General Certificate of Education Advanced Subsidiary Examination January 2011
Mathematics
MS/SS1B
Unit Statistics 1B
Statistics Unit Statistics 1B Friday 14 January 2011
d
1.30 pm to 3.00 pm
e
For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
s
n
Time allowed * 1 hour 30 minutes
e
Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Fill in the boxes at the top of this page. * Answer all questions. * Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. * You must answer the questions in the spaces provided. Do not write outside the box around each page. * Show all necessary working; otherwise marks for method may be lost. * Do all rough work in this book. Cross through any work that you do not want to be marked. * The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures.
d
n
o
C
Information * The marks for questions are shown in brackets. * The maximum mark for this paper is 75. * Unit Statistics 1B has a written paper only. Advice * Unless stated otherwise, you may quote formulae, without proof, from the booklet.
P33232/Jan11/MS/SS1B 6/6/
MS/SS1B
2
Estimate, without undertaking any calculations, the value of the product moment correlation coefficient between the variables x and y for each of the two scatter diagrams.
1 (a)
(i)
(ii)
y
y
x
x (2 marks)
The table gives the circumference, x centimetres, and the weight, y grams, of each of 12 new cricket balls.
(b)
x
22.5
y
160.3 159.4 157.8 158.0 157.3 159.8 158.3 159.6 161.3 156.4 162.5 161.2 (i)
22.7
22.6
22.4
22.5
22.8
22.6
22.7
22.8
22.4
22.9
22.6
Calculate the value of the product moment correlation coefficient between x and y. (3 marks)
(ii) Assuming that the 12 balls may be considered to be a random sample, interpret your
value in context.
(2 marks)
P33232/Jan11/MS/SS1B
3
The number of MPs in the House of Commons was 645 at the beginning of August 2009. The genders of these MPs and the political parties to which they belonged are shown in the table.
2
Political Party
Gender
Labour
Conservative
Liberal Democrat
Other
Total
Male
255
175
54
35
519
Female
94
18
9
5
126
Total
349
193
63
40
645
One MP was selected at random for an interview. Calculate, to three decimal places, the probability that the MP was:
(a)
(i)
a male Conservative;
(1 mark)
(ii) a male;
(1 mark)
(iii) a Liberal Democrat;
(1 mark)
(iv) Labour, given that the MP was female;
(2 marks)
(v) male, given that the MP was not Labour.
(3 marks)
(b)
Two female MPs were selected at random for an enquiry. Calculate, to three decimal places, the probability that both MPs were Labour. (2 marks)
(c)
Three MPs were selected at random for a committee. Calculate, to three decimal places, the probability that exactly one MP was Labour and exactly one MP was Conservative. (4 marks)
s
Turn over
P33232/Jan11/MS/SS1B
4
The volume, X litres, of orange juice in a 1-litre carton may be modelled by a normal distribution with unknown mean m .
3
The volumes, x litres, recorded to the nearest 0.01 litre, in a random sample of 100 cartons are shown in the table. Volume (x litres)
Number of cartons ( f )
0.95 – 0.97
2
0.98 – 1.00
7
1.01 – 1.03
15
1.04 – 1.06
32
1.07 – 1.09
22
1.10 – 1.12
14
1.13 – 1.15
7
1.16 – 1.18
1
Total
100
For the group ‘0.98 – 1.00’:
(a) (i)
show that it has a mid-point of 0.99 litres;
(ii) state the minimum and the maximum values of x that could be included in this group.
(2 marks) (b)
Calculate, to three decimal places, estimates of the mean and the standard deviation of these 100 volumes. (3 marks)
(c) (i)
Construct an approximate 99% confidence interval for m .
(4 marks)
(ii) State why use of the Central Limit Theorem was not required when calculating this
confidence interval.
(1 mark)
(iii) Give a reason why the confidence interval is approximate rather than exact.
(1 mark) Give a reason in support of the claim that:
(d) (i)
m > 1;
(ii) Pð0:94 < X < 1:16Þ is approximately 1 .
(2 marks)
P33232/Jan11/MS/SS1B
5
Clay pigeon shooting is the sport of shooting at special flying clay targets with a shotgun.
4
Rhys, a novice, uses a single-barrelled shotgun. The probability that he hits a target is 0.45 , and may be assumed to be independent from target to target.
(a)
Determine the probability that, in a series of shots at 15 targets, he hits: at most 5 targets;
(1 mark)
(ii) more than 10 targets;
(2 marks)
(iii) exactly 6 targets;
(2 marks)
(iv) at least 5 but at most 10 targets.
(3 marks)
(i)
Sasha, an expert, uses a double-barrelled shotgun. She shoots at each target with the gun’s first barrel and, only if she misses, does she then shoot at the target with the gun’s second barrel.
(b)
The probability that she hits a target with a shot using her gun’s first barrel is 0.85 . The conditional probability that she hits a target with a shot using her gun’s second barrel, given that she has missed the target with a shot using her gun’s first barrel, is 0.80 . Assume that Sasha’s shooting is independent from target to target. (i)
Show that the probability that Sasha hits a target is 0.97 .
(2 marks)
(ii) Determine the probability that, in a series of shots at 50 targets, Sasha hits at least
48 targets.
(3 marks)
(iii) In a series of shots at 80 targets, calculate the mean number of times that Sasha
shoots at targets with her gun’s second barrel.
(2 marks)
s
Turn over
P33232/Jan11/MS/SS1B
6
Craig uses his car to travel regularly from his home to the area hospital for treatment. He leaves home at x minutes after 7.30 am and then takes y minutes to arrive at the hospital’s reception desk.
5
His results for 11 mornings are shown in the table. x
0
5
10
15
20
25
30
35
40
45
50
y
31
42
32
58
47
56
79
68
89
95
85
(a)
Explain why the time taken by Craig between leaving home and arriving at the hospital’s reception desk is the response variable. (1 mark)
(b)
Calculate the equation of the least squares regression line of y on x, writing your answer in the form y ¼ a þ bx . (5 marks)
(c)
On a particular day, Craig needs to arrive at the hospital’s reception desk no later than 9.00 am. He leaves home at 7.45 am. Estimate the number of minutes before 9.00 am that Craig will arrive at the hospital’s reception desk. Give your answer to the nearest minute. (5 marks)
(d) (i)
Use your equation to estimate y when x ¼ 85 .
(1 mark)
(ii) Give one statistical reason and one reason based on the context of this question as to (2 marks) why your estimate in part (d)(i) is unlikely to be realistic.
The volume of shampoo, V millilitres, delivered by a machine into bottles may be modelled by a normal random variable with mean m and standard deviation s .
6
Given that m ¼ 412 and s ¼ 8 , determine:
(a)
PðV < 400Þ ;
(3 marks)
(ii) PðV > 420Þ ;
(2 marks)
(iii) PðV ¼ 410Þ .
(1 mark)
(i)
A new quality control specification requires that the values of m and s are changed so that
(b)
PðV < 400Þ ¼ 0:05 and (i)
PðV > 420Þ ¼ 0:01
Show, with the aid of a suitable sketch, or otherwise, that 400 m ¼ 1:6449s
(ii) Hence calculate values for m and s .
and
420 m ¼ 2:3263s
(3 marks) (3 marks)
Copyright ª 2011 AQA and its licensors. All rights reserved. P33232/Jan11/MS/SS1B
Version 1.0
General Certificate of Education (A-level) January 2011
Mathematics
MS/SS1B
(Specification 6360) Statistics 1B
Mark Scheme
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 events which all examiners participate in and is the scheme which was used by them in this examination. The standardisation process 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 standardisation each examiner analyses a number of candidates’ scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, examiners encounter unusual answers which have not been raised 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 from: aqa.org.uk Copyright © 2011 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.
Mark Scheme – General Certificate of Education (A-level) Mathematics – Statistics 1B – January 2011
Key to mark scheme abbreviations 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 c sf dp
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 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. 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
Mark Scheme – General Certificate of Education (A-level) Mathematics – Statistics 1B – January 2011
MS/SS1B Q 1(a)(i)
(ii)
Solution
Marks
Total
AWFW ( ≈ 0.8) If answers are not labelled, assume order is (a)(i) then (a)(ii)
r = 0.6 to 0.98
B1
r = −0.5 to − 0.02
B1
2
AWFW ( ≈ −0.3) Eg: (a)(i) 0.7 to 0.9 ⇒ B1 (a)(ii) −0.6 to − 0.4 ⇒ B0
r = 0.757 r = 0.75 to 0.77 r = 0.65 to 0.85 or
B3 (B2) (B1)
3
AWRT (0.75708) AWFW AWFW
Attempt at ∑ x ∑ x 2 ∑ y ∑ y 2 and ∑ xy
(M1)
Accept answers as ranges if and only if contained entirely within given ranges (b)(i)
Comments
271.5 6142.97 1911.9 304650.01 and 43259.17 (all 5 attempted)
or
0.2825 36.5425 and 2.4325 (all 3 attempted)
Attempt at S xx S yy and S xy Attempt at substitution into correct corresponding formula for r r = 0.757
(m1) (A1)
(ii) Strong/fairly strong/moderate positive (linear) correlation/relationship/ association/link (but not ‘trend’)
AWRT Dependent on 0.65 < r < 0.85 Or equivalent; must qualify strength and indicate positive Bdep0 for very strong/high/average/ medium/some etc.
Bdep1
between Circumference/size and weight of (cricket) balls
B1 Total
2 7
4
Context; providing 0 < r < 1
Mark Scheme – General Certificate of Education (A-level) Mathematics – Statistics 1B – January 2011
MS/SS1B (cont) Q Solution 2(a)(i)
Marks
Total
175 35 = = 0.271 645 129
B1
1
AWRT; accept either correct fraction
519 173 = = 0.804 to 0.805 645 215
B1
1
AWFW; accept either correct fraction
1
AWFW; accept either correct fraction
P ( M ∩ C) =
(ii)
P(M) =
(iii)
P ( LD ) =
63 21 = = 0.097 to 0.098 645 215
B1
(iv)
P ( L F) =
94 47 = 126 63
M1
= 0.746
(v)
A1
P ( M L') = =
519 − 255 175 + 54 + 35 = 645 − 349 193 + 63 + 40
(c)
Accept 2
94 126 ÷ 645 645
AWRT Allow one arithmetic slip Allow one arithmetic slip
M1 M1
264 132 66 33 = = = 296 148 74 37
Any of these implies M1 M1 A1
= 0.891 to 0.893 (b)
Comments
8742 ⎛ 94 93 ⎞ P ( L ∩ L F) = ⎜ × ⎟ or 15750 126 125 ⎝ ⎠
B1
= 0.555
B1
3
AWFW
4371 1457 ⎛ 47 93 ⎞ Or ⎜ × or ⎟ or 7875 2625 63 125 ⎝ ⎠ 2
AWRT
P ( L ∩ C ∩ ( LD + O ) ) 349 193 63 + 40 × × 645 644 643
M1 M1
Correct numerator Correct denominator
× 6 or 3
M1
⎛ 645 ⎞ Note that a denominator of ⎜ ⎟ ⎝ 3 ⎠ ⇒ M2 (second and third M1 marks)
= 0.155 to 0.157
A1
=
SC The three correct fractions identified but not multiplied ⇒ M1 M0 M0 A0
NB:
4
0.026 with no working ⇒ M1 only 0.026 × 6 = 0.156 with no working ⇒ 4 marks Total
14
5
AWFW
Mark Scheme – General Certificate of Education (A-level) Mathematics – Statistics 1B – January 2011
MS/SS1B (cont) Q Solution 3(a)(i) 0.98 + 1.00 0.975 + 1.005 or or 2 2 0.02 0.03 0.98 + = 0.99 or 0.975 + 2 2 (ii)
0.97 + 0.98 = 0.975 and 2 1.00 + 1.01 = 1.005 2 SC In (a)(i) and (a)(ii) allow 1.0049 or 1.0049… etc
(b) Mean, x = 1.062 Standard deviation, s or σ = 0.043
(c)(i)
99% ( 0.99 ) ⇒ z = 2.57 to 2.58 CI for μ is x ± ( z or t ) × Thus 1.062 ± 2.5758 ×
( s or σ )
Marks
Total
AG (At least) one correct expression seen Ignore contradictions Accept any valid equivalent
B1
B1
2
Hence 1.06 ± 0.01 or (1.05, 1.07) (ii) Volumes/ X / (parent) population may be modelled by a normal distribution / is normally distributed (Ignore contradictions) (iii) Sample data grouped Exact sample values unknown / midpoints used x and s calculated from grouped data (d)(i) CI for μ or CI in (c)(i) > 1 LCL of CI for μ or LCL of CI in (c)(i) > 1 (ii) 99 or 100 or all sample/ table/ data volumes/ values/ x-values/ cartons are within this range (or none/0 or 1 volumes outside) Total
Both CAO Can not be implied from (a)(i) Similar forms for lower boundary
B1 B2
CAO 3
∑ fx = 106.2 Ignore notation
∑ fx 2 = 112.9662 AWRT If B0 B0, M1 can be awarded for attempt ∑ fx at 100
B1 (B1)
AWFW (2.5758) t99 ( 0.995 ) = 2.626 AWRT
M1
Used Must have
A1F
F on x , s / σ and z/t
n
0.043 100 or 99
Comments
A1
B1
n with n > 1
4
AWRT; award even if previous inaccuracies in x , s/ σ or z/t Dependent on A1F
1
Or equivalent; not distribution, data, values (in table), sample, n large, nor simply ‘It is stated in question’
σ unknown B1
1
2 13
6
x (not μ ) and s are estimates NOT data values rounded
Or equivalent; must compare CI to 1 Dependent on CI in (c)(i) > 1
B1
B1
s calculated from a sample
Mark Scheme – General Certificate of Education (A-level) Mathematics – Statistics 1B – January 2011
MS/SS1B (cont) Q Solution 4(a) R ∼ B (15, 0.45 ) (i)
P ( R ≤ 5 ) = 0.26 ( 0 ) to 0.261
(ii)
P ( R > 10 ) = 1 − P ( R ≤ 10 )
(iii)
Total
B1
1
AWFW
2
Requires ‘1 –’ Accept 3dp rounding or truncation Can be implied by 0.025 to 0.026 but not by 0.0769 to 0.077 AWFW (0.0255)
= 1 − ( 0.9745 or 0.9231)
M1
= 0.025 to 0.026
A1
Comments
(0.2608)
P ( R = 6 ) = 0.4522 − (a)(i) ⎛ 15 ⎞ 6 9 or = ⎜ ⎟ ( 0.45 ) ( 0.55 ) ⎝ 6⎠ = 0.191 to 0.192
(iv)
Marks
M1 A1
P ( 5 ≤ R ≤ 10 ) = 0.9745 or 0.9231
( p1 )
Can be implied by a correct answer 2
M1
AWFW
(0.1914)
Accept 3dp rounding or truncation p2 − p1 ⇒ M0 M0 A0
(1 − p2 ) − p1 ⇒ M0 M0 A0 p1 − (1 − p2 ) ⇒ M1 M0 A0 only providing result > 0
Minus 0.1204 or 0.2608 = 0.853 to 0.855
( p2 )
M1 A1
Or B (15, 0.45) terms stated for at least 3 values within 4 ≤ R ≤ 11 gives probability = 0.853 to 0.855 (b)(i)
P ( S ) = 0.85 plus
( 0.15 × 0.80 )
3
(M1)
Can be implied by a correct answer
(A2)
AWFW
1 minus
B1
( 0.15 × 0.20 )
B1
(ii)
(0.8541)
CAO; requires ‘plus’ or ‘minus’ 2
CAO; not simply 0.12 or 0.03 AG
= 0.97 NB:
Accept 3dp rounding or truncation AWFW (0.8541)
( 0.85 × 0.20 ) + 0.80 ⇒ B0 B0 ( 0.85 × 0.20 ) + ( 0.85 × 0.80 ) + ( 0.15 × 0.80 ) ⇒ B0 B1
P ( S ≥ 48 ) = 0.81 to 0.82 or 0.5553
M2
or 0.9372 =0.81(0) to 0.811 NB: Answer = 0.4447 or 0.1892 or 0.0628 ⇒ M1 only
A1
(iii) p = 1 – 0.85 = 0.15
3
Accept 3dp rounding or truncation M2 for the three correctly expressed terms for B (50, 0.03) or B (50, 0.97) added AWFW (0.8108)
2
CAO; may be implied by correct answer or correct expression for mean CAO
B1
Mean, μ = 80 × 0.15 = 12 SC Mean = 9.6 ⇒ B1 only
B1 Total
15 7
Mark Scheme – General Certificate of Education (A-level) Mathematics – Statistics 1B – January 2011
MS/SS1B (cont) Q Solution 5(a) Time taken is dependent upon leaving time (b) b (gradient) = 1.28 (or 141/110) b (gradient) = 1.25 to 1.35
a (intercept) = 29.95 to 30 (or 659/22) a (intercept) = 29 to 31 Thus y = 30 + 1.28 x or
Marks
Total
B1
1
Or equivalent
B2 (B1)
AWRT; (CAO or equivalent) (1.28182) AWFW Treat rounding of correct answers as ISW
B2 (B1) B1F
AWFW; (CAO or equivalent) (29.95455) AWFW F on a and b
5
Attempt at ∑ x ∑ x 2 ∑ y and ∑ xy ( ∑ y 2 ) or
Comments
275 9625 682 and 20575 (47494) (All four attempted) (M1)
Attempt at S xx and S xy ( S yy )
2750 and 3525 (5210) (Both attempted)
Attempt at correct formula for b gradient b (gradient) = 1.28 (or 141/110) a (intercept) = 29.95 to 30 (or 659/22)
(m1) (A1) (A1)
AWRT; (CAO or equivalent) AWFW; (CAO or equivalent)
Thus y = 30 + 1.28 x
(B1F)
F on a and b
Accept a and b interchanged only if identified correctly by a clearly shown equation
If a and b are not identified anywhere in the question, then: 1.25 to 1.35 ⇒ B1 29 to 30 ⇒ B1
(c) 7.45 am ⇒ x = 15 ⇒ y15 = 30 + 1.28 × 15 = 47 to 52
B1 M1 A1
CAO; stated, used or implied Use of 10 < x < 20 AWFW (49.2)
Time before 9.00 am = 9.00 – (7.45 + c’s y15) = 23 to 28
M1
May be implied
A1
5
SC Answer of 17 CAO (use of c’s y15 = 58) gains 2 marks (d)(i)
y85 = 30 + 1.28 × 85 = 135 to 146
(ii) Extrapolation/ outside/ above range of x-values Implies leaves home at 8.55 so different traffic conditions Total
AWFW NB:
B1
1
B1 B1
An answer of 8.32 to 8.37 gains B1 M1 A1 M0 A0 (138.9)
Or equivalent 2 14
8
AWFW
(25.8)
Or equivalent; 8.55 may be implied by 5 minutes
Mark Scheme – General Certificate of Education (A-level) Mathematics – Statistics 1B – January 2011
MS/SS1B (cont) Q Solution 2 6(a) Volume, V ∼ N ( 412, 8 ) (i)
(ii)
(iii)
Marks
Total
Comments
M1
Standardising 400 with 412 and 8 and/or ( 412 − x )
= P ( Z < −1.5 ) = 1 − P ( Z < 1.5 )
M1
Area change May be implied by a correct answer or an answer < 0.5
= 1 − 0.93319 = 0.066 to 0.067
A1
P (V > 420 ) = P ( Z > 1)
B1
= 1 − P ( Z < 1) = 1 − 0.84134 = 0.158 to 0.159
B1
2
P (V = 410 ) = 0 or zero or impossible
B1
1
400 − 412 ⎞ ⎛ P (V < 400 ) = P ⎜ Z < ⎟ 8 ⎝ ⎠
3
(0.06681)
CAO but ignore inequality and sign May be implied by a correct answer
AWFW
(0.15866)
Ignore any working B0 for ‘impossible to calculate’ or ‘no answer’ Simple statement that z = ±1.6449 and/or z = ±2.3263 or sketch of normal curve with at least one z-value marked
(b)(i) A statement/indication that
(–) 1.6449 and/or 2.3263 are z-values
AWFW
B1
Do not allow Φ ( 0.99 ) = 2.3263 , etc but
allow Φ −1 ( 0.99 ) = 2.3263 Do not award for z-value(s) simply embedded in standardisation statement(s)
A clear use of z =
v−μ
σ
or v = μ + 2σ
SC Immediate algebraic use of v − μ = zσ ⇒ B1 M1 A0
M1
with 400 and/or 420 (condone sign errors) The two given equations correctly derived (ii) Thus 20 = 2.3263 + 1.6449 σ ( )
A1
3
AG;
watch for sign inconsistencies
M1
A sensible (one that would lead to values required if completed correctly) attempt at solving the two given equations by eliminating μ or σ Do NOT allow MC or MR
σ = 5.04
A1
AWRT (5.03626)
μ = 408
A1 Total TOTAL
9
3 12 75
AWRT (408.284)
Scaled mark unit grade boundaries - January 2011 exams A-level Max. Scaled Mark
Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E
Code
Title
MD01 MFP1
GCE MATHEMATICS UNIT D01 GCE MATHEMATICS UNIT FP1
75 75
-
61 63
43 43
37 37
MM1A MM1B MPC1
GCE MATHEMATICS UNIT M1A GCE MATHEMATICS UNIT M1B GCE MATHEMATICS UNIT PC1
100 75 75
-
no candidates were entered for this unit 61 53 46 39 56 49 42 36
32 30
GCE MATHEMATICS UNIT S1A GCE MATHEMATICS UNIT S1A - WRITTEN GCE MATHEMATICS UNIT S1A - COURSEWORK
100 75 25
-
84 64 20
74
64
54
44 34 10
MS1B MD02 MFP2 MM2B MPC2 MS2B XMCA2 MFP3 MPC3 MFP4 MPC4
GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT S2B GCE MATHEMATICS UNIT XMCA2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT PC4
75 75 75 75 75 75 125 75 75 75 75
69 67 63 66 105 66 66 63 68
61 63 60 55 61 59 93 59 59 55 61
53 56 51 47 54 52 81 52 52 47 54
46 50 42 40 47 45 70 45 45 40 47
39 44 34 33 40 38 59 38 38 33 41
32 38 26 26 33 31 48 31 31 26 35
MEST1 MEST2 MEST3 MEST4
GCE MEDIA STUDIES UNIT 1 GCE MEDIA STUDIES UNIT 2 GCE MEDIA STUDIES UNIT 3 GCE MEDIA STUDIES UNIT 4
80 80 80 80
69 72
56 63 58 65
49 54 47 53
42 45 37 42
35 36 27 31
28 28 17 20
PHIL1 PHIL2
GCE PHILOSOPHY UNIT 1 GCE PHILOSOPHY UNIT 2
90 90
-
54 62
48 56
42 50
37 44
32 38
MS1A MS/SS1A/W MS/SS1A/C
55 56
49 49
