General Certificate of Education Advanced Subsidiary Examination June 2012
Mathematics
MS/SS1B
Unit Statistics 1B
Statistics Unit Statistics 1B Friday 18 May 2012
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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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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 each question in the space provided for that question. If you require extra space, use an AQA supplementary answer book; do not use the space provided for a different question. * 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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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. * You do not necessarily need to use all the space provided.
P50188/Jun12/MS/SS1B 6/6/6/
MS/SS1B
2
A production line in a rolling mill produces lengths of steel.
1
A random sample of 20 lengths of steel from the production line was selected. The minimum width, x centimetres, and the minimum thickness, y millimetres, of each selected length was recorded. The following summarised information was then calculated from these records. Sxx ¼ 2:030
Syy ¼ 1:498
Sxy ¼ 0:410
(a)
Calculate the value of the product moment correlation coefficient between x and y. (2 marks)
(b)
Interpret your value in the context of the question.
2
(2 marks)
Katy works as a clerical assistant for a small company. Each morning, she collects the company’s post from a secure box in the nearby Royal Mail sorting office. Katy’s supervisor asks her to keep a daily record of the number of letters that she collects. Her records for a period of 175 days are summarised in the table. Daily number of letters (x) 0–9
5
10 –19
16
20
23
21
27
22
31
23
34
24
16
25 –29
10
30 –34
5
35 –39
3
40 –49
4
50 or more
1
Total
(02)
Number of days (f )
175
P50188/Jun12/MS/SS1B
3
For these data:
(a) (i)
state the modal value;
(1 mark)
(ii) determine values for the median and the interquartile range.
(3 marks)
(b)
The most letters that Katy collected on any of the 175 days was 54. Calculate estimates of the mean and the standard deviation of the daily number of letters collected by Katy. (4 marks)
(c)
During the same period, a total of 280 letters was also delivered to the company by private courier firms. Calculate an estimate of the mean daily number of all letters received by the company during the 175 days. (2 marks)
The table shows the maximum weight, yA grams, of Salt A that will dissolve in 100 grams of water at various temperatures, x C.
3
x
10
15
20
25
30
35
40
45
50
60
70
80
yA
20
35
48
57
77
92
101 111 121 137 159 182
(a)
Calculate the equation of the least squares regression line of yA on x.
(b)
The data in the above table are plotted on the scatter diagram on page 4. Draw your regression line on this scatter diagram.
(4 marks)
(2 marks)
For water temperatures in the range 10 C to 80 C, the maximum weight, yB grams, of Salt B that will dissolve in 100 grams of water is given by the equation
(c)
yB ¼ 60:1 þ 0:255x (i)
Draw this line on the scatter diagram.
(2 marks)
(ii) Estimate the water temperature at which the maximum weight of Salt A that will
dissolve in 100 grams of water is the same as that of Salt B.
(1 mark)
(iii) For Salt A and Salt B, compare the effects of water temperature on the maximum
weight that will dissolve in 100 grams of water. Your answer should identify two distinct differences. (2 marks)
s
(03)
Turn over
P50188/Jun12/MS/SS1B
4
Temperatures and Maximum Weights y~ 200 ±
6
180 ±
6
160 ±
6
120 ±
100 ±
6
60 ±
6 40 ±
10
20
30
40
50
60
70
±
±
±
±
±
6 ±
±
6
~
0± 0
6
6
±
20 ±
6
6
6
80 ±
±
Maximum weight (grams)
140 ±
80 x
Temperature (C)
(04)
P50188/Jun12/MS/SS1B
5
A survey of the 640 properties on an estate was undertaken. Part of the information collected related to the number of bedrooms and the number of toilets in each property.
4
This information is shown in the table. Number of toilets 2
1
46
14
0
0
60
2
24
67
23
0
114
3
7
72
99
16
194
4
0
19
123
48
190
5 or more
0
0
11
71
82
Total
77
172
256
135
640
Number of bedrooms
3
4 or more
1
Total
A property on the estate is selected at random.
(a)
Find, giving your answer to three decimal places, the probability that the property has: (i)
exactly 3 bedrooms;
(1 mark)
(ii) at least 2 toilets;
(2 marks)
(iii) exactly 3 bedrooms and at least 2 toilets;
(2 marks)
(iv) at most 3 bedrooms, given that it has exactly 2 toilets.
(3 marks)
(b)
Use relevant answers from part (a) to show that the number of toilets is not independent of the number of bedrooms. (2 marks)
(c)
Three properties are selected at random from those on the estate which have exactly 3 bedrooms. Calculate the probability that one property has 2 toilets, one has 3 toilets and the other has at least 4 toilets. Give your answer to three decimal places. (4 marks)
s
(05)
Turn over
P50188/Jun12/MS/SS1B
6
A general store sells lawn fertiliser in 2.5 kg bags, 5 kg bags and 10 kg bags.
5
The actual weight, W kilograms, of fertiliser in a 2.5 kg bag may be modelled by a normal random variable with mean 2.75 and standard deviation 0.15 .
(a)
Determine the probability that the weight of fertiliser in a 2.5 kg bag is: (i)
less than 2.8 kg;
(ii) more than 2.5 kg.
(5 marks)
The actual weight, X kilograms, of fertiliser in a 5 kg bag may be modelled by a normal random variable with mean 5.25 and standard deviation 0.20 .
(b)
(i)
Show that Pð5:1 < X < 5:3Þ ¼ 0:372 , correct to three decimal places.
(2 marks)
(ii) A random sample of four 5 kg bags is selected. Calculate the probability that none
of the four bags contains between 5.1 kg and 5.3 kg of fertiliser.
(2 marks)
The actual weight, Y kilograms, of fertiliser in a 10 kg bag may be modelled by a normal random variable with mean 10.75 and standard deviation 0.50 .
(c)
A random sample of six 10 kg bags is selected. Calculate the probability that the mean weight of fertiliser in the six bags is less than 10.5 kg. (4 marks)
A bin contains a very large number of paper clips of different colours. The proportion of each colour is shown in the table.
6
Colour
White
Yellow
Green
Blue
Red
Purple
Proportion
0.15
0.15
0.20
0.15
0.22
0.13
Packets are filled from the bin. Each filled packet contains exactly 30 paper clips which may be considered to be a random sample.
(a)
Use binomial distributions to determine the probability that a filled packet contains: (i)
exactly 2 purple paper clips;
(3 marks)
(ii) a total of more than 10 red or purple paper clips;
(3 marks)
(iii) at least 5 but at most 10 green paper clips.
(3 marks)
(06)
P50188/Jun12/MS/SS1B
7
Jumbo packets are also filled from the bin. Each filled jumbo packet contains exactly 100 paper clips.
(b)
(i)
Assuming that the number of paper clips in a jumbo packet may be considered to be a random sample, calculate the mean and the variance of the number of red paper clips in a filled jumbo packet. (2 marks)
(ii) It is claimed that the proportion of red paper clips in the bin is greater than 0.22 and
that jumbo packets do not contain random samples of paper clips. An analysis of the number of red paper clips in each of a random sample of 50 filled jumbo packets resulted in a mean of 22.1 and a standard deviation of 4.17 . Comment, with numerical justification, on each of the two claims.
(3 marks)
The volume of bleach in a 5-litre bottle may be modelled by a random variable with a standard deviation of 75 millilitres.
7
The volume, in litres, of bleach in each of a random sample of 36 such bottles was measured. The 36 measurements resulted in a total volume of 181.80 litres and exactly 8 bottles contained less than 5 litres. (a)
Construct a 98% confidence interval for the mean volume of bleach in a 5-litre bottle. (5 marks)
(b)
It is claimed that the mean volume of bleach in a 5-litre bottle exceeds 5 litres and also that fewer than 10 per cent of such bottles contain less than 5 litres. Comment, with numerical justification, on each of these two claims.
(c)
(3 marks)
State, with justification, whether you made use of the Central Limit Theorem in (1 mark) constructing the confidence interval in part (a).
Copyright ª 2012 AQA and its licensors. All rights reserved.
(07)
P50188/Jun12/MS/SS1B
Version 1.0
General Certificate of Education (A-level) June 2012
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 students’ 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 students’ 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 students’ 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 © 2012 AQA and its licensors. All rights reserved. Copyright AQA retains the copyright on all its publications. However, registered schools/colleges 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 schools/colleges 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.
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.
MS/SS1B - AQA GCE Mark Scheme 2012 June Series
MS/SS1B Q 1 (a)
Solution
r
S xy S xx S yy
Marks M1
0.410
=
Total
Correct substitution into correct formula May be implied by a correct answer
= –0.235
2.030 1.498
A1
2
between width and thickness of lengths of steel SC
Adep1
B1
(–0.235115)
2
Context; do not allow ‘cms’ or ‘mms’
r = (+)0.235 M1 A0 Adep0 B1 max
Total
Q 2 (a)(i)
AWRT
Dependent on –0.235 or –0.24 OE; must qualify strength and state negative Ignore extra words unless contradict Not ‘no’, ‘low’, ‘small’, ‘unlikely’ or ‘trend’
(b) Some / (very) weak / (very) little / (very)slight negative correlation/relationship/association/link
Comments
Solution
(ii)
4
Marks
Total
Comments
Mode = 23
B1
1
Median (88th value) = 22
B1
CAO
B1
CAO; either May be implied by IQR = 3
CAO
Upper quartile (132nd value) = 23 Lower quartile (44 value) = 20 th
Interquartile range = 3 (b)
SC
(c)
Mean =
22.3
B1
3
CAO; but only award B1 (22.3) if incorrect mid-points or fx seen AWFW (fx = 3902.5)
B2
Mean = 21 to 23
(B1)
Standard deviation = 6.37 or 6.39 Standard deviation = 5 to 7
B2 (B1)
CAO; do not award if seen to be not based on 23 and 20
4
AWRT AWFW
(s = 6.391 = 6.372) (fx2 = 94132.25)
Only if B0 B0 or B1 B0 then award as follows but only up to a maximum total part mark of 2 1 At least 2 correct mid-points 4.5, 14.5, 27, 32, 37, 44.5, 54 seen M1 2 Clear use of fx/(175 or 174) M1
Mean = (c’s mean from (b)) +
280 175
Adding (1.6 or equivalent) CAO to (c’s mean from (b)) or to (c’s new mean)
M1
= 22.3 + 1.6 Mean = 23.9 Total
AF1
2 10
F on (c’s mean from (b)) or on (c’s new mean)
MS/SS1B - AQA GCE Mark Scheme 2012 June Series
MS/SS1B (cont) Q 3 (a)
Solution
Marks
Total
Comments
b (gradient) = 2.27 b (gradient) = 2.2 to 2.3
B2 (B1)
AWRT AWFW
a (intercept) = 4.16 to 4.2 a (intercept) = 3 to 7
B2 (B1)
AWFW AWFW
(2.27075)
Treat rounding of correct answers as ISW
Attempt at
2 x x y & xy
or Attempt at S xx & S xy
y
(b)
480 24500 1140 & 57635 (135908) (all 4 attempted)
2
(M1)
S
5300 & 12035 (27608) (both attempted)
yy
Attempt at correct formula for b (gradient) b (gradient) = 2.27 a (intercept) = 4.16 to 4.2 Notes
(m1) (A1) (A1)
4
AWRT AWFW
1 Values of a and b interchanged and equation y = ax + b stated in (a) max of 4 marks 2 Values of a and b interchanged and equation y = a + bx stated in (a) 0 marks 3 Values are not identified or simply a = # and b = #, then 2.2 to 2.3 B1 and 3 to 7 B1 but accept, for example, as identification, [a = #, b = # with y = a + bx but no substitution for a & b] or [intercept(a) = #, gradient(b) = #] 4 b = 2407/1060 CAO B2, otherwise B1 if fraction equates to 2.2 to 2.3 (Notes 1, 2 & 3 also apply) a = 221/53 CAO B2, otherwise B1 if fraction equates to 3 to 7 (Notes 1, 2 & 3 also apply) 5 Some/all of marks can be scored in (b), and in c(ii) & (iii), even if some/all of marks are lost in (a) but marks lost in (a) cannot be recouped by subsequent working in (b)
Correct straight line drawn on scatter diagram Correct shortened and/or freehand line drawn on scatter diagram
B2 (B1)
2
Line must go from x ≤ 20 to x ≥ 70 and fall between the following 2 lines: Lower: (10, 25) (80, 180) Upper: (10, 30) (80, 190)
Notes
1 If B0 but seen correct attempt at ≥2 points even if incorrectly evaluated M1 2 If B0 but no seen evidence to support ≥2 points (correct or incorrect) marked on scatter diagram M0
(c)(i)
Correct straight line drawn on scatter diagram Correct shortened and/or freehand line drawn on scatter diagram
Notes
(4.16981)
B2 (B1)
2
Line must go from x ≤ 20 to x ≥ 70 and fall between the following 2 lines: Lower: (10, 60) (80, 75) Upper: (10, 65) (80, 85)
1 If B0 but seen correct attempt at ≥2 points even if incorrectly evaluated M1 2 If B0 but no seen evidence to support ≥2 points (correct or incorrect) marked on scatter diagram M0
(ii)
27 to 29
B1
(iii)
At low temperatures more B (than A) dissolves At high temperatures more A (than B) dissolves
B1
Amount increases more rapidly for A (than B) Amount increases more slowly for B (than A)
B1
Total
1
AWFW (calculation 27.75) Must clearly identify x-value Thus (27 to 29, y-value) B0 Either; OE (eg a comparison using lines and/or data at a specific temperature but not at 0C)
2
11
Either; OE
Any comments about b or a B0 Comment about ‘rate’ must relate to temp
MS/SS1B - AQA GCE Mark Scheme 2012 June Series
MS/SS1B (cont) Q 4 (a)(i)
(ii)
Solution
Marks
Total
P(B = 3) = 194/640 or 97/320 or 0.303 or 30.3% P(T ≥ 2) =
172 256 135 77 563 or 1 or 640 640 640 = 563/640
B1
1
= 187/640 or 0.292 or 29.2% (iv)
P(B ≤ 3 | T = 2) = 14 67 72
172
or
172 19 153 or 172 172
(0.303125)
CAO A1
P(B = 3 & T ≥ 2) = 72 99 16 194 7 187 or or 640 640 640
CAO or AWRT
M1 2
or (0.879 to 0.88) or (87.9% to 88%) (iii)
Comments Ratios (eg 194:640) are only penalised by 1 accuracy mark at first correct answer
AWFW
(0.879688)
CAO or AWRT
(0.292188)
M1 A1
2
Correct numerator (accept both 640) Correct denominator
M1 M1
= 153/172
CAO A1
3 AWFW
or (0.888 to 0.89) or (88.8% to 89%)
(0.889535)
Answers as fractions, percentages or ratios lose accuracy (A & B) marks in (b) & (c)
(a)(i) × (a)(ii) (a)(iii)
(b)
M1
Attempted
since 0.303 × 0.88 = 0.265 to 0.27 0.292 SC
(c)
Notes
A1
2
AWFW & AWRT
Any correct fully-explained reasoning, using other than answers from part (a), which results in an inequality () with both sides as numerically correct decimals (to 3 dp) B1 (eg P(B = 3) = 0.303 P(B = 3| T = 2) = 72/172 = 0.419) but no/unclear/incomplete reasoning or no/incorrect/incomplete numerical work B0
M1 M1
Correct 3 values multiplied in numerator Correct 3 values multiplied in denominator 0.371 × 0.513 × 0.083 (all AWRT) M1 M1 (OE products)
abc multiplied by 6 or 3
M1
0 < (a, b & c) < 1
= 0.095 to 0.0952
A1
72 99 16 P(2T 3T ≥4T | B = 3) = 194 193 192
4
AWFW
1 Incorrect answer with no working 0 marks 2 The 3 correct fractions/decimals identified but not multiplied (eg added) M1 M0 M0 A0 3 The 3 correct fractions/decimals identified together with 0.016 (AWRT) M1 M1 M0 A0 4 A denominator of
194
C3 = 1198144 M2 (2nd & 3rd M1 marks) Total
14
(0.095187)
MS/SS1B - AQA GCE Mark Scheme 2012 June Series
MS/SS1B (cont) Q 5 (a) (i)
(ii)
Solution
Marks
Weight, W ~ N(2.75, 0.152) 2.8 2.75 P(W < 2.8) = P Z 0.15
M1
Standardising 2.8 with 2.75 and 0.15; allow (2.75 – 2.8)
= P(Z < 0.33 or 1/3)
A1
AWRT/CAO; ignore inequality and sign May be implied by a correct answer
= 0.629 to 0.633
A1
AWFW
M1
Correct area change May be implied by a correct answer or an answer > 0.5
P(W > 2.5) = P(Z > –1.67) = P(Z < +1.67)
A1
(b)
Weight, X ~ N(5.25, 0.202)
(i)
P(5.1 < X < 5.3) = P(Z < 0.25) – P(Z < –0.75) = 0.59871 MINUS [(1 – 0.77337) or 0.22663] = 0.372(08)
B1 B1
P(0 in 4) = [1 – 0.372]4
M1
= 0.6284 = 0.155 to 0.156 (c)
Comments In (a)(i) & (c), ignore the inclusion of a lower limit of 0; it has no effect on either answer
= 0.951 to 0.953
(ii)
Total
A1
5
2
AWFW
(0.63056)
(0.95221)
Must have diff of 2 probs for each B1 Accept 0.599 Accept 0.773 or 0.227 AG; do not mark simply on answer Accept [1 – c’s (b)(i)]4
2
AWFW
(0.15554)
Weight, Y ~ N(10.75, 0.502) Variance of Y6 = 0.52/6 = 0.0416 to 0.0417 or Sd of Y6 = 0.5/6 = 0.204
B1
CAO or AWFW Stated or used CAO or AWRT
10.5 10.75 P Y6 10.5 = P Z = 0.0416
M1
Standardising 10.5 with 10.75 and 0.0416 OE; allow (10.75 – 10.5)
m1
Correct area change May be implied by a correct answer or an answer < 0.5; but do not award for use of z = 0.22
P(Z < –1.22) = 1 – P(Z < 1.22) =
1 – (0.88877 to 0.89065) = 0.109 to 0.112 Total
A1
4 13
AWFW (0.11034) (1 – answer) B1 M1 max
MS/SS1B - AQA GCE Mark Scheme 2012 June Series
MS/SS1B (cont) Q 6 (a)(i)
Solution
Marks
U ~ B(30, 0.13, 0.35 or 0.20)
M1
Used correctly anywhere in (a)
A1
Can be implied by a correct answer
30 2 28 P(P = 2) = 0.13 0.87 2 = 0.148 to 0.15
p = 0.35
AWFW
(0.1489)
CAO
P(R P > 10) = 1 – (0.5078 or 0.3575)
M1
Requires ‘1 –’ Accept 3 dp rounding or truncation Can be implied by 0.49 to 0.493 but not by 0.642 to 0.643
= 0.49 to 0.493
A1
3
AWFW
(0.4922)
P(5 ≤ G ≤ 10) = 0.9744 or 0.9389
(p1)
M1
Accept 3 dp rounding or truncation
MINUS 0.2552 or 0.4275
(p2)
M1
Accept 3 dp rounding or truncation
= 0.719 to 0.72 (p3)
A1
Notes
1 p3 ≤ 0 or p3 ≥ 1 M0 M0 A0 2 p2 – p1 M0 M0 A0 3 (1 – p2) – p1 M0 M0 A0
(b)(i)
Mean or = 100 × 0.22 Variance or 2 = 100 × 0.22 × 0.78
(ii)
3
Comments
B1
(ii)
(iii)
A1
Total
3
AWFW
(0.7192)
4 p1 – (1 – p2) M1 M0 A0 5 p1 × p2 M1 M0 A0 6 (1 – p2) – (1 – p1) M1 M1 (A1)
= 22
B1
= 17.1 to 17.2
B1
22.1 /= 22 or means similar/equal or 0.221 /= 0.22 or proportions similar/equal so reject claim (that p > 0.22) or accept that p = 0.22
CAO 2
AWFW (ignore notation) ISW all subsequent working
(17.16)
Dependent on 22 seen in (b)(i) or (ii) Accept diff = 0.1 CAO Correct (numerical) comparison with correct conclusion (even if at end and stated as ‘reject (both) claims’)
B1
17.1 to 17.2 = 4.13 to 4.15 /= 4.17
Comparison using two values or one value + diff (0.02 to 0.04 AWFW) B1
or 17.1 to 17.2 /= 17.3 to 17.4 so reject claim that not random samples or accept that are random samples Total
Comparison using two values or one value + diff (0.1 to 0.3 AWFW)
Bdep1
3
14
Dependent on previous B1 Correct conclusion regarding randomness of sample
MS/SS1B - AQA GCE Mark Scheme 2012 June Series
MS/SS1B (cont) Q 7 (a)
Solution
x
CI for is
Thus
Hence
181.8 = 36
Marks
Total
5.05 or 5050
B1
CAO
98% (0.98) z = 2.32 to 2.33
B1
AWFW
M1
Used with z (2.05 to 2.58), x (5.05, 5050 or 181.8), (0.0075, 0.075, 0.75, 7.5 or 75) and n with n > 1
A1
z (2.05 to 2.06 or 2.32 to 2.33 or 2.57 to 2.58), x (5.05) & (0.075) or x (5050) & (75) and 36 or 35
x z
n
5.05 2.3263
0.075 36
5.05 0.03 or 5050 30
Adep1
OR
5
(5.02, 5.08) or (5020, 5080) Note (b)
Clear correct comparison of 5 or 5000 with LCL or CI so agree with (first) claim (about mean) or
CAO/AWRT Dependent on previous A1 so can be scored with z 2.32 to 2.33 Ignore (absence of) quoted units AWRT to 3sf accuracy
Adep1
Dependent on Adep1 in (a) Must use consistent units
B1
Mention of a value on LHS and a value on RHS
8 v 3.6 (3 to 4)
so 8/36 OE >/≠ 1/10 OE so disagree with (second) claim (about individuals)
(c)
(2.3263)
Use of t(2.43 to 2.72) B1 B0 M1 A0 A0 max
(8/36 or 0.22 or 22%) v (1/10 or 0.10 or 10%)
Notes
Comments
Bdep1
3
Dependent on B1 Explicit comparison of values and correct conclusion
1 It/(claimed) mean/(claimed) value < LCL/CI Adep0 Must indicate 5 or 5000 2 98% have (mean) weights between CLs so ... Adep0 3 Any reference to CI for second claim B0 Bdep0 Claim refers to individual bottles
Yes because volumes/bleach/litres/bottles/ (parent) population are not (stated as) normally distributed Total TOTAL
B1
1
9 75
OE; but do not accept ‘data’ or ‘sample’ or ‘it’ Reference to sample size only B0 (eg n > 25 or n > 30)
Scaled mark unit grade boundaries - June 2012 exams A-level
Code
Title
Max. Scaled Mark
Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E
MS/SS1A/W
MATHEMATICS UNIT S1A - WRITTEN
75
56
33
MS/SS1A/C
MATHEMATICS UNIT S1A - COURSEWORK
25
20
10
MS1B
MATHEMATICS UNIT MS1B
75
-
55
49
43
37
32
MS2B
MATHEMATICS UNIT MS2B
75
68
62
55
48
41
34
MS03
MATHEMATICS UNIT MS03
75
66
59
52
45
38
31
MS04
MATHEMATICS UNIT MS04
75
67
59
51
43
35
28
MEST1
MEDIA STUDIES UNIT 1
80
-
54
47
41
35
29
MEST2
MEDIA STUDIES UNIT 2
80
-
63
54
45
36
28
MEST3
MEDIA STUDIES UNIT 3
80
71
62
51
40
29
19
MEST4
MEDIA STUDIES UNIT 4
80
74
68
56
45
34
23
MHEB1
MODERN HEBREW UNIT 1
100
-
64
58
52
46
40
MHEB2
MODERN HEBREW UNIT 2
100
78
70
62
54
47
40
MUSC1
MUSIC UNIT 1
80
-
53
47
41
36
31
MUS2A
MUSIC UNIT 2A
60
-
48
42
36
30
25
MUS2B
MUSIC UNIT 2B
60
-
49
43
37
32
27
MUS2C
MUSIC UNIT 2C
60
-
50
44
39
34
29
MUSC3
MUSIC UNIT 3
80
-
71
65
59
54
49
MUSC4
MUSIC UNIT 4
100
78
68
62
56
50
45
