General Certificate of Education January 2009 Advanced Subsidiary Examination
MATHEMATICS Unit Statistics 1B
MS/SS1B
STATISTICS Unit Statistics 1B Friday 9 January 2009
9.00 am to 10.30 am
For this paper you must have: * an 8-page answer book * the blue AQA booklet of formulae and statistical tables * an insert for use in Question 6 (enclosed). You may use a graphics calculator.
Time allowed: 1 hour 30 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MS/SS1B. Answer all questions. Show all necessary working; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Fill in the boxes at the top of the insert. * *
* * *
*
Information The maximum mark for this paper is 75. The marks for questions are shown in brackets. Unit Statistics 1B has a written paper only. * * *
Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet.
*
P10449/Jan09/MS/SS1B 6/6/6/
MS/SS1B
2
Answer all questions.
1 Ms N Parker always reads the columns of announcements in her local weekly newspaper. During each week of 2008, she notes the number of births announced. Her results are summarised in the table. Number of births
1
2
3
4
5
6
7
8
Number of weeks
1
2
9
13
7
13
6
1
(a) Calculate the mean, median and modes of these data.
(5 marks)
(b) State, with a reason, which of the three measures of average in part (a) you consider to be the least appropriate for summarising the number of births. (2 marks)
2 A greengrocer sells bunches of 9 carrots at his Saturday market stall. Tom and Geri are two Statistics students who work on the stall. Each selects a bunch of carrots at random. (a) At home, Tom measures the length, x centimetres, and the maximum diameter, y centimetres, of each carrot in his selected bunch with the following results. x
16.2
13.1
10.4
12.1
14.6
9.7
11.8
13.6
17.3
y
4.2
3.9
4.7
3.3
3.7
2.4
3.1
3.5
2.7
(i) Calculate the value of the product moment correlation coefficient.
(3 marks)
(ii) Interpret your value in context.
(2 marks)
(b) At her home, Geri measures the length, in centimetres, and the weight, in grams, of each carrot in her selected bunch and then obtains a value of 0.986 for the product moment correlation coefficient. Comment, with a reason, on the likely validity of Geri’s value.
P10449/Jan09/MS/SS1B
(2 marks)
3
3 UPVC facia board is supplied in lengths labelled as 5 metres. The actual length, X metres, of a board may be modelled by a normal distribution with a mean of 5.08 and a standard deviation of 0.05 . (a) Determine: (i) PðX < 5Þ ;
(3 marks)
(ii) Pð5 < X < 5:10Þ .
(2 marks)
(b) Determine the probability that the mean length of a random sample of 4 boards: (i) exceeds 5.05 metres; (ii) is exactly 5 metres.
(4 marks) (1 mark)
(c) Assuming that the value of the standard deviation remains unchanged, determine the mean length necessary to ensure that only 1 per cent of boards have lengths less than 5 metres. (4 marks)
4 Gary and his neighbour Larry work at the same place. On any day when Gary travels to work, he uses one of three options: his car only, a bus only or both his car and a bus. The probability that he uses his car, either on its own or with a bus, is 0.6 . The probability that he uses both his car and a bus is 0.25 . (a) Calculate the probability that, on any particular day when Gary travels to work, he: (i) does not use his car;
(1 mark)
(ii) uses his car only;
(2 marks)
(iii) uses a bus.
(3 marks)
(b) On any day, the probability that Larry travels to work with Gary is 0.9 when Gary uses his car only, is 0.7 when Gary uses both his car and a bus, and is 0.3 when Gary uses a bus only. (i) Calculate the probability that, on any particular day when Gary travels to work, Larry travels with him. (4 marks) (ii) Assuming that option choices are independent from day to day, calculate, to three decimal places, the probability that, during any particular week (5 days) when Gary travels to work every day, Larry never travels with him. (2 marks)
P10449/Jan09/MS/SS1B
s
Turn over
4
5 The times taken by new recruits to complete an assault course may be modelled by a normal distribution with a standard deviation of 8 minutes. A group of 30 new recruits takes a total time of 1620 minutes to complete the course. (a) Calculate the mean time taken by these 30 new recruits.
(1 mark)
(b) Assuming that the 30 recruits may be considered to be a random sample, construct a 98% confidence interval for the mean time taken by new recruits to complete the course. (4 marks) (c) Construct an interval within which approximately 98% of the times taken by individual new recruits to complete the course will lie. (2 marks) (d) State where, if at all, in this question you made use of the Central Limit Theorem. (1 mark)
6 [Figure 1, printed on the insert, is provided for use in this question.] For a random sample of 10 patients who underwent hip-replacement operations, records were kept of their ages, x years, and of the number of days, y, following their operations before they were able to walk unaided safely. Patient
A
B
C
D
E
F
G
H
I
J
x
55
51
62
66
72
59
78
55
62
70
y
34
33
39
49
48
43
51
41
46
51
(a) On Figure 1, complete the scatter diagram for these data.
(2 marks)
(b) Calculate the equation of the least squares regression line of y on x.
(4 marks)
(c) Draw your regression line on Figure 1.
(2 marks)
(d) In fact, patients H, I and J were males and the other 7 patients were females. (i) Calculate the mean of the residuals for the 3 male patients.
(4 marks)
(ii) Hence estimate, for a male patient aged 65 years, the number of days following his hip-replacement operation before he is able to walk unaided safely. (3 marks)
P10449/Jan09/MS/SS1B
5
7 The proportion of passengers who use senior citizen bus passes to travel into a particular town on ‘Park & Ride’ buses between 9.30 am and 11.30 am on weekdays is 0.45 . It is proposed that, when there are n passengers on a bus, a suitable model for the number of passengers using senior citizen bus passes is the distribution Bðn, 0:45Þ . (a) Assuming that this model applies to the 10.30 am weekday ‘Park & Ride’ bus into the town: (i) calculate the probability that, when there are 16 passengers, exactly 3 of them are using senior citizen bus passes; (3 marks) (ii) determine the probability that, when there are 25 passengers, fewer than 10 of them are using senior citizen bus passes; (2 marks) (iii) determine the probability that, when there are 40 passengers, at least 15 but at most 20 of them are using senior citizen bus passes; (3 marks) (iv) calculate the mean and the variance for the number of passengers using senior citizen bus passes when there are 50 passengers. (2 marks) (b)
(i) Give a reason why the proposed model may not be suitable.
(1 mark)
(ii) Give a different reason why the proposed model would not be suitable for the number of passengers using senior citizen bus passes to travel into the town on the 7.15 am weekday ‘Park & Ride’ bus. (1 mark)
END OF QUESTIONS
P10449/Jan09/MS/SS1B
Surname
Other Names
Centre Number
Candidate Number
Candidate Signature
General Certificate of Education January 2009 Advanced Subsidiary Examination
MATHEMATICS Unit Statistics 1B
MS/SS1B
STATISTICS Unit Statistics 1B
Insert Insert for use in Question 6. Fill in the boxes at the top of this page. Fasten this insert securely to your answer book.
Turn over for Figure 1
P10449/Jan09/MS/SS1B 6/6/6/
s
Turn over
2
Figure 1 (for use in Question 6)
Hip-Replacement Operations
y
~
60 –
55 – G
50 –
D
E
45 – F
Days 40 –
C
35 –
A
B
30 –
–
–
–
–
–
–
50
55
60
65
70
75
80
Age (years)
Copyright Ó 2009 AQA and its licensors. All rights reserved.
P10449/Jan09/MS/SS1B
~
–
–
0– 0
x
Version 1.0: 0109
hij General Certificate of Education
Mathematics 6360 Statistics 6380 MS/SS1B Statistics 1B
Mark Scheme 2009 examination - January series
Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation meeting attended by all examiners and is the scheme which was used by them in this examination. The standardisation meeting ensures that the mark scheme covers the candidates’ responses to questions and that every examiner understands and applies it in the same correct way. As preparation for the standardisation meeting each examiner analyses a number of candidates’ scripts: alternative answers not already covered by the mark scheme are discussed at the meeting and legislated for. If, after this meeting, examiners encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.
Further copies of this Mark Scheme are available to download from the AQA Website: www.aqa.org.uk Copyright © 2009 AQA and its licensors. All rights reserved. COPYRIGHT AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance.
The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester M15 6EX Dr Michael Cresswell Director General
MS.SS1B - AQA GCE Mark Scheme 2009 January series
Key to mark scheme and abbreviations used in marking M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA
mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach
MC MR RA FW ISW FIW BOD WR FB NOS G c sf dp
mis-copy mis-read required accuracy further work ignore subsequent work from incorrect work given benefit of doubt work replaced by candidate formulae book not on scheme graph candidate significant figure(s) decimal place(s)
No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.
3
MS.SS1B - AQA GCE Mark Scheme 2009 January series
MS/SS1B Q 1 (a)
Solution
Mean =
∑ fx ∑x
=
247 = 4.75 or 4¾ 52
If B0 but evidence of
∑ fx 52
Median (26, 26½) = 5 If B0 but evidence of cumulative frequencies F: (0) 1 3 12 25 32 45 51 52 or If data assumed continuous so use of
4+
Marks
Total
247 ⇒ B1 52 CAO (4.75 = 5 ⇒ ISW) 39 ⇒ B2 4 52
B2
(M1) CAO
B2 (B1)
Stated identification of 26 or 26½ Need to see attempt at ≥ 4 F–values
(M1) (4 < median < 4.29)
x where 0 < x < 2 7
Mode(s) = 4 and 6 (b) Mode(s) More than one mode/value Two modes/values No unique mode/value Notes: If data treated as two separate sets, then only marks available are B1 B1dep in (b) If averages confused then mark (a) as stated eg median = 4 and 6 ⇒ B0 in (a) and in (b) “median, as two values” ⇒ B0 B0
Comments
B1
5
B1
CAO both (so mode = 5 ⇒ B0) CAO Or equivalent; eg not unique
B1dep Dep only on previous B1 scored Modes = 1 and 13 ⇒ B0 in (a) but B1 B1dep available in (b)
Total
4
2 7
MS.SS1B - AQA GCE Mark Scheme 2009 January series
MS/SS1B (cont) Q 2 (a)(i)
Solution r = 0.022 to 0.023 r = 0.02 to 0.03 r = –0.1
OR Attempt at
Marks B3 (B2)
to 0.1
∑ x ∑ x2 ∑ y ∑ y 2 ∑ xy
Total
AWFW AWFW AWFW
(B1) &
or
118.8 1619.36 31.5 114.43 & 416.13 (all 5 attempted) (M1) 51.2 4.18 & 0.33 (all 3 attempted)
Attempt at S xx S yy & S xy Attempt at correct formula for r r = 0.022 to 0.023
(m1) (A1)
3
AWFW Or equivalent qualification of NO strength; do not follow-through from (i) B0 for very weak/weak/some/ little/slight/positive/hardly any/etc unless correct qualification also stated
2
Context; providing –1 < r < 1
(ii)
(Almost/virtually) no/zero (linear) correlation (relationship/association/link) between length and (maximum) diameter of carrots (b) Unlikely/wrong/incorrect/invalid
Would expect a positive value or Would expect weight to increase with length or Would imply shorter carrots are heavier
Comments (0.022557)
B1
B1 B1
Or equivalent
B1
Or equivalent reason
Total
5
2 7
MS.SS1B - AQA GCE Mark Scheme 2009 January series
MS/SS1B (cont) Q Solution 3 X ~ N(5.08, 0.052) (a)(i) 5 − 5.08 ⎞ ⎛ P ( X < 5) = P ⎜ Z < ⎟ = 0.05 ⎠ ⎝ P ( Z < −1.6 ) = 1 – P ( Z < 1.6 ) = 1 – 0.9452 (ii)
Marks
Total
Standardising (4.5, 4.95, 5, 5.05 or 5.5) with 5.08 and ( 0.05 , 0.05 or 0.052) and/or (5.08 – x)
M1 m1
= 0.0545 to 0.055
A1
P ( 5 < X < 5.10 ) = P ( X < 5.10 ) – (i)
M1
3
= P ( Z < 0.4 ) – (i) = 0.65542 – 0.0548 = 0.6 to 0.601 (b)(i) Variance of X 4 = 0.052/4 = 0.000625
A1
Comments
2
Area change; may be implied AWFW (0.0548) (1 – answer) ⇒ M1 max Or equivalent; must be clear correct method if answer incorrect and answer > 0
AWFW
(0.60062)
B1
CAO; stated or used
5.05 − 5.08 ⎞ ⎛ P ( X 4 > 5.05 ) = P ⎜ Z > ⎟ 0.025 ⎠ ⎝
M1
Standardising 5.05 with 5.08 and 0.025; allow (5.08 – 5.05)
= P ( Z > −1.2 ) = P ( Z < 1.2 )
m1
SD of X 4 = 0.05/2 = 0.025
= 0.884 to 0.886
A1
4
(ii) Zero
B1
1
(c) 1% (0.01) ⇒ z = –2.33 to –2.32
B1
z=
5−μ 0.05 = –2.3263
μ = 5.11 to 5.12
M1 A1
Area change; may be implied AWFW (0.88493) (1 – answer) ⇒ B1 M1 max CAO; or equivalent (ignore any working) AWFW; ignore sign
(–2.3263)
Standardising 5 with 0.025; allow (μ – 5) Only allow: ±2.05 ±2.32 ±2.57
μ and 0.05 or
AWFW
A1
Note: 5−μ = 2.3263 ⇒ 5.116 0.05 ⇒ B1 M1 A1 A0
to ±2.06 to ±2.33 to ±2.58 (5.1163)
Or equivalent inconsistent signs
Total
6
4 14
MS.SS1B - AQA GCE Mark Scheme 2009 January series
MS/SS1B (cont) Q Solution 4 P(C) = 0.6 P(C ∩ B) = 0.25
Marks
Total
P(C′) = 1 – P(C) = 1 – 0.6 = 0.4
B1
1
(ii) P(C ∩ B′) = 0.6 – 0.25 = 1 – (0.4 + 0.25) = 0.35
M1
(iii) P(B) = (i) + p = (i) + 0.25
M1 A1 A1
Can be implied by correct answer Can be implied by correct answer CAO; or equivalent
(M2) (A1)
Can be implied by correct answer
{P(C only) = 0.35 (a) (i)
P(B only) = 0.4}
with p < 0.6 = 0.65
OR P(B) = 1 – (ii)
= 0.65 OR 1 = P(C) + P(B) – P(C ∩ B) Thus P(B) = 1 – (0.6 – 0.25) = 0.65 (b)
A1
(M1) (A1) (A1)
Comments In (a), ratios (eg 4 : 10) are only penalised by 1 mark at first correct answer
CAO; or equivalent Can be implied by correct answer
2
3
CAO; or equivalent
Can be implied by correct answer Can be implied by correct answer CAO; or equivalent
P(L | GC) = 0.9 P(L | GCB) = 0.7 P(L | GB) = 0.3 (0.315)
M1
0.25 × 0.7
(0.175)
M1
[(a)(iii) – 0.25] × 0.3
(0.12)
M1
(i) P(G ∩ L) ⇒ (a)(ii) × 0.9
Follow through or correct
Follow through or correct Ignore any multiplying factors Ignore any additional terms
Note: Each pair of multiplied probabilities must be > 0 to score the corresponding method mark ⇒ 0.315 + 0.175 + 0.12 = 0.61 (ii) Probability = {1 – (b)(i)}5
= 0.395 = 0.009
A1
4
Allow 5 × {1 – (b)(i)}5
M1 A1 Total
7
CAO
2 12
AWRT
(0.00902)
MS.SS1B - AQA GCE Mark Scheme 2009 January series
MS/SS1B (cont) Q 5 (a)
(b)
Solution
Marks
Total
1620 = 54 30
B1
1
98% (0.98) ⇒ z = 2.32 to 2.33
B1
Mean =
CI for μ is Thus
Hence or
x ± z×
σ
n 8 54 ± 2.3263 × 30 54 ± (3.38 to 3.42)
CAO; cannot be gained in (b) AWFW Used Must have
M1
(2.3263) n with n > 1
F on x (but not 1620) and z only Allow x = 54 even if B0 in (a) CAO & AWFW (54 & 3.4)
A1F A1
Comments
4
(50.58 to 50.62, 57.38 to 57.42)
AWFW
(50.6, 57.4)
Notes: Use of n = 1 in (b) must not be deemed as answer to (c) Use of n = 1 in (b) followed by use of n = 1 in (c) ⇒ (b) B1, (c) M1 A1 max Use of n = 1 with (b) or (c) not identified ⇒ (b) B1, (c) 0 max (c) Repeat of structure in (b) but with n = 1 and 1.96 ≤ z ≤ 3.03
M1
Or equivalent CAO & AWFW
54 ± (18.56 to 18.64)
Thus
If z-value incorrect, then must use 54 ± 8 × ⎡⎣ z from ( b ) ⎤⎦
A1F
or
(35.36 to 35.44, 72.56 to 72.64)
(54 & 18.6)
2
AWFW
B1
1
CAO; or equivalent (ignore any reasoning)
Total
8
(35.4, 72.6)
Note: Accept sensible non-symmetric intervals such as: (0, 54 + 2.0537 × 8) = (0, 70.4 to 70.5) (d)
Nowhere or
No
8
MS.SS1B - AQA GCE Mark Scheme 2009 January series
MS/SS1B (cont) Q 6(a) Figure 1: (b)
Solution 3 correct labelled points 2 correct labelled points
b (gradient) = 0.685 b (gradient) = 0.68 to 0.69 a (intercept) = 0.344 a (intercept) = 0.34 to 0.35
Marks B2 (B1)
Total
2
B2 (B1) B2 (B1)
Comments
Deduct 1 mark if not labelled
AWRT
(0.68502)
AWFW AWRT AWFW
(0.34404)
OR
Attempt at
∑ x ∑ x2 ∑ y
&
630 40344 435 & 27853 (all 4 attempted)
∑ xy
or Attempt at S xx & S xy
(M1)
Attempt at correct formula for b (gradient) b (gradient) = 0.685 a (intercept) = 0.344 Accept a & b interchanged only if then identified correctly by a stated or used equation in (c) or (d)
(m1) (A1) (A1)
654 & 448 (both attempted)
(c) Figure 1: Correct line (50, 34 to 35) (60, 40½ to 42) (70, 47¼ to 49) (80, 54 to 56) If B0 but evidence of use of line for ≥ 2 points within range 50 ≤ x ≤ 80 (d)(i) Residual = y – (a + bx) [or (a + bx) – y]
H I 2.5 to 4(.0) 2.5 to 4(.0)
J 2(.0) to 4(.0)
Mean = 2.3 to 4(.0) (ii) or
y65 = a + b × 65 y65 = 44 to 45.5 + [(d)(i)] or [2.95 to 2.97]
= 46 to 50 Special Cases: Line drawn/calcd on H, I & J or linear interpn using I & J = 47 to 49
AWRT AWRT 4 Dep on ≥ B1 B1 or ≥A1 A0 in (b) At least from x ≈ 55 to 70 Any two
B2dep
(M1)
2
Used or implied; or equivalent
M1
(using graph); ≥ 1 residual correct (2.98) AWFW; ignore signs only (3.19) providing all the same (2.70)
A2,1 (–1 EE)
A1dep
4
A1
yM = 4.51 + 0.666 x ⇒ 47.8 OR no evidence of method {from (d)(i) and/or (d)(ii)} Evidence of incorrect method ⇒ B0
(B2)
9
(44.9)
Use shown or AWFW; ignore sign of mean residual AWFW (47.8)
m1
Total
AWFW; do not ignore sign (2.96) Dep on previous A2 scored Use shown or AWFW
M1
44 to 45.5 seen with no evidence ⇒ B1
Calcn or points shown on graph
3 15
MS.SS1B - AQA GCE Mark Scheme 2009 January series
MS/SS1B (cont) Q 7 (a)(i)
Solution B(16 or 25 or 40, 0.45) ⎛16 ⎞ 3 13 P(S = 3) = ⎜ ⎟ ( 0.45 ) ( 0.55 ) 3 ⎝ ⎠ = 0.021 to 0.022
(ii)
P(S < 10) = 0.3843 or 0.2424 = 0.242 to 0.243
(iii) P(15 ≤ S ≤ 20) = 0.7870 or 0.6844
Marks M1
Total
May be implied by correct answer Ignore any additional terms
A1 A1
3
B1 B1
2
OR B(40, 0.45) expressions stated for at least 3 terms within 14 ≤ S ≤ 20 gives probability = 0.654 to 0.655 (iv) Mean, μ = np = 50 × 0.45 = 22.5 or 22½
Variance, σ 2 = np(1 – p) = 50 × 0.45 × 0.55 = 12.3 to 12.4 (b)(i) Non-independence of senior citizens travel Senior citizens tend to travel in pairs/groups (ii) 7.15 am is outside 9.30 am to 11.30 am Cannot use SCPs before 9.30 am Cannot use SCPs @ 7.15 am Cannot use SCPs during morning ‘rush hour’ Value of p likely to be smaller/different/zero Data not available Senior citizens not out at this time Passengers likely to be workers/school children
AWFW (0.0215) Accept 3 dp accuracy from tables or calculation AWFW (0.2424)
M1
Accept 3 dp accuracy
M1 A1
p2 – p1 ⇒ M0 M0 A0 p1 – (1 – p2) ⇒ M1 M0 A0 Accept 3 dp accuracy / truncation AWFW (0.6544)
(p1)
minus 0.1326 or 0.2142 (p2) = 0.654 to 0.655
Comments Used at least once in (a)(i) to (iii)
(M1) (A2)
Or implied by a correct answer 3
AWFW CAO (22.5 = 22 or 23 ⇒ ISW)
B1
Accept 12 3 8 or
99 8
B1
2
AWFW
(12.375)
B1
1
Or equivalent; but must be a clear indication of non-independent events
Or equivalent Accept other sensible reasons
B1
Distribution of types of passenger different Total Paper
10
1 12 75
klm
Scaled mark component grade boundaries - January 2009 exams GCE (Legacy) Component Code Component Title LAW5 MD01 MD02 MFP1 MFP2 MFP3 MFP4 MM1A/C MM1A/W MM1B MM2B MPC1 MPC2 MPC3 MPC4 MS/SS1A/W MS/SS1A/C MS1B MS2B XMCAS XMCA2 MED1 MED2 MED3 MED4
Maximum Scaled Mark
Scaled Mark Grade Boundaries B C D
A
GCE LAW UNIT 5
85
57
53
49
45
41
GCE MATHEMATICS UNIT D01 GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP1 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M1A - COURSEWORK GCE MATHEMATICS UNIT M1A - WRITTEN GCE MATHEMATICS UNIT M1B GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC1 GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT PC4 GCE MATHS/STATISTICS UNIT 1A - WRITTEN GCE MATHS/STATISTICS UNIT 1A - COURSEWORK GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT S2B GCE MATHEMATICS UNIT XMCAS GCE MATHEMATICS UNIT XMCA2
75 75 75 75 75 75 25 75 75 75 75 75 75 75 75 25 75 75 125 125
60 61 58 60 59 56 20 64 64 61 62 65 59 60 60 20 59 56 101 100
52 53 51 52 51 49 17 56 56 53 54 57 51 52 52 17 51 49 89 88
45 46 44 44 44 42 14 48 49 45 46 49 43 44 44 14 43 42 77 76
38 39 37 36 37 35 12 40 42 37 39 41 36 36 37 12 35 36 66 64
31 32 31 29 30 28 10 33 35 30 32 34 29 29 30 10 28 30 55 53
GCE MEDIA STUDIES UNIT 1 GCE MEDIA STUDIES UNIT 2 GCE MEDIA STUDIES UNIT 3 GCE MEDIA STUDIES UNIT 4
60 60 100 60
38 41 72 42
33 35 64 38
28 30 56 34
23 25 48 30
19 20 40 26
E
