General Certificate of Education Advanced Subsidiary Examination January 2010
Mathematics
MS/SS1B
Unit Statistics 1B
Statistics Unit Statistics 1B Wednesday 13 January 2010
1.30 pm to 3.00 pm
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 7 (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 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.
P20793/Jan10/MS/SS1B 6/6/6/
MS/SS1B
2
Answer all questions.
1 Draught excluder for doors and windows is sold in rolls of nominal length 10 metres. The actual length, X metres, of draught excluder on a roll may be modelled by a normal distribution with mean 10.2 and standard deviation 0.15 . (a) Determine: (i) PðX < 10:5Þ ;
(3 marks)
(ii) Pð10:0 < X < 10:5Þ .
(3 marks)
(b) A customer randomly selects six 10-metre rolls of the draught excluder. Calculate the probability that all six rolls selected contain more than 10 metres of draught excluder. (3 marks)
2 Lizzie, the receptionist at a dental practice, was asked to keep a weekly record of the number of patients who failed to turn up for an appointment. Her records for the first 15 weeks were as follows. 20
26
32
a
37
14
27
34
15
18
b
25
37
29
25
Unfortunately, Lizzie forgot to record the actual values for two of the 15 weeks, so she recorded them as a and b. However, she did remember that a < 10 and that b > 40 . (a) Calculate the median and the interquartile range of these 15 values.
(4 marks)
(b) Give a reason why, for these data: (i) the mode is not an appropriate measure of average; (ii) the standard deviation cannot be used as a measure of spread.
(2 marks)
(c) Subsequent investigations revealed that the missing values were 8 and 43. Calculate the mean and the standard deviation of the 15 values.
P20793/Jan10/MS/SS1B
(2 marks)
3
3 The table shows, for each of a random sample of 7 weeks, the number of customers, x, who purchased fuel from a filling station, together with the total volume, y litres, of fuel purchased by these customers. x
230
184
165
147
241
174
210
y
4551
3410
3252
3756
3787
4024
4254
(a) Calculate the equation of the least squares regression line of y on x.
(4 marks)
(b) Estimate the volume of fuel sold during a week in which 200 customers purchase fuel. (2 marks) (c) Comment on the likely reliability of your estimate in part (b), given that, for the regression line calculated in part (a), the values of the 7 residuals lie between approximately 415 litres and þ430 litres. (2 marks)
4 Each school-day morning, three students, Rita, Said and Ting, travel independently from their homes to the same school by one of three methods: walk, cycle or bus. The table shows the probabilities of their independent daily choices. Walk
Cycle
Bus
Rita
0.65
0.10
0.25
Said
0.40
0.45
0.15
Ting
0.25
0.55
0.20
(a) Calculate the probability that, on any given school-day morning: (i) all 3 students walk to school;
(2 marks)
(ii) only Rita travels by bus to school;
(2 marks)
(iii) at least 2 of the 3 students cycle to school.
(4 marks)
(b) Ursula, a friend of Rita, never travels to school by bus. The probability that: Ursula walks to school when Rita walks to school is 0.9 ; Ursula cycles to school when Rita cycles to school is 0.7 . Calculate the probability that, on any given school-day morning, Rita and Ursula travel to school by: (i) the same method;
(3 marks)
(ii) different methods.
(1 mark)
P20793/Jan10/MS/SS1B
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4
5 In a random sample of 12 bags of flour, the weight, in grams, of flour in each bag was recorded as follows. 1011
995
1018
1022
1014
1005
1017
1015
993
1018
992
1020
(a) It may be assumed that the weight of flour in a bag is normally distributed with a standard deviation of 10.5 grams. (i) Construct a 98% confidence interval for the mean weight, m grams, of flour in a bag, giving the limits to four significant figures. (5 marks) (ii) State why, in constructing your confidence interval, use of the Central Limit Theorem was not necessary. (1 mark) (iii) If the distribution of the weight of flour in a bag was unknown, indicate a minimum number of weights that you would consider necessary for a confidence interval for m to be valid. (1 mark) (b) The statement ‘1 kg’ is printed on each bag. Comment on this statement using both the confidence interval that you constructed in part (a)(i) and the weights of the given sample of 12 bags. (3 marks) (c) Given that m ¼ 1000, state the probability that a 98% confidence interval for m will not contain 1000 . (1 mark)
P20793/Jan10/MS/SS1B
5
6 During the winter, the probability that Barry’s cat, Sylvester, chooses to stay outside all night is 0.35 , and the cat’s choice is independent from night to night. (a) Determine the probability that, during a period of 2 weeks (14 nights) in winter, Sylvester chooses to stay outside: (i) on at most 7 nights;
(2 marks)
(ii) on at least 11 nights;
(2 marks)
(iii) on more than 5 nights but fewer than 10 nights.
(3 marks)
(b) Calculate the probability that, during a period of 3 weeks in winter, Sylvester chooses to stay outside on exactly 4 nights. (3 marks) (c) Barry claims that, during the summer, the number of nights per week, S, on which Sylvester chooses to stay outside can be modelled by a binomial distribution with 5
n ¼ 7 and p ¼ 7 . (i) Assuming that Barry’s claim is correct, find the mean and the variance of S. (2 marks) (ii) For a period of 13 weeks during the summer, the number of nights per week on which Sylvester chose to stay outside had a mean of 5 and a variance of 1.5 . Comment on Barry’s claim.
(2 marks)
Turn over for the next question
P20793/Jan10/MS/SS1B
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6
7 [Figure 1, printed on the insert, is provided for use in this question.] Harold considers himself to be an expert in assessing the auction value of antiques. He regularly visits car boot sales to buy items that he then sells at his local auction rooms. Harold’s father, Albert, who is not convinced of his son’s expertise, collects the following data from a random sample of 12 items bought by Harold.
Item
Purchase price (£ x)
Auction price (£ y)
A
20
30
B
35
45
C
18
25
D
50
50
E
45
38
F
55
45
G
43
50
H
81
90
I
90
85
J
30
190
K
57
65
L
112
25
(a) Calculate the value of the product moment correlation coefficient between x and y. (3 marks) (b) Interpret your value in the context of this question.
(2 marks)
(c)
(i) On Figure 1, complete the scatter diagram for these data.
(3 marks)
(ii) Comment on what this reveals.
(2 marks)
(d) When items J and L are omitted from the data, it is found that Sxx ¼ 4854:4
Syy ¼ 4216:1
Sxy ¼ 4268:8
(i) Calculate the value of the product moment correlation coefficient between x and y for the remaining 10 items. (2 marks) (ii) Hence revise as necessary your interpretation in part (b).
END OF QUESTIONS P20793/Jan10/MS/SS1B
(1 mark)
Centre Number
Candidate Number
Surname Other Names Candidate Signature
General Certificate of Education Advanced Subsidiary Examination January 2010
Mathematics
MS/SS1B
Unit Statistics 1B
Statistics Unit Statistics 1B
Insert Instructions * Insert for use in Question 7. * Fill in the boxes at the top of this page. * Fasten this insert securely to your answer book at the end of the examination.
Turn over for Figure 1
P20793/Jan10/MS/SS1B 6/6/6/
s
Turn over
2
Figure 1 (for use in Question 7)
Prices of Antiques y
~
200 –
180 –
160 –
140 –
120 – Auction price (£) 100 –
80 –
60 – D
B
E
40 –
F
A
C 20 –
–
–
–
–
–
20
40
60
80
100
120
Purchase price (£) Copyright Ó 2010 AQA and its licensors. All rights reserved.
P20793/Jan10/MS/SS1B
~
–
–
0– 0
x
Version 1.0: 0110
klm General Certificate of Education
Mathematics 6360 Statistics 6380 MS/SS1B/W Statistics 1B
Mark Scheme 2010 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 © 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 Dr Michael Cresswell Director General
2
MS/SS1B/W - AQA GCE Mark Scheme 2010 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/W - AQA GCE Mark Scheme 2010 January series
MS/SS1B Q
Solution
Marks
Total
Comments
1(a)(i) X ~ N(10.2, 0.152)
M1
Standardising (10.45, 10.5 or 10.55) with 10.2 and ( 0.15 , 0.15 or 0.152) and/or (10.2 – x)
= P(Z < 2)
A1
CAO; ignore inequality and sign May be implied by a correct answer
= 0.977
A1
10.5 − 10.2 ⎞ ⎛ P ( X < 10.5 ) = P ⎜ Z < ⎟ 0.15 ⎠ ⎝
(ii)
P (10.0 < X < 10.5 )
3
= (a)(i) – P ( Z < −1.33)
Method correct using – 1.3 gives 0.88 to 0.881 ⇒ M1 m1 A0
= (a)(i) – (1 – p)
= 0.885 to 0.887
(b)
(0.97725)
Or equivalent; must be clear correct method if answer incorrect and answer > 0
M1
= [C’s (a)(i)] – P ( X < 10.0 )
= 0.97725 – (1 – 0.90824)
AWRT
Area change May be implied by a correct answer or answer > 0.5
m1 A1
3
AWFW (0.88604) M1 m1 A1 for 0.90824 – [1 – (a)(i)] = 0.886 M1 m0 A0 for (a)(i) – 0.90824 = 0.0685 M0 mo A0 for answer < 0
B1F
Correct value or F on value used or implied in (a)(ii) providing > 0.5 Use of – 1.3 gives 0.9032
P(6 rolls > 10) = 0.908246
M1
Accept any probability to power 6
0.56 to 0.565
A1
3
Total
9
P(X > 10) = p[from (a)(ii)] = 0.908 to 0.909
Note: B0F M1 A0 is possible
4
AWFW
MS/SS1B/W - AQA GCE Mark Scheme 2010 January series
MS/SS1B (cont) Q
Solution
Marks
Total
Comments
2(a) (a) 14 15 18 20 25 25 26 27 29 32 34 37 37 (b)
M1
May be implied by correct median or correct IQR Ignore any reference to a and b
Median = 26
A1
CAO
IQR = 34 – 18 = 16
A2
CAO
Ordering values gives:
Special Case: Identification that LQ = 18 and UQ = 34
(A1)
(b)(i) Two values (25 and 37) of mode No unique value Sparse data Many different values
B1
(ii) a and b (two values) unknown Impossible to calculate Cannot be calculated
B1
(c)
Mean =
∑x n
=
390 = 26 15
4
Both CAO
Or equivalent
2
Or equivalent
CAO
B1
If not identified, assume order is x then s SD
(∑ x
2
= 11472 ) = 9.4 to 9.8
B1
2
Special Case:
Evidence of
∑x
Can only be awarded if no marks scored elsewhere in (c)
(M1)
15
Total
5
AWFW (9.423 & 9.754) Treat rounding of a correct stated answer to an integer as ISW
8
MS/SS1B/W - AQA GCE Mark Scheme 2010 January series
MS/SS1B (cont) Q 3(a)
Solution
Marks
b (gradient) = 7.05 b (gradient) = 7(.00) to 7.1(0)
B2 (B1)
a (intercept) = 2500 to 2502 a (intercept) = 2490 to 2510
B2 (B1)
or Attempt at ∑ x ∑ x2
Total
4
Comments
AWRT (7.05134) AWFW Treat rounding of correct stated answers as ISW AWFW (2501.091) AWFW 1351 268047 27034 & 5269065 (105653202) (all 4 attempted)
∑ y & ∑ xy ( ∑ y 2 ) (M1)
or
Attempt at S xx & S xy
(S )
7304 & 51503 (1247894) (both attempted)
yy
Attempt at correct formula for b (gradient) b (gradient) = 7.05 a (intercept) = 2500 to 2502
(m1) (A1) (A1)
AWRT AWFW If a and b are not identified anywhere in solution, then: 7.05 ⇒ B1 2500 to 2502 ⇒ B1
Accept a & b interchanged only if identified correctly by a clearly shown equation (stated answers are not sufficient) in (b) (b) y200 = a + b × 200
M1
= 3890 to 3930
A1
2
B1 B1dep
2
(c) Large residuals / residual range suggest estimate may be unreliable or Largest residuals only small in relation to y-values (10%) so estimate may be reliable (unreliable) Special Case: If B0 B0dep then: Involves interpolation Does not involve extrapolation Within observed range
Used May be implied by correct answer
B1 B1dep
6
(3911.36)
(unreliable) requires (10% or equivalent)
(B1)
Total
AWFW
Any one; or equivalent
8
MS/SS1B/W - AQA GCE Mark Scheme 2010 January series
MS/SS1B (cont) Q
Solution
4(a)(i) P(all 3 walk) = 0.65 × 0.40 × 0.25
= 65/1000 = 13/200 = 0.065 (ii) P(Rita by bus) = 0.25×(1– 0.15)×(1– 0.20)
= 17/100 = 0.17
Marks
Total
Ratios (eg 65:1000) are only penalised by 1 mark at first correct answer Can be implied by correct answer
M1 A1
2
CAO; do not confuse with 0.65 Can be implied by correct answer
M1 A1
Comments
2
CAO
(iii) P(2 cycle) = 0.10 × 0.45 × (0.25 + 0.20) = 0.02025 + 0.10 × (0.40 + 0.15) × 0.55 = 0.03025 + (0.65 + 0.25) × 0.45 × 0.55 = 0.22275 (0.27325)
B1
CAO at least 1 of these 3 terms or equivalent but allow a ‘× 3’
P(3 cycle) = 0.10 × 0.45× 0.55 = 0.02475
B1
CAO
P(≥ 2 cycle) = P(2 cycle) + P(3 cycle)
M1
Sum of 4 or 7 terms each a product of 3 probabilities but not ‘× 3’
= 0.298 or P(0 cycle) = 0.90 × 0.55 × 0.45 = 0.22275
(b)(i)
A1
4
CAO
(B1)
CAO
P(1 cycles) = 0.10 × 0.55 × 0.45 = 0.02475 + 0.90 × 0.45 × 0.45 = 0.18225 (0.47925) + 0.90 × 0.55 × 0.55 = 0.27225 P(≥ 2 cycle) = 1 – [P(0 cycle) + P(1 cycles)]
(B1)
CAO at least 1 of these 3 terms but allow a ‘× 3’
(M1)
1 – [sum of 4 terms each a product of 3 probabilities but not ‘× 3’]
1 – 0.702 = 0.298
(A1)
CAO
P(WW) = (0.65 × 0.90) = 0.585
CAO either
B1
P(CC) = (0.10 × 0.70) = 0.070 P(WW or CC) = 0.585 + 0.070 = 0.655 (ii)
P(different) = 1 – (b)(i) = 0.345
A1
3
Sum of 2 terms each a product of 2 probabilities CAO; or equivalent
B1F
1
F on (b)(i) providing 0 < p < 1
Total
12
M1
7
MS/SS1B/W - AQA GCE Mark Scheme 2010 January series
MS/SS1B (cont) Q 5(a)(i)
Solution
Marks
Total
Comments
12120 = 1010 12
B1
CAO
98% (0.98) ⇒ z = 2.32 to 2.33
B1
AWFW
M1
Used Must have
A1F
F on x and z only
Mean =
σ
CI for μ is
x ± z×
Thus
1010 ± 2.3263 ×
n 10.5 12
(2.3263)
n with n > 1
A1dep
5
CAO & AWFW (accept 7) Dependent on A1F AWRT
(ii) Weight of flour in a bag (may be assumed to be) is normally distributed
B1
1
Or equivalent; must refer to weight
(iii) Any number such that 20 ≤ number ≤ 50
B1
1
Must be a single integer value Ignore any reasoning
Hence 1010 ± (7(.0) to 7.1) or (1003, 1017) Notes: Use of t11 ( 0.99 ) = 2.718 ⇒ maximum of B1 B0 M1 A0F A0 Use of a ‘corrected’ 10.5 ⇒ maximum of B1 B1 M1 A0F A0
(b) 1 kg or 1000 grams is outside / below CI or From CI, (population) mean weight is greater than 1kg or 1000 grams 3 or 3/12 or 25% of bags in sample weigh less than 1kg or 1000 grams
Statement appears dubious/incorrect/invalid (c)
2/100 or 1/50 or 0.02 or 2%
B1F
Or equivalent F on (a)(i) Any reference to 1010 ⇒ B0F
B1
Or equivalent; but not ‘some’
B1dep
3
Dependent on both B1F and B1
B1
1
CAO; not 0.02%
Total
11
8
MS/SS1B/W - AQA GCE Mark Scheme 2010 January series
MS/SS1B (cont) Q Solution 6(a)(i) R ~ B(14, 0.35) P(R ≤ 7) = 0.924 to 0.925
Marks M1 A1
(ii) P(R ≥ 11) = 1 – P(R ≤ 10) = 1 – (0.9989 or 0.9999)
minus 0.6405 or 0.4227
A1 (p1)
(p2)
= 0.353 to 0.354 or B(14, 0.35) expressions stated for at least 3 terms within 4 ≤ R ≤ 11 gives probability = 0.353 to 0.354 (b) R ~ B(21, 0.35) ⎛ 21⎞ 4 17 P(R = 4) = ⎜ ⎟ ( 0.35 ) ( 0.65 ) 4 ⎝ ⎠ = 0.059 to 0.0595
(c)(i)
S ~ B(7, 5/7) Mean = np = 7 × 5/7 = 5 If not identified, assume order is μ then σ2 Variance = np(1 – p) = 7 × 5/7 × 2/7 = 10/7 or 1.42 to 1.43
(ii)
2
2
M1
AWRT
(0.001106)
Accept 3 dp accuracy p2 – p1 ⇒ M0 M0 A0 (1 – p2) – p1 ⇒ M0 M0 A0 p1 – (1 – p2) ⇒ M1 M0 A0 only providing result > 0 Accept 3 dp accuracy
M1 A1
Comments Used somewhere in (a); may be implied AWFW (0.92466)
Requires ‘1 –’and ≥ 4 dp accuracy
M1
= 0.0011 (iii) P(5 < R < 10) = 0.9940 or 0.9989
Total
3
AWFW
(0.35346)
(M1)
Can be implied by correct answer
(A2)
AWFW
M1
Implied from correct stated formula; do not accept misreads
A1
Can be implied by a correct answer Ignore any additional terms
A1
3
AWFW
(0.35346)
(0.059274)
B1
CAO
B1
Must clearly state variance value if standard deviation (also) stated CAO / AWFW 2
Means are the same and (both comparisons clearly stated) Variances/standard deviations are similar Do not accept statements involving correct/incorrect/exact/etc
B1dep
Barry’s claim appears/is sound/valid/correct/likely
B1dep
2
Total
14
9
Must have scored B1 B1 in (i) or B1 B0 plus 10/7 v 1.5 or 10 / 7 v 1.5 stated
Must have scored previous B1dep
MS/SS1B/W - AQA GCE Mark Scheme 2010 January series
MS/SS1B (cont) Q 7(a)
Solution
r = – 0.0355 to – 0.035 r = – 0.036 to – 0.034 r = – 0.04 to + 0.04 or Attempt at ∑ x ∑ x 2 ∑ y ∑ xy or
Marks
Total
B3
3
(B2) (B1)
∑ y2
(M1)
Attempt at substitution into correct corresponding formula for r
(m1)
r = – 0.0355 to – 0.035
(A1)
(c)(i) Figure 1:
6 correct labelled points 5 or 4 correct labelled points 3 correct labelled points
(ii) (Two) outlier/anomaly/unusual or identification of J and L
(Otherwise) a positive/linear correlation
8994 22907 & –509 (all 3 attempted)
AWFW Dependent on – 0.1 < r < 0.1 Or equivalent; must qualify strength as ‘zero’; B0dep for very weak/weak/etc unless then qualified correctly
B1dep
B1
2
B3 (B2) (B1)
3
B1 B1
2
r = 0.943 to 0.944
A1
2
(ii) Very strong/strong positive (linear) correlation/relationship/association/link
B1dep
1
Previous calculation of r was not appropriate (due to outliers)
(B1) Total TOTAL
10
Deduct 1 mark if > 1 point not labelled or labelled incorrectly
Or equivalent; ignore any qualification of ‘strength’
Used Award B2 for a correct answer without/with different method
M1
r=
Context; providing –1 < r < 1
Or equivalent
4268.8 4854.4 × 4216.1
(d)(i)
(– 0.03546)
636 42702 738 68294 &38605 (all 5 attempted)
Attempt at S xx S yy & S xy
between purchase and auction prices of antiques
AWFW AWFW AWFW
&
(b) Almost/virtually/practically no / zero (linear) correlation / relationship / association / link (but not ‘no trend’)
Comments
13 75
AWFW
(0.94359)
Dependent on 0.9 < r < 1 Or equivalent; must qualify strength and indicate positive; B0dep for high/etc
klm
Scaled mark component grade boundaries - January 2010 exams A2 units (legacy) Component Code Component Title
Maximum Scaled Mark
A
Scaled Mark Grade Boundaries B C D
E
SC5W SCY6
GCE SOCIOLOGY UNIT 5 WRITTEN GCE SOCIOLOGY UNIT 6
60 60
39 40
36 36
33 32
30 29
28 26
SP04 SP5W SP6T
GCE SPANISH UNIT 4 GCE SPANISH UNIT 5 WRITTEN GCE SPANISH UNIT 6 CENTRE ORAL
140 60 70
107 43 60
97 38 54
87 33 48
77 28 42
67 23 37
PED4 PED5
GCE SPORT & PE UNIT 4 GCE SPORT & PE UNIT 5
64 70
46 44
40 39
35 35
30 31
25 27
75 25 75 75 75 75
59 20 62 58 60 60
51 17 55 51 52 52
44 14 48 44 44 44
37 12 41 37 36 37
30 10 34 30 29 30
MS/SS1A/W GCE STATISTICS UNIT 1A WRITTEN MS/SS1A/C GCE STATISTICS UNIT 1A CWK SS1B GCE STATISTICS UNIT 1B SS02 GCE STATISTICS UNIT 2 SS03 GCE STATISTICS UNIT 3 SS04 GCE STATISTICS UNIT 4
Version 1.0
11 March 2010
Copyright © 2010 AQA and its licensors. All rights reserved. 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.
