Centre Number
For Examiner’s Use
Candidate Number
Surname Other Names
Examiner’s Initials
Candidate Signature Question
General Certificate of Education Advanced Subsidiary Examination June 2009
Mark
1 2 3
Mathematics
MS/SS1B
Unit Statistics 1B
4 5
Statistics
6
Unit Statistics 1B
7
Specimen paper for examinations in June 2010 onwards
TOTAL
For this paper you must have: * the blue AQA booklet of formulae and statistical tables. 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. * 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 space 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. 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.
MS/SS1B
Do not write outside the box
2
Answer all questions in the spaces provided.
A large bookcase contains two types of book: hardback and paperback. The number of books of each type in each of four subject categories is shown in the table.
1
Subject category Crime
Romance
Science fiction
Thriller
Total
Hardback
8
16
18
18
60
Paperback
16
40
14
30
100
Total
24
56
32
48
160
Type
A book is selected at random from the bookcase. Calculate the probability that the book is:
(a)
(i)
a paperback;
(1 mark)
(ii) not science fiction;
(2 marks)
(iii) science fiction or a hardback;
(2 marks)
(iv) a thriller, given that it is a paperback.
(2 marks)
(b)
Three books are selected at random, without replacement, from the bookcase. Calculate, to three decimal places, the probability that one is crime, one is romance and one is science fiction. (4 marks)
QUESTION PART REFERENCE
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(02)
Do not write outside the box
4
Hermione, who is studying reptiles, measures the length, x cm, and the weight, y grams, of a sample of 11 adult snakes of the same type. Her results are shown in the table.
2
Snake
A
B
C
D
E
F
G
H
I
J
K
x
46
39
54
79
47
58
73
35
43
51
36
y
55
48
58
88
61
55
82
51
50
66
57
(a)
Calculate the value of the product moment correlation coefficient, r, between x and y. (3 marks)
(b)
Interpret your value in context.
(2 marks)
(c)
Complete the scatter diagram, opposite, for these data.
(2 marks)
(d)
Subsequently it is found that, of the 11 adult snakes, 9 are male and 2 are female. (i)
Given that female adult snakes are generally larger than male adult snakes, identify the 2 snakes which are most likely to be female. (1 mark)
(ii) Hence, without further calculation, estimate the value of r for the 9 male snakes (2 marks) and revise, as necessary, your interpretation in part (b). QUESTION PART REFERENCE
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(04)
Do not write outside the box
5 QUESTION PART REFERENCE
y
(c)
~
90 –
D
80 –
70 – Weight (grams)
E
60 –
C
F
A
50 –
B
40 –
–
–
–
–
–
30
40
50
60
70
80
~
–
–
0– 0
x
Length (cm)
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(05)
Turn over
Do not write outside the box
6
The weight, X grams, of talcum powder in a tin may be modelled by a normal distribution with mean 253 and standard deviation s .
3
Given that s ¼ 5 , determine:
(a) (i)
PðX < 250Þ ;
(ii) Pð245 < X < 250Þ ; (iii) PðX ¼ 245Þ . (b)
(3 marks) (2 marks) (1 mark)
Assuming that the value of the mean remains unchanged, determine the value of s necessary to ensure that 98% of tins contain more than 245 grams of talcum powder. (4 marks)
QUESTION PART REFERENCE
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(06)
Do not write outside the box
8
As part of an investigation, a chlorine block is immersed in a large tank of water held at a constant temperature. The block slowly dissolves, and its weight, y grams, is noted x days after immersion. The results are shown in the table.
4
x days
5
10
15
20
30
40
50
60
75
y grams
47
44
42
38
35
27
23
16
9
(a)
Calculate the equation of the least squares regression line of y on x.
(4 marks)
(b)
Hence estimate, to the nearest gram, the initial weight of the block.
(1 mark)
(c)
A company which markets the chlorine blocks claims that a block will usually dissolve completely after about 13 weeks. Comment, with justification, on this claim. (3 marks)
QUESTION PART REFERENCE
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(08)
Do not write outside the box
10
A survey of all the households on an estate is undertaken to provide information on the number of children per household.
5
The results, for the 99 households with children, are shown in the table. Number of children (x)
1
2
3
4
5
6
7
Number of households ( f )
14
35
25
13
9
2
1
For these 99 households, calculate values for:
(a) (i)
the median and the interquartile range;
(ii) the mean and the standard deviation.
(3 marks) (3 marks)
In fact, 163 households were surveyed, of which 64 contained no children.
(b) (i)
For all 163 households, calculate the value for the mean number of children per household. (2 marks)
(ii) State whether the value for the standard deviation, when calculated for all
163 households, will be smaller than, the same as, or greater than that calculated in (1 mark) part (a)(ii). (iii) It is claimed that, for all 163 households on the estate, the number of children per
household may be modelled approximately by a normal distribution. Comment, with justification, on this claim. Your comment should refer to a fact other than the discrete nature of the data. (2 marks) QUESTION PART REFERENCE
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(10)
Do not write outside the box
12 6 (a)
The time taken, in minutes, by Domesat to install a domestic satellite system may be modelled by a normal distribution with unknown mean, m , and standard deviation 15 . The times taken, in minutes, for a random sample of 10 installations are as follows. 47
39
25
51
47
36
Construct a 98% confidence interval for m . (b)
63
41
78
43 (5 marks)
The time taken, Y minutes, by Teleair to erect a TV aerial and then connect it to a TV is known to have a mean of 108 and a standard deviation of 28 . Estimate the probability that the mean of a random sample of 40 observations of Y is more than 120 . (4 marks)
(c)
Indicate, with a reason, where, if at all, in this question you made use of the Central Limit Theorem. (2 marks)
QUESTION PART REFERENCE
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(12)
Do not write outside the box
14
Mr Alott and Miss Fewer work in a postal sorting office.
7
The number of letters per batch, R, sorted incorrectly by Mr Alott when sorting batches of 50 letters may be modelled by the distribution Bð50, 0:15Þ .
(a)
Determine: (i)
PðR < 10Þ ;
(ii) Pð5 4 R 4 10Þ .
(4 marks)
It is assumed that the probability that Miss Fewer sorts a letter incorrectly is 0.06 , and that her sorting of a letter incorrectly is independent from letter to letter.
(b)
(i)
Calculate the probability that, when sorting a batch of 22 letters, Miss Fewer sorts exactly 2 letters incorrectly. (3 marks)
(ii) Calculate the probability that, when sorting a batch of 35 letters, Miss Fewer sorts at
least 1 letter incorrectly.
(2 marks)
(iii) Calculate the mean and the variance for the number of letters sorted correctly by
Miss Fewer when she sorts a batch of 120 letters.
(2 marks)
(iv) Miss Fewer sorts a random sample of 20 batches, each containing 120 letters. The
number of letters sorted correctly per batch has a mean of 112.8 and a variance of 56.86 . Comment on the assumptions that the probability that Miss Fewer sorts a letter incorrectly is 0.06 , and that her sorting of a letter incorrectly is independent from letter to letter. (3 marks) QUESTION PART REFERENCE
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(14)
Surname
Other Names
Centre Number
Candidate Number
Candidate Signature
General Certificate of Education June 2009 Advanced Subsidiary Examination
MATHEMATICS Unit Statistics 1B
MS/SS1B
STATISTICS Unit Statistics 1B
Insert Insert for use in Question 2. Fill in the boxes at the top of this page. Fasten this insert securely to your answer book.
Turn over for Figure 1
P15525/Jun09/MS/SS1B 6/6/6/
s
Turn over
2
Figure 1 (for use in Question 2)
Lengths and Weights of Snakes
y
~
90 –
D
80 –
70 – Weight (grams)
E
60 –
C
F
A
50 –
B
40 –
–
–
–
–
–
30
40
50
60
70
80
Length (cm)
Copyright Ó 2009 AQA and its licensors. All rights reserved.
P15525/Jun09/MS/SS1B
~
–
–
0– 0
x
Version : 1.0 0609
klm General Certificate of Education
Mathematics 6360 Statistics 6380
MS/SS1B/W Statistics 1B
Mark Scheme 2009 examination - June 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/W - AQA GCE Mark Scheme 2009 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/W - AQA GCE Mark Scheme 2009 June series
MS/SS1B/W Q 1(a) (i)
Solution
Total
B1
1
Comments In (a), ratios (eg 100:160) are only penalised by 1 mark at first correct answer
P(P) = 100/160 = 50/80 = 25/40 = 10/16 = 5/8 = 0.625
(ii)
Marks
P(S′) = 1 −
32 160
or
P(S) =
32 160
CAO
Or equivalent Ignore labels of S′ & S Can be implied by correct answer
M1
= 128/160 = 64/80 = 32/40 = 16/20 = 8/10 = 4/5 = 0.8 (iii)
P(S or H) = P(S ∪ H) = 60 + 32 − 18 60 + 14 32 + 8 + 16 + 18 or or 160 160 160 = 74/160 = 37/80 = 0.462 to 0.463
(iv)
A1
160 (i)
A1
CAO
Or equivalent Can be implied by correct answer 2
CAO/AWFW
(0.4625)
Or equivalent Can be implied by correct answer or 48 But watch for 18 160 160
M1
= 3/100 = 3/10 = 0.3 (b)
2
M1
30
P(T | P) =
A1
2
CAO
P(1C & 1R & 1S) =
M1
Multiplication of any 3 different given subject totals Multiplication of 160, 159 & 158
M1
Accept 3dp accuracy Award for 3 ≤ multiplier ≤ 6
M1
24 56 32 × × 160 159 158 (0.15 × 0.35220 × 0.20253)
× 6
= 0.064 to 0.0644
Special Case: (Any given subject total) ÷ 160 seen anywhere in (b)
4
AWFW (0.0642) Do not accept a fraction as answer A correct answer can imply 4 marks
A1
(M1)
4
Total
11
Can award if no marks scored in (b) Accept a decimal equivalent
MS/SS1B/W - AQA GCE Mark Scheme 2009 June series
MS/SS1B/W (cont) Q Solution 2(a) r = 0.893 to 0.8933
Marks B3
Total
Comments
AWFW
r = 0.89
to 0.896
(B2)
AWFW
r = 0.8
to 0.95
(B1)
AWFW
(0.89319)
or
Attempt at
∑ x ∑ x2 ∑ y ∑ y 2 ∑ xy
&
561 30667 671 42613 & 35882 (all 5 attempted) (M1)
or
2056 1682 & 1661 (all 3 attempted)
Attempt at S xx S yy & S xy Attempt at correct corresponding formula for r r = 0.893 to 0.8933 (b)
(m1) (A1)
Fairly strong / strong / very strong positive (linear) correlation / relationship / association / link (but not trend)
3
AWFW Or equivalent; must qualify strength and indicate positive Dependant on 0.8 ≤ r ≤ 0.95 B0 for some/average/medium/etc
B1dep
between length and weight of adult snakes (c)
Figure 1:
5 correct labelled points 4 or 3 correct labelled points D and G
(d)(i) (ii)
B1
2
B2 (B1)
2
B1
1
B1
B1dep
2
Total
10
Do not accept comparison with value in (a) or statement in (b)
5
Deduct 1 mark if points not labelled Both CAO AWFW (0.48790) No penalty for calculation Accept a range only if whole of it falls within 0.25 to 0.75
r = 0.25 to 0.75
Fairly weak / weak / some / moderate positive (linear) correlation / relationship / association / link
Context; providing 0 < r < 1
Or equivalent; must qualify strength and indicate positive Dependant on 0.25 ≤ r ≤ 0.75 B0 for very weak/little/slight/hardly any/fair/average/medium/anything involving strong/etc
MS/SS1B/W - AQA GCE Mark Scheme 2009 June series
MS/SS1B/W (cont) Q 3(a) (i)
(ii)
Solution X ~ N(253, 52)
Marks
Total
Comments
250 − 253 ⎞ ⎛ P ( X < 250 ) = P ⎜ Z < ⎟ = 5 ⎝ ⎠
M1
P ( Z < −0.6 ) = 1 – P ( Z < 0.6 ) = 1 – 0.72575
Standardising (249.5, 250 or 250.5) with 253 and ( 5 , 5 or 52) and/or (253 – x)
m1
Area change; may be implied
= 0.274 to 0.275
A1
P ( 245 < X < 250 ) = [ C's(a)(i) ] – P ( X < 245 )
M1
3
AWFW (0.27425) (1 – answer) ⇒ M1 max Or equivalent; must be clear correct method if answer incorrect and answer > 0
= (i) – P ( Z < −1.6 ) = 0.27425 – 0.0548 AWFW = 0.219 to 0.22(0)
(iii)
P ( X = 245 ) = 0 or zero or impossible 98% (0.98) ⇒
(b)
z=
z = –2.05 to –2.06
σ
2
B1
1
σ
Ignore any working B0 for ‘for impossible to calculate’
B1
AWFW; ignore sign
M1
Standardising 245 with 253 and σ ; allow (253 – 245)
245 − 253
Only allow:
= –2.0537
Note: 245 − 253
A1
A1
σ = 3.88 to 3.9(0)
(0.21945)
M1 A0 for [1 – (i)] – 0.0548 = 0.67095 M0 A0 for 0.9452 – [(i)] = 0.67095 M1 A1 for 0.9452 – [1 – (i)] = 0.21945
A1
AWFW
(–2.0537)
±2.05 to ±2.06 ±2.32 to ±2.33 (3.8954)
= 2.0537 ⇒ σ = 3.8954 ⇒ B1 M1 A1 A0 Total
6
4 10
Or equivalent inconsistent signs
MS/SS1B/W - AQA GCE Mark Scheme 2009 June series
MS/SS1B/W (cont) Q Solution 4(a) b (gradient) = –0.5485 to – 0.5475 b (gradient) = –0.55 to – 0.54 Omission of –ve sign
Marks B2 (B1) (B0)
a (intercept) = 49.7 to 49.9 a (intercept) = 49 to 50
Total
B2 (B1)
Comments
AWFW AWFW
(–0.54814)
AWFW AWFW
(49.7982)
or
Attempt at
∑ x ∑ x2 ∑ y
&
∑ xy ( ∑ y 2 )
305 14975 281 & 6980 (10173) (all 4 attempted) (M1)
or
4638.89 & –2542.78 (both attempted)
Attempt at S xx & S xy Attempt at correct formula for b (gradient) b (gradient) = –0.5485 to –0.5475 a (intercept) = 49.7 to 49.9
(m1) (A1) (A1)
4
If a and b not identified anywhere in question, then: –0.5485 to –0.5475 ⇒ B1 49.7 to 49.9 ⇒ B1
Accept a & b interchanged only if identified correctly by a clearly shown equation (stated answers are not sufficient) in (b) or (c) (b)
C’s value of intercept from (a) providing > 47 or Value 50 stated even if (a) incorrect or not attempted
(c)
AWFW AWFW
B1F
13 weeks ⇒ 91 days
B1
y = –1.1 to + 1.1
B1
1
Accept value rounded to nearest integer (50) Stated or used Accept a descriptive answer that includes 91 and a value in range AWFW (–0.08254)
or
y = 0 ⇒ x = 89 to 93
(B1)
⇒ 13 weeks (approximately) Note: B1 B1 or (B1) (B1) are available even if (a) not attempted
AWFW (90.84942) Accept a descriptive answer that includes a value in range and 13 Stated
(B1)
Or equivalent; ignore reasoning unless contradictory Dependent upon 2nd B1 in (c) or 2nd (B1) in (c)
Thus claim appears justified B1 dep
or
Thus tablet likely to have dissolved or
Extrapolation required so cannot comment
(B1)
Note: If (B1) for extrapolation maximum mark is 2; other mark available is for 91
Not dependent
3
7
MS/SS1B/W - AQA GCE Mark Scheme 2009 June series
Total
8
8
MS/SS1B/W - AQA GCE Mark Scheme 2009 June series
MS/SS1B/W (cont) Q 5(a) (i) Median (50) = 3
Solution
Marks B1
Total
CAO Do not award marks if correct answers are based on shown incorrect method; eg accept use of 99/2, etc but not 276/2, etc CAO; but 25th value ⇒ IQR = 2 ⇒ B0
If not identified, then assume order is median then IQR IQR (75 – 25) = 4 – 2 = 2
Special Cases: Identification that LQ = 2 and UQ = 4
B2
Both CAO
(B1)
Statement of ≥ 4 cumulative frequencies F: 14 49 74 87 96 98 99
(M1) 3
(ii)
Mean =
∑ fx = 275 = 2.77 to 2.78 99 ∑f
If not identified, assume order is x then s SD ( ∑ fx 2 = 933) = 1.3(0) to 1.32
Comments
Can award if no marks scored in (i) even if then applied to continuous data
AWFW
B1
(2.778)
Treat rounding to integers as ISW B2
AWFW
(1.307 & 1.314)
Special Case:
Evidence of (b)(i)
(ii)
Mean163 =
∑ fx
(M1)
3
Can award if no marks scored in (ii)
99 99 × Mean 99 or 163
∑ fx from(a)(ii)
Or equivalent; may be implied by an answer within range
M1
163 = 1.68 to 1.69
A1
Increase
B1
2
AWFW
(1.687)
CAO; or equivalent Ignore any working
(1.696) (1.702)
1 (iii)
Data is (positively/negatively) skewed / not symmetric / bimodal / not bell-shaped from frequency distribution / given table or
Or equivalent
B1
[C’s mean in (b)(i)] – 2 × [C’s SD in (a)(ii)] < 0
or [C’s mean in (b)(i)] – 2 × [1.69 to 1.71] < 0 Thus claim appears not valid Total
9
(–1.75 to –0.90) B1 dep
2 11
Or equivalent Dependent upon previous B1
MS/SS1B/W - AQA GCE Mark Scheme 2009 June series
MS/SS1B/W (cont) Q Solution 6(a) 470 Mean = = 47 10
Marks
98% (0.98) ⇒ z = 2.32 to 2.33
σ
CI for μ is
x ± z×
Thus
47 ± 2.3263 ×
Hence
n 15 10
Total
B1
CAO
B1
AWFW
M1
Used Must have
A1F
F on x and z only
47 ± 11.0 to 11.1
n with n > 1
5 AWRT (accept 36 & 58)
(35.9 to 36.0, 58.0 to 58.1)
Y ~ N(108, 282) Variance of Y40 = 282/40 = 19.6
B1
CAO Stated or used AWFW
⎛ 120 − 108 ⎞ P (Y40 > 120 ) = P ⎜⎜ Z > ⎟ 28 40 ⎟⎠ ⎝
M1
Standardising 120 with 108 and 19.6 or (4.425 to 4.43) or equivalent; allow (108 – 120)
= P ( Z > 2.71) = 1 – P ( Z < 2.71)
m1
Area change; may be implied
√SD of Y40 = 28√40 = 4.425 to 4.43
= 1 – 0.99664 = 0.0033 to 0.0034 (c)
(2.3263)
CAO & AWRT (accept 11) A1
Or (b)
Comments
A1
4
AWFW (0.00336) (1 – answer) ⇒ B1 M1 max
Part (b) or Teleair times
B1
Or equivalent; ignore reasoning
Distribution of Y not known
B1
Or equivalent; must be clear reference to Y or population B0 for n > 30
Note: To score B1 B1 there must be both a clear indication of where in question and a valid reason
Any reference to part (a) ⇒ B0 B0
Total
10
2 11
MS/SS1B/W - AQA GCE Mark Scheme 2009 June series
MS/SS1B/W (cont) Q 7(a)
Marks
P(R < 10) = 0.791
(i) (ii)
Solution R ~ B(50, 0.15)
Total
Comments
B1
AWRT
(0.7911)
P(5 ≤ R ≤ 10) = 0.8801 or 0.7911
( p1 )
M1
minus 0.1121 or 0.2194
(p2)
M1
Accept 3 dp accuracy (1 – p2) – p1 ⇒ M0 M0 A0 p1 – (1 – p2) ⇒ M1 M0 A0 only providing result > 0 Accept 3 dp accuracy
= 0.768
A1
AWRT
(0.7680)
or
B(50, 0.15) expressions stated for at least 3 terms within 4 ≤ R ≤ 10 gives probability = 0.768 (b) (i)
(iii)
(iv)
(A2)
4
Confusion of 22, 35, 120 and/or 0.15, 0.06
AWRT Do not treat as misreads
S ~ B(22, 0.06) ⎛ 22 ⎞ 2 20 P(S = 2) = ⎜ ⎟ ( 0.06 ) ( 0.94 ) ⎝2⎠ = 0.24 to 0.242
(ii)
Can be implied by correct answer
(M1)
P(S ≥ 1) = 1 – q 35 where 0.84 ≤ q ≤ 0.96
M1
Used in (b)(i) as evidenced by any correct binominal term for S > 0
A1
Can be implied by correct answer Ignore any additional terms
A1
3
A1
Mean = np = 120 × 0.94 = 112.8 or 113 If not identified, assume order is μ then σ 2 Variance = np(1 – p) = 120 × 0.94 × 0.06 = 6.76 to 6.78
B1
B1
(0.24125)
Can be implied by correct answer Award for (0.94)35 seen in an expression but not if accompanied by a multiplier ≠ 1
M1 (B1)
= 0.885 to 0.89
AWFW
2
AWFW
(0.88532)
Either
2
Must clearly state variance value AWFW (6.768) Must have scored 1st B1 in (iii)
Means are (approximately) the same stated or Variances are (very) different stated
B1
Must have scored 2nd B1 in (iii)
Agree with P(sorts letter incorrectly) = 0.06
B1 dep
Dependent on ‘means same’ stated
Disagree with independent from letter to letter
B1 dep
Dependent on ‘variances different’ stated
Total Paper
11
3 14 75
MS/SS1B/W - AQA GCE Mark Scheme 2009 June series
12
klm
Scaled mark component grade boundaries - June 2009 exams GCE (legacy) Component Code Component Title
Maximum Scaled Mark
A
MATHEMATICS UNIT MS03 MATHEMATICS UNIT MS04 MATHEMATICS UNIT MS1A WRITTEN MATHEMATICS UNIT MS1A CWK MATHEMATICS UNIT MS1B MATHEMATICS UNIT MS2B MATHEMATICS UNIT XMCA2 MATHEMATICS UNIT XMCAS
75 75 75 25 75 75 125 125
61 61 60 20 62 59 88 101
53 53 52 17 54 51 76 89
45 45 44 14 46 44 64 77
38 38 36 12 39 37 53 65
31 31 29 10 32 30 42 53
MED1 MED2 MED3 MED4 MED5 MED6
MEDIA STUDIES UNIT 1 MEDIA STUDIES UNIT 2 MEDIA STUDIES UNIT 3 MEDIA STUDIES UNIT 4 MEDIA STUDIES UNIT 5 MEDIA STUDIES UNIT 6
60 60 100 60 60 60
40 42 72 41 50 43
35 36 64 36 41 37
31 31 56 32 32 31
27 26 48 28 23 25
23 21 40 24 15 20
HEB1 HEB2
MODERN HEBREW UNIT 1 MODERN HEBREW UNIT 2
100 100
65 70
57 61
50 52
43 44
36 36
MUS1 MUS2 MUS3 MUS4 MUS5 MUS6
MUSIC UNIT 1 MUSIC UNIT 2 MUSIC UNIT 3 MUSIC UNIT 4 MUSIC UNIT 5 MUSIC UNIT 6
100 60 60 120 80 40
63 46 49 78 67 36
54 40 44 70 63 33
45 34 39 62 59 30
36 28 34 55 55 27
27 23 29 48 51 25
PAN1 PAN2
PANJABI UNIT 1 PANJABI UNIT 2
100 100
74 77
64 66
55 55
46 45
37 35
MS03 MS04 MS/SS1A/W MS/SS1A/C MS1B MS2B XMCA2 XMCAS
Scaled Mark Grade Boundaries B C D
E
