General Certificate of Education Advanced Subsidiary Examination June 2010
Mathematics
MS/SS1B
Unit Statistics 1B
Statistics Unit Statistics 1B Thursday 27 May 2010
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9.00 am to 10.30 am
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For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
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n
Time allowed * 1 hour 30 minutes
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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.
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n
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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.
P28274/Jun10/MS/SS1B 6/6/6/
MS/SS1B
2
The weight, x kg, and the engine power, y bhp, of each car in a random sample of 10 hatchback cars are shown in the table.
1
x
1196
1062
1335
1429
1012
1355
1145
1417
1275
1284
y
123
88
150
158
69
120
94
143
107
128
(a)
Calculate the value of the product moment correlation coefficient between x and y. (3 marks)
(b)
Interpret your value in the context of the question.
(2 marks)
Before leaving for a tour of the UK during the summer of 2008, Eduardo was told that the UK price of a 1.5-litre bottle of spring water was about 50p.
2
Whilst on his tour, Eduardo noted the prices, x pence, which he paid for 1.5-litre bottles of spring water from 12 retail outlets. He then subtracted 50p from each price and his resulting differences, in pence, were 18 (a) (i)
11
1
15
7
1
17
16
18
3
0
Calculate the mean and the standard deviation of these differences.
9 (2 marks)
(ii) Hence calculate the mean and the standard deviation of the prices, x pence, paid by
Eduardo.
(2 marks)
Based on an exchange rate of a1.22 to £1, calculate, in euros, the mean and the standard deviation of the prices paid by Eduardo. (3 marks)
(b)
Each day, Margot completes the crossword in her local morning newspaper. Her completion times, X minutes, can be modelled by a normal random variable with a mean of 65 and a standard deviation of 20 .
3
Determine:
(a) (i)
PðX < 90Þ ;
(ii) PðX > 60Þ .
(5 marks)
Given that Margot’s completion times are independent from day to day, determine the probability that, during a particular period of 6 days:
(b)
(i)
she completes one of the six crosswords in exactly 60 minutes;
(1 mark)
P28274/Jun10/MS/SS1B
3 (ii) she completes each crossword in less than 60 minutes;
(3 marks)
(iii) her mean completion time is less than 60 minutes.
(4 marks)
In a certain country, 15 per cent of the male population is left-handed.
4
Determine the probability that, in a random sample of 50 males from this country:
(a) (i)
at most 10 are left-handed;
(2 marks)
(ii) at least 5 are left-handed;
(2 marks)
(iii) more than 6 but fewer than 12 are left-handed.
(3 marks)
In the same country, 11 per cent of the female population is left-handed.
(b)
Calculate the probability that, in a random sample of 35 females from this country, exactly 4 are left-handed. (3 marks) A sample of 2000 people is selected at random from the population of the country. The proportion of males in the sample is 52 per cent.
(c)
How many people in the sample would you expect to be left-handed?
(4 marks)
Hugh owns a small farm.
5
He has two sows, Josie and Rosie, which he feeds at a trough in their field at 8.00 am each day.
(a)
The probability that Josie is waiting at the trough at 8.00 am on any given day is 0.90 . The probability that Rosie is waiting at the trough at 8.00 am on any given day is 0.70 when Josie is waiting at the trough, but is only 0.20 when Josie is not waiting at the trough. Calculate the probability that, at 8.00 am on a given day: both sows are waiting at the trough;
(2 marks)
(ii) neither sow is waiting at the trough;
(2 marks)
(i)
(iii) at least one sow is waiting at the trough.
(1 mark)
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Turn over
P28274/Jun10/MS/SS1B
4
Hugh also has two cows, Daisy and Maisy. Each day at 4.00 pm, he collects them from the gate to their field and takes them to be milked.
(b)
The probability, PðDÞ , that Daisy is waiting at the gate at 4.00 pm on any given day is 0.75 . The probability, PðMÞ , that Maisy is waiting at the gate at 4.00 pm on any given day is 0.60 . The probability that both Daisy and Maisy are waiting at the gate at 4.00 pm on any given day is 0.40 . (i)
In the table of probabilities, D 0 and M 0 denote the events ‘not D’ and ‘not M’ respectively. M D
M0
Total
0.40
0.75
0.60
1.00
D0 Total
Complete the table above.
(2 marks)
(ii) Hence, or otherwise, find the probability that, at 4.00 pm on a given day: (A) neither cow is waiting at the gate;
(1 mark)
(B) only Daisy is waiting at the gate;
(1 mark)
(C) exactly one cow is waiting at the gate.
(2 marks)
P28274/Jun10/MS/SS1B
5
During a study of reaction times, each of a random sample of 12 people, aged between 40 and 80 years, was asked to react as quickly as possible to a stimulus displayed on a computer screen.
6
Their ages, x years, and reaction times, y milliseconds, are shown in the table. Person
Age (x years) 41 54 66 72 71 57 60 47 77 65 51 59
A B C D E F G H I J K L
Reaction time (y ms) 520 750 650 920 280 620 740 950 970 780 550 730
(a)
Calculate the equation of the least squares regression line of y on x.
(4 marks)
(b) (i)
Draw your regression line on the scatter diagram on page 6.
(2 marks)
(ii) Comment on what this reveals.
(2 marks)
It was later discovered that the reaction times for persons E and H had been recorded incorrectly. The values should have been 820 and 590 respectively.
(c)
After making these corrections, computations gave Sxx ¼ 1272 (i)
Sxy ¼ 14 760
x ¼ 60
y ¼ 720
Using the symbol , plot the correct values for persons E and H on the scatter (1 mark) diagram on page 6.
(ii) Recalculate the equation of the least squares regression line of y on x, and draw this (3 marks) regression line on the scatter diagram on page 6. (iii) Hence revise as necessary your comments in part (b)(ii).
(2 marks)
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Turn over
P28274/Jun10/MS/SS1B
6
Reaction Times y
~
1000 –
I
H
D
900 –
800 – B
J G L
700 – Reaction time (ms)
C
F 600 – A
K
500 –
400 –
300 –
E
200 – –
–
–
–
40
50
60
70
80
~
–
–
0– 0
x
Age (years)
P28274/Jun10/MS/SS1B
7
An ambulance control centre responds to emergency calls in a rural area. The response time, T minutes, is defined as the time between the answering of an emergency call at the centre and the arrival of an ambulance at the given location of the emergency.
7
Response times have an unknown mean m T and an unknown variance. Anita, the centre’s manager, asked Peng, a student on supervised work experience, to record and summarise the values of T obtained from a random sample of 80 emergency calls. Peng’s summarised results were Variance (unbiased estimate), s 2 ¼ 19:3
Mean, t ¼ 6:31
Only 1 of the 80 values of T exceeded 20 Anita then asked Peng to determine a confidence interval for m T . Peng replied that, from his summarised results, T was not normally distributed and so a valid confidence interval for m T could not be constructed.
(a)
(i)
Explain, using the value of t 2s , why Peng’s conclusion that T was not normally distributed was likely to be correct. (2 marks)
(ii) Explain why Peng’s conclusion that a valid confidence interval for m T could not be
constructed was incorrect.
(2 marks)
(b)
Construct a 98% confidence interval for m T .
(4 marks)
(c)
Anita had two targets for T. These were that m T < 8 and that PðT 4 20Þ > 95% . Indicate, with justification, whether each of these two targets was likely to have been met. (3 marks) END OF QUESTIONS
Copyright Ó 2010 AQA and its licensors. All rights reserved.
P28274/Jun10/MS/SS1B
Version 1.0
klm General Certificate of Education June 2010 Mathematics Statistics
MS1B SS1B
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 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 © 2010 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
MS/SS1B - AQA GCE Mark Scheme 2010 June 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 2010 June series
MS/SS1B Q 1(a)
Solution r = 0.915
Marks B3
r = 0.91 to 0.92 r = 0.88 to 0.95 OR Attempt at ∑ xy or
(B2) (B1)
(0.91504)
AWFW AWFW 12510 15835890 1180 146616 and 1510062 (all 5 attempted)
Attempt at S xx S yy and S xy
(M1)
Attempt at substitution into correct corresponding formula for r
(m1)
Very strong / strong / fairly strong positive (linear) correlation / relationship / association / link (but not ‘ trend’)
Comments AWRT
∑ x ∑ x2 ∑ y ∑ y 2 and
r = 0.915 (b)
Total
(A1)
185880 7376 and 33882 (all 3 attempted)
3
AWRT Dependent on 0.88 < r < 0.95 Or equivalent; must qualify strength and indicate positive B0dep for (almost) perfect / high / average / medium / some / etc
B1dep
between weight and (engine) power/bhp of (hatchback) cars
B1
2
Examples: The more weight/heavier the more/greater power ⇒ B0dep B1 Strong correlation and as weight/kg increases so does engine power / bhp ⇒ B0dep B1
Context; providing 0 < r < 1
No mention of strength Mention of strength but implied suggestion of positive not sufficient
Total
5
4
MS/SS1B - AQA GCE Mark Scheme 2010 June series
MS/SS1B (cont) Q Solution 2 –18 –11 1 15 7 –1 17 –16 18 –3 0 9 (a)(i) Mean,
Marks
= 1.5
B1
Standard deviation, σ d or sd = 11.7 to 12.3
B1
(ii) Mean,
d
x = 50 + d = 51.5
Total
Comments
2
CAO ∑ d = 18 Ignore notation and units (11.737 or 12.259) AWFW ∑ d 2 = 1680
B1F
F on (a)(i) or correct
∑x
= 618 Ignore notation and units
x: 32 39 51 65 57 49 67 4 68 47 50 59 Standard deviation, σ x or sx = 11.7 to 12.3 (b) [Values, mean or sd in (a)(i) or (a)(ii)] 1.22 or 1.22 × 100
B1F
2
∑ x2 = 33480
F on (a)(i) providing > 0 or correct
M1
Award if use seen or implied by ≥ 1 Subsequent correct or (correct × 100) answer
Mean = 0.628 to 0.63
A1
AWFW
(0.6283)
Standard deviation = 0.14 to 0.151
A1
AWFW
(0.1432 or 0.1496)
3
Special Cases: At least one answer correct with no stated units or incorrect stated units ⇒ M1 A1 A1 max At least one answer × 100 with its units stated as ‘cents’ ⇒ M1 A1 A1 max At least one answer × 100 with no units stated or units stated as euros / pence / £ ⇒ M1 only
‘cents’ attached to ≥ 1 answer × 100
Total
7
5
MS/SS1B - AQA GCE Mark Scheme 2010 June series
MS/SS1B (cont) Q Solution Time, X ~ N(65, 202) 3 (a) (i) 90 − 65 ⎞ ⎛ P(X < 90) = P ⎜ Z < ⎟– 20 ⎠ ⎝ ⎡ ⎛ 0 − 65 ⎞ ⎤ ⎢ P ⎜ Z < 20 ⎟ = P ( Z < −3.25 ) = 0.00058⎥ ⎠ ⎣ ⎝ ⎦
(ii)
Total
Comments
M1
Standardising (89.5, 90 or 90.5 or 59.5, 60 or 60.5) with 65 and ( 20 , 20 or 202) and/or (65 – x) May be gained in (a)(i) or (a)(ii)
= P(Z < 1.25)
A1
CAO; ignore inequality and sign May be implied by a correct answer
= 0.893 to 0.895
A1
AWFW
M1
Area change May be implied by a correct answer or answer > 0.5
(0.89435)
P(X > 60) = P(Z > –0.25) = P(Z < 0.25)
= 0.598 to 0.599 (b) (i)
Marks
P(1 in 6 = 60) = 0 or zero or impossible
(ii) P(X < 60) = 1 – [(a)(ii)] or (0.401 to 0.402) P(6 in 6 < 60) = p6
with
0 < p 0 Accept 3 dp accuracy
= 0.576 OR B(50, 0.15) expressions stated for at least 3 terms within 5 ≤ R ≤ 12 gives probability = 0.576
A1
AWRT
(M1)
F ~ B(35, 0.11)
M1
Implied from correct stated formula; do not accept misreads
A1
Can be implied by a correct answer Ignore any additional terms
P(6 < R < 12) = 0.9372 or 0.9699
minus 0.3613 or 0.2194
(b)
(A2)
⎛ 35 ⎞ 4 31 P(F = 4) = ⎜ ⎟ ( 0.11) ( 0.89 ) ⎝4⎠ = 0.206 to 0.208
(c)
A1
(0.5759)
Can be implied by correct answer 3
3
AWFW
AWFW
(0.5759)
(0.20685)
or
P(M and LH) = 0.52 × 0.15 = 0.078 ) N(M) = 2000 × 0.52 = 1040 )
M1
≥1 of these 2 probabilities or ≥1 of these 2 numbers attempted; may be implied
or
P(F and LH) = 0.48 × 0.1 = 0.0528) N(F) = 2000 × 0.48 = 960 )
A1
2 probabilities or 2 numbers evaluated correctly
N(M and LH) = 2000 × 0.078 = 1040 × 0.15 = 156 ) N(F and LH) = 2000 × 0.0528 = 960 × 0.11 = 105.6 ) or P(LH) = 0.078 + 0.0528 = 0.1308 )
Evaluation of ≥1 of these 2 numbers A1
N(LH) = 156 + 105.6 = 2000 × 0.1308 = 261 to 262
A1
Total
or Addition of these 2 probabilities 4 14
7
262/2000 ⇒ A0 AWFW
(261.6)
MS/SS1B - AQA GCE Mark Scheme 2010 June series
MS/SS1B (cont) Q 5 (a)
Solution
P(J) = 0.9
Marks
Total
Comments Ratios (eg 63:100) are only penalised by1 mark at first correct answer F marks can only be awarded if 0 < p < 1
P(R | J) = 0.7 P(R | J′ ) = 0.2
(i) P(both at trough) = 0.9 × 0.7 = 0.63 = 63/100
M1 A1
(ii) P(neither at trough) = (1 – 0.9) × (1 – 0.2) = 0.1 × 0.8
M1
= 0.08 = 8/100 = 4/50 = 2/25
A1
2
CAO
B1F
1
F on (ii) or correct answer
(iii)
2
Can be implied by correct answer CAO Can be implied by correct answer
P(at least one at trough) = (1 – (ii)) = 0.92 = 92/100 = 46/50 = 23/25
(b)(i) D D′ Total
M 0.40 0.20 0.60
M′ 0.35 0.05 0.40
Total 0.75 0.25 1.00
B1
Both row and column totals ie 0.25 and 0.40; CAO
B1
2
P(neither at gate) = 0.05
B1F
1
F on table or correct answer by ‘otherwise’
P(only Daisy at gate) = 0.35
B1F
1
F on table or correct answer by ‘otherwise’
Notes: Use of Venn or tree diagrams without table completion ⇒ B0 B0 Table not completed on page 13 but completed on page 10 ⇒ max of B1 B1
Three table values ie 0.35 and 0.20 and 0.05; CAO
(ii) Accept answers ÷ 1.00 (A)
(B)
(C) P(exactly one at gate) = P ( D ∩ M ′) + P ( D′ ∩ M ) 0.35 + 0.20 = 0.55
Only correct two values from c’s table shown and added Can be implied by correct answer
M1
A1F
Total
2
11
8
F on table or correct answer by ‘otherwise’
MS/SS1B - AQA GCE Mark Scheme 2010 June series
MS/SS1B (cont) Q 6 (a)
Solution
OR Attempt at
(∑ y )
Marks
b (gradient) = 3.25 to 3.26 b (gradient) = 3.2 to 3.3
B2 (B1)
a (intercept) = 509 to 510 a (intercept) = 507 to 513
B2 (B1)
Total
AWFW (3.25472) AWFW Treat rounding of correct answers as ISW AWFW (509.71698) AWFW
∑ x ∑ x2 ∑ y and ∑ xy
720 44472 8460 and 511740 (6399400) (all 4 attempted)
2
or Attempt at S xx and S xy
(S )
(M1) 1272 and 4140 (435100) (both attempted)
yy
Attempt at correct formula for b (gradient) b (gradient) = 3.25 to 3.26 a (intercept) = 509 to 510
(m1) (A1) (A1)
4
Accept a and b interchanged only if identified correctly by a clearly shown or drawn equation (b) (i) Correct line drawn on graph (40, 630 to 650) (80, 750 to 790) If B0 but evidence of use of line for ≥ 2 points within range 0 ≤ x ≤ 80 or ‘intercept’ and means
Comments
Dep on ≥ B1 B1 or ≥ A1 A0 in (a) From x ≈ 40 to 80
B2dep (M1)
AWFW AWFW If a and b are not identified anywhere in question, then: 3.25 to 3.26 ⇒ B1 509 to 510 ⇒ B1
2
Calcn or points shown on graph Allow point (‘0’, 500 to 520) Graph
(ii) Outliers / at least E and H identified / wide scatter (of points) / large residuals Evidence of a (+ ve) relationship or correlation /model is not appropriate (c) (i) Correct two points marked on graph (ii)
b (gradient) = 11.6 a (intercept) = 23 to 24 Correct line on graph (40, 480 to 500) (80, 930 to 970)
(iii) No outliers / less scatter / small residuals Strong(er)/more evidence of a positive link/ relationship or more rapid increase (of reaction time with age) or model is more appropriate Total
B1
Or equivalent words Or equivalent words; none of strong/ negative/trend/etc or unreliable/invalid
B1
2
B1
1
Labels are not required; nor is ☼ Graph AWRT (11.60377) AWFW (23.77358)
3
Graph Dependent on B1 B1
B1 B1 B1dep B1 B1
Or equivalent words 2 14
9
Or equivalent words; must indicate change from (b)(ii) in context; not some/weak/etc or reliable/valid References to correlation alone ⇒ B0
MS/SS1B - AQA GCE Mark Scheme 2010 June series
MS/SS1B (cont) Q Solution 7(a)(i) t − 2 s = 6 .31 − 2 19 .3 = –2.48 to –2.47 Negative value is impossible for a measurement of time (ii) Sample size, n = 80 is large / > 25 Thus sample mean (T ) ~ approximately normal due to CLT (b)
98% (0.98) ⇒ z = 2.32 to 2.33
s
CI for μ is
t ± z/t ×
Thus
6.31 ± 2.3263 ×
Hence or
n 19.3 80
6.31 ± (1.13 to 1.15)
Marks B1
Total
B1
2
AWRT
(c)
⇒ Yes
Or equivalent; allow if negative value incorrect or not stated Indication that given sample is ‘large’
B1dep
Dependent on previous B1 Requires ‘mean’ and ‘normal’ and ‘CLT’
2
B1 (B1)
AWFW t79 ( 0.99) = 2.37
M1
Used Must have
A1F
F on z/t only
(2.3263) AWRT
n with n > 1
CAO and AWFW A1 AWFW (A1)
4
μT < 8 Since CI / UCL < 8
(–2.4764)
B1
(5.16 to 5.18, 7.44 to 7.46) Note: Use of t gives 6.31 ± (1.17) or (5.14, 7.48)
Comments
(5.17, 7.45)
AWRT
B1F
F on (b); must clearly compare 8 with CI/UCL and state a correct followthrough conclusion
B1
CAO/AWFW; accept eg ‘1 in 80’ B0 for use of normal distribution CAO/AWFW; accept eg ’79 in 80’
P(T ≤ 20) > 95% P(T > 20) = 1/80 = 0.01 to 0.013 or P(T ≤ 20) = 79/80 = 0.987 to 0.99 P(T > 20) < 0.05 or 5% or P(T ≤ 20) > 0.95 or 95%
⇒ Yes
B1dep
Total TOTAL
3
11 75
10
Dependent on previous B1 A correct comparison must be clearly stated together with clear conclusion Do not accept use of 2% or 98% OE
Scaled mark unit grade boundaries - June 2010 exams A-level
Code
Title
Max. Scaled Mark
Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E
MS1B MD02 MFP2 MM2B MPC2 MS2B XMCA2 MFP3 MM03 MPC3 MS03 MFP4 MM04 MPC4 MS04 MM05
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 M03 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT S03 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M04 GCE MATHEMATICS UNIT PC4 GCE MATHEMATICS UNIT S04 GCE MATHEMATICS UNIT M05
75 75 75 75 75 75 125 75 75 75 75 75 75 75 75 75
69 70 63 65 102 69 71 67 60 67 68 67 68 63
59 64 65 55 61 58 92 64 68 62 54 60 61 62 61 56
52 56 57 47 54 51 80 56 61 54 48 52 53 56 53 49
45 48 49 39 47 44 68 49 54 46 42 44 45 50 45 42
39 40 41 31 41 37 57 42 48 39 36 36 37 44 38 35
33 32 33 23 35 31 46 35 42 32 31 29 29 39 31 29
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
52 63 59 65
45 54 48 53
38 45 38 42
31 36 28 31
25 28 18 20
MHEB1 MHEB2
GCE MODERN HEBREW UNIT 1 GCE MODERN HEBREW UNIT 2
100 100
80
66 71
59 62
52 53
45 45
38 37
MUSC1 MUS2A MUS2B
GCE MUSIC UNIT 1 GCE MUSIC UNIT 2A GCE MUSIC UNIT 2B
80 60 60
-
57 46 48
51 41 42
45 36 36
39 31 31
34 26 26
