plotted points B2 allow B1 for 6 or 7 correctly plotted

(ii) not in the set of four lowest x values

or any indication of need to order data by x values first B1 (iii) method for calculating either semi-average M1

plot of (2, 41) A1 plot of (4.5, 69) A1 method for calculating overall mean M1 plot of (3.25, 55) A1

(iv) straight line through at least two of their plotted points in (iii) B1

correct method for gradient, m, of their line M1 correct method for c M1 m = 11.0 – 11.4 and c = 18 – 19 A1

(v) 52 B1 (vi) because its line has the greatest gradient oe M1

Science (or Statistics if their m > 13.8) A1 (vii) difficult to know if pupils perform well because they like a subject,

or they like a subject because they perform well in it B1 11 (i) (a) (0.9)2 M1

0.81 or equiv fraction A1

(b) 1 – their 0.81 or (0.1 × 0.9 × 2) + (0.1)2 M1 0.19 or equiv fraction A1

(ii) 0.1 × 0.4 M1

× (0.9)3 M1

× 4 M1 0.11664 or 0.1166 or 0.117 or 0.12 or equiv fraction (729/6250) A1

(iii) (a) (0.1 × 0.6)2 M1

× 0.9 × 3 M1 0.00972 or 0.0097 or equiv fraction (243/25000) A1

(b) (0.1 × 0.4) × (0.9)2 × 3 (L, not O, not O) (0.0972) M1

(0.1 × 0.6) × (0.9)2 × 3 (S, not O, not O) (0.1458) M1 (0.9)3 (not O, not O, not O) (0.729) M1 addition of their 0.00972, their 0.0972, their 0.1458, their 0.729 M1 0.98172 or 0.9817 or 0.982 or 0.98 or equiv fraction (24543/25000) A1





4040 STATISTICS








1 8 is the mode M1 The value which occurs most frequently. A1 9 is the median M1 Obtained by arranging the values in ascending or descending order and selecting the 'middle' one. A1 11 is the (Arithmetic) mean M1 Obtained by summing the numbers and then dividing by 13. A1

2 (i) X is discrete B1*

Because it only takes integer values (or equivalent comment) B1dep (ii) 0 and 4 (B1 for each) B2 (iii)

x 0 1 2 3 4 5 6 7

Frequency 0 5 15 10 0 7 6 7

(–1 each independent error) B2 3 (a) Similar in that both would sample proportionately from the different age groups. B1

In stratified sampling interviewers would be given a list of specific people to interview, in quota sampling the interviewer selects the individuals. B1

(b) (i) Because the last page of a chapter is less likely than all other others to be

filled with words, B1 the sample is likely to be biased. B1

(ii) A systematic sample is a form of random sampling B1

and so unless there is some pattern in the pages which matches the sampling interval the sample will be unbiased. B1

4 (i) 0 8 18 35 46 50 (all correct) B1 (ii) All points plotted correctly both horizontally and vertically B1 Plotted points connected by a suitable smooth curve B1 (iii) (a) Correct reading from graph of a point between cum. freqs. 12 and 13 B1 (b) Clear attempt to use appropriate point on the graph and any valid method

to find the required percentage. M1 14%–16% A1




5 (i) Advantage: it shows actual amounts of wood. B1 Disadvantage: it only shows information about individual sizes. B1

(ii) The total amount of wood of all sizes produced. B1 (iii) Pie chart B1

Sectional (component) bar chart B1 (iv) Change chart B1 6 (i) Attempt to sum the values in the diagram and subtract the total from 70. M1

5 A1 (ii) None of the people in the sample speak all three languages. B1 (iii) (a) No, because this person will still only speak two languages. B1 (b) Yes, because the person now speaks all three languages. B1 (c) No, as this person only speaks one of the three languages. B1




7 (a) Sight of 3/7 used B1 EITHER 1 – sum of two two-factor products M1

1 – [(4/7 × 1/5) + (3/7 × 1/9)] A1 88/105 A1 OR Sight of 4/5 and 8/9 used M1

(4/7 × 4/5) + (3/7 × 8/9) A1 88/105 A1

(b) (i) EITHER 3/7 × 2/6 × 1/5 OR 1/7 × 1/6 × 1/5 × 3! M1 1/35 A1

(ii) Any appreciation of the fact that it is irrelevant which two are the

brother and sister. B1

EITHER 1/7 × 1/6 (× 1) × 3! OR 5/7 × 1/6 × 1/5 × 3! M1 1/7 A1

(c) (i) Clear attempt at both two blue and two white M1

(2/8 × 3/8) + (6/8 × 5/8) A1 9/16 A1

(ii) Given first balls were the same colour, P(both were blue) = 1/6,

P(both were white) = 5/6 B1

Attempt to add probabilities relating to whether first balls were blue or white M1

(1/6)[(3/9 × 5/7) + (6/9 × 2/7)] + (5/6)[(2/9 × 4/7) + (7/9 × 3/7)] A1 86/189 = 0.455 A1




8

(1) (2) (3) (4) (5) (6)

Time (x) (minutes)

Frequency (f) Mid-pts (m) y fy fy²

0 – under 30 6 15 –12 –72 864

30 – under 35 11 32.5 –5 –55 275

35 – under 40 4 37.5 –3 –12 36

40 – under 50 40 45 0 0 0

50 – under 60 26 55 4 104 416

60 – under 70 14 65 8 112 896

70 – under 100 4 85 16 64 1024

TOTAL 105 141 3511

(i) Mid-points correct B1 (ii) Values of y found correctly M1

y values correct A1 (iii) fy values found correctly M1 (iv) fy² values found correctly M1 (v) Summations correct A1 (vi) Use of their values in a correct method for mean of y M1

Mean of y = 1.34 A1 (vii) Use of their values in a correct formula for variance or s.d. of y M1 s.d. of y = 5.62 A1

(viii) (a) (Their y mean × 2.5) + 45 M1

48.4 A1

(b) (Their y s.d. × 2.5) only M1 14.1 A1

(ix) The distribution is reasonably symmetrical with relatively

few extreme values, (or similar comment), M1 and so the s.d. is preferable to the IQR. A1




9 (i) 36 32 in second and third cells B1 Any appreciation of area being proportional to frequency M1 24 28 in first and last cells A1 21 18 22 19 in remaining cells A1

(ii) Correct classes, 15–17, 17–19 etc. M1

Correct frequencies 24 68 80 28 A1 Their results presented in a suitable table B1

(iii) Four rectangles of equal width M1

Vertical axis correctly annotated M1 Rectangles of correct heights A1

(iv) Use of ‘diagonal line’ on histogram or equivalent numerical method seen M1

19.35 cm A1 (v) Proportions of first and last classes found correctly M1

Total cakes which can be sold found correctly M1 Percentage expressed correctly M1 84% A1

10 (i) (3 × 7) or (3 × 7000)/1000 or equivalent seen AG B1 (ii) Total deaths 25 + 21 + 47 + 83 ( = 176) M1

Total population 4500 + 7000 + 6000 + 7000 (= 24500) M1

CDR = (Total deaths / Total population) × 1000 M1 = 7.18 A1

(iii) (Deaths/Population) × 1000 seen for any age group (or can be implied by one correct result) M1

5.56 7.83 11.86 all correct A1

(iv) Rate × SP% seen for any age group (or can be implied by one correct result) M1

Attempt to sum results for all age groups M1

5.56 × 0.2 + 3 × 0.35 + 7.83 × 0.25 + 11.86 × 0.2 A1 6.49 A1

(v) Rate × SP% added for four groups M1 7.90 A1

(vi) Any valid comment relating to the towns having different age structures B1 (vii) Because the SDR is lower M1

Eastbury has the healthier environment. A1




11 (i) Correct plots (–1 each error) B2 Correct labels B1

(ii) (37.5,104.5) (B1 each coordinate) B2

Correct plot B1 (iii) Correct SA plots (B1 for each) B2

Line of best fit through at least two averages B1 (iv) A and B results are both approximately linear. B1

C results are completely inconsistent. B1 (v) Correct plot B1 (vi) Experienced technician’s result totally consistent with those of B, B1

suggesting that B’s observations are accurate. B1 (vii) Line drawn through results of B and the experienced technician B1 (viii) 135 kg, with clear indication value found from use of the revised line B1





4040 STATISTICS








1 (i) A variable whose outcomes can only take specific values, or can be counted or listed.

B1

(ii) Correct example e.g. height, weight… B1

(iii) A variable which has non-numerical outcomes. B1

(iv) Correct example e.g. shoe size, number of people on a bus…

B1

2 (i) 6 B1

(ii) 15th value or (29 + 1)/2 5 www

M1 A1

(iii) Any attempt to work with a cumulative frequency of 18 or sight of total of 35

(29 + n + 1)/2 = 18 n = 6 S. C. B1 for 0 following an answer of 4.5 in (ii)

M1

A1

3 (i) (a) 5/20 × 4/19 (n/m × (n – 1)/(m – 1)) or 5/20 × any probability 1/19 oe or 0.053 or better

M1 A1

(b) (5/20 × 15/19) × 2 Product of two probabilities × 2 oe or (5/20 × 15/19) oe 15/38 oe or 0.39 or better

M1 A1

(ii) 8/20 + 12/20 × 8/20 p/m + (m – p)/m × p/m oe (accept additional terms for this mark) 16/25 or 0.64

M1 A1

4 (i) Median IQR The data contains extreme values (if these are specified they must be the large values of m) or data is not symmetrical. (There must be a single/the same reason.) (B3 for 2 correct measures and correct single/the same reason B2 at least 1 correct measure and correct reason for that measure B1 for median and IQR with incorrect reason or error in reasoning)

B3

(ii) 100/150 × 19 oe (or accept 50/150 × 19 oe) 100/150 × 19 + 12 only oe 25

M1 M1 A1




5 (i) Percentage sectional/component/composite bar chart B1

(ii) 42 36 22 36 24 60

B1 B1

(iii) Scale from 0, going up in equal intervals to at least their max freq with label ‘no. of students’ or ‘frequency’ (may appear in title) Three pairs of bars and correct labelling on horizontal axis Bars correctly shaded and drawn to correct heights (ft their (ii))

B1

B1 B1

(iv) ‘It shows actual numbers/original data, (rather than percentages)’ or ‘It allows for easy comparison of numbers of males and females (taking each option)’.

B1

6 (i) (a) A and B, A and C, A and D (–1 each error or omission) B2

(b) B and C, C and D (–1 each error or omission) B2

(ii) EITHER 1/6 (awrt 0.17) and 1/2(oe) seen P(A ∪ B) = P(A) + P(B) – P(A ∩ B) and P(A ∩ B) = P(A) × P(B) oe 7/12 oe (or awrt 0.58) OR Find that there are 21 outcomes in A ∪ B M1 Find that there are 36 outcomes in total M1 7/12 oe (or awrt 0.58) A1

B1 M1 A1

(iii) 0 and 5/6 (awrt 0.83) B1




7 (a) (i) Two from: To smooth out/eliminate the variation To look for the trend To find seasonal components To make predictions

B1 B1

(ii) 3 B1

(iii) As n is odd (must ft their n) Moving average values correspond to original data points (must ft their n) No (dependent on one M)

M1 M1

A1

(b) (i) a = 75 b = 70.75 c = 82

B1 B1 B1

(ii) 62 – 71.5 (= –9.5) or 58 – 67.75 (= –9.75) (ignore sign errors) (sum of two differences)/2 –9.6 (accept –9.63 or –9.625 or –9600 tonnes etc.)

M1 M1 A1

(iii) Correctly plotted points Suitable trend line

B1 B1

(iv) Attempt at a reading from trend line (even if in wrong place) + ‘their (ii)’ Ans in range 53.9 to 55 (or 53 900 to 55 000) www

M1 A1

8 (i) Electricity = 0.09 × 5000 or 450 Wages = 6.5 × 4000 or 26 000 15 600 : 450 : 26 000 is equivalent to given ratio (÷ 50)

M1 M1

A1(AG)

(ii) 100s in first column Ingredients for 2012: 108 Electricity for 2012: 0.11/0.09 × 100 or 0.02/0.09 × 100 122 (allow 122.2 or 122.2…) Wages for 2012: 97

B1 B1 M1 A1 B1

(iii) (312 × ‘108’ + 9 × ‘122’ + 520 × ‘97’) Sum of 3 products / (312 + 9 + 520) 101.3 (or 101.4 from 122.2…) (must be 1 dp)

M1 M1

A1

(iv) (15 600 + ‘450’ + ‘26 000’) × (‘101.3’ / 100) 42 600 (must be 3sf)

M1* M1dep

A1

(v) Two from: Amount of electricity used may have changed Number of staff/hours may have changed Amount of ingredients may have changed Weights/quantities may have changed There may be other expenses/an additional category is suggested

B1 B1




9 (i) (a) 1/4 or (1/2 × 1/2) seen 1/4 × 1/4 × 1/4 or (1/2 × 1/2) x (1/2 × 1/2) x (1/2 × 1/2) = 1/64 (working essential)

B1 B1(AG)

(b) Evidence that this can happen in three ways: 110, 101, 011 or 3 × 1/4 × 1/4 × 3/4 9/64 (accept 0.14 here) OR 9 ways listed HH HH HT, HH HH TH, HH HH TT,HH HT HH, HH TH HH, HH TT HH, HT HH HH, TH HH HH, TT HH HH or 9 × M1* 1/4 × 1/4 × 1/4 M1dep*

M1 M1 A1

(ii) P(0 points) = 3/4 × 3/4 × 3/4 P(1 point) = 3 × 1/4 × 3/4 × 3/4 one correct method 27/64 and 27/64 both correct (accept 0.42 here) Table with X = 0, 1, 2, 3 Their probabilities sum to 1

M1 A1 B1 B1

(iii) 58 × ‘1/64’ + x × ‘their (i)(b)’ = 4 22 OR if profit used i.e. $4 subtracted from winnings then 54 × ‘1/64’ + –4 × ‘27/64’ + –4 × ‘27/64’ + (x – 4) × ‘9/64’ = 0 M2 OR 54 × ‘1/64’ + –4 × ‘27/64’ + –4 × ‘27/64’ + y × ‘9/64’ = 0 M1* y + 4 M1dep

M1* M1dep

A1

(iv) 1/5 (oe) seen 50 × ‘their 1/5’ + 12.5 × (1 – ‘their 1/5’) $20 Correct decision based on ‘their (iii)’ and ‘their $20’ OR 1/5 (oe) seen B1 (50 – ‘22’) × ‘1/5’ + (12.5 – ‘22’) × (1 – ‘1/5’) M1 – $2 A1 Correct decision based on –ve or +ve result A1

B1 M1 A1

A1




10 (a) (i) (0 × 13) + 1 × 11 + 2 × 7 + 3 × 6 + 4 × 4 + 5 × 1 64

M1 A1

(ii) Total days = 13 + 11 + 7 + 6 + 4 + 1 (=42) ‘64’/’42’ 1.5 (allow 1.52 or 1.524)

M1* M1dep

A1

(b) 10 3 5.5 1.5 58 15 11 3 (B4 for 8 correct, B3 for 6/7 correct, B2 for 4/5 correct, B1 for 2/3 correct)

B4

(c) (i) Marks in Algebra generally higher (oe) Marks in Algebra generally more varied (oe)

B1 B1

(ii) Better in Geometry together with a comparison of her mark with the class mean in terms of the class standard deviation for at least one of Algebra or Geometry. Correct comparison for both e.g. 1 standard deviation above the mean in Algebra and 2 standard deviations above the mean in Geometry (may be seen as a calculation of standardised scores)

M1

A1

(iii) (±) 10

5587 −

or (±) σ

60100 −

10

5587 −

= σ

60100 −

12.5=σ

M1

M1

A1




11 (i) 47 00 51 32 85 11 67 05 10 (–1 each independent error)

B2

(ii) 01 followed by numbers at equal intervals (not nec ints of 10) 11 21 31 41 51 61 71 81 Intervals of 10 (even if insufficient values or values out of range) 9 values at intervals of 10 all in range (wrap around if necessary)

B1

B1 B1

(iii) Attempt at machine totals (20, 30, 40) 2 3 4

M1

A1

(iv) Asad’s sample over represents A (or under represents B or C) Or Asad’s sample does not accurately represent the jars as he has 4 from machine A (or 2 from machine B or 3 from machine C) Omar’s sample accurately represents jars filled by each machine

B1

B1

(v) 44 03 59 14 27 20 78 60 81 (–1 each independent error)

B3

(vi) 39/10 or 51/10 4 and 5

M1 A1

(vii) Because the mass of jam (in each jar) is being checked A sample stratified by machine is more appropriate

M1 A1





4040 STATISTICS








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




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




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




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




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

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