General Certificate of Education
Advanced Subsidiary Examination
June 2011
Mathematics
MS/SS1B
Unit Statistics 1B
Statistics
Unit Statistics 1B
Friday 20 May 2011
d
1.30 pm to 3.00 pm
e
For this paper you must have:
*
the blue AQA booklet of formulae and statistical tables.
You may use a graphics calculator.
s
n
Time allowed
*
1 hour 30 minutes
e
Instructions
*
Use black ink or black ball-point pen. Pencil should only be used for
drawing.
*
Fill in the boxes at the top of this page.
*
Answer all questions.
*
Write the question part reference (eg (a), (b)(i) etc) in the left-hand
margin.
*
You must answer the questions in the spaces provided. Do not write
outside the box around each page.
*
Show all necessary working; otherwise marks for method may be
lost.
*
Do all rough work in this book. Cross through any work that you do
not want to be marked.
*
The final answer to questions requiring the use of tables or
calculators should normally be given to three significant figures.
d
n
o
C
Information
*
The marks for questions are shown in brackets.
*
The maximum mark for this paper is 75.
*
Unit Statistics 1B has a written paper only.
Advice
*
Unless stated otherwise, you may quote formulae, without proof,
from the booklet.
P38946/Jun11/MS/SS1B 6/6/6/
MS/SS1B
2
The number of matches in each of a sample of 85 boxes is summarised in the table.
1
Number of matches
Number of boxes
Less than 239
1
239–243
1
244–246
2
247
3
248
4
249
6
250
10
251
13
252
16
253
20
254
5
255–259
3
More than 259
1
Total
85
For these data:
(a)
(i)
state the modal value;
(ii) determine values for the median and the interquartile range.
(1 mark)
(3 marks)
Given that, on investigation, the 2 extreme values in the above table are 227
and 271 :
(b)
(i)
calculate the range;
(ii) calculate estimates of the mean and the standard deviation.
(c)
(02)
(1 mark)
(4 marks)
For the numbers of matches in the 85 boxes, suggest, with a reason, the most
appropriate measure of spread.
(2 marks)
P38946/Jun11/MS/SS1B
3
The diameter, D millimetres, of an American pool ball may be modelled by a normal
random variable with mean 57.15 and standard deviation 0.04 .
2
Determine:
(a)
(i)
PðD < 57:2Þ ;
(3 marks)
(ii) Pð57:1 < D < 57:2Þ .
(2 marks)
A box contains 16 of these pool balls. Given that the balls may be regarded as a
random sample, determine the probability that:
(b)
(i)
all 16 balls have diameters less than 57.2 mm;
(2 marks)
(ii) the mean diameter of the 16 balls is greater than 57.16 mm.
(4 marks)
During a particular summer holiday, Rick worked in a fish and chip shop at a seaside
resort.
3 (a)
He suspected that the shop’s takings, £y, on a weekday were dependent upon the
forecast of that day’s maximum temperature, x C, in the resort, made at 6.00 pm on
the previous day.
To investigate this suspicion, he recorded values of x and y for a random sample
of 7 weekdays during July.
(i)
x
23
18
27
19
25
20
22
y
4290
3188
5106
3829
5057
4264
4485
Calculate the equation of the least squares regression line of y on x.
(4 marks)
(ii) Estimate the shop’s takings on a weekday during July when the maximum
temperature was forecast to be 24 C.
(2 marks)
(iii) Explain why your equation may not be suitable for estimating the shop’s takings on a
weekday during February.
(1 mark)
(iv) Describe, in the context of this question, a variable other than the maximum
temperature, x, that may affect y.
(1 mark)
Seren, who also worked in the fish and chip shop, investigated the possible linear
relationship between the shop’s takings, £z, recorded in £000s, and each of two other
explanatory variables, v and w.
(b)
(i)
She calculated correctly that the regression line of z on v had a z-intercept of 1 and
a gradient of 0.15 .
Draw this line, for values of v from 0 to 40, on Figure 1 on page 4.
s
(03)
Turn over
P38946/Jun11/MS/SS1B
4
(ii) She also calculated correctly that the regression line of z on w had a z-intercept of 5
and a gradient of 0.40 .
Draw this line, for values of w from 0 to 10, on Figure 2 below.
(3 marks)
Figure 1
z~
6–
5–
4–
3–
2–
1–
10
20
30
40
~
0–
0
v
1 –
Figure 2
z~
6–
5–
4–
3–
2–
1–
2
4
6
8
10
~
0–
0
w
1 –
(04)
P38946/Jun11/MS/SS1B
5
Rice that can be cooked in microwave ovens is sold in packets which the
manufacturer claims contain a mean weight of more than 250 grams of rice.
4
The weight of rice in a packet may be modelled by a normal distribution.
A consumer organisation’s researcher weighed the contents, x grams, of each of a
random sample of 50 packets. Her summarised results are:
X
x ¼ 251:1 and
ðx xÞ2 ¼ 184:5
(a)
Show that, correct to two decimal places, s ¼ 1:94 , where s 2 denotes the unbiased
estimate of the population variance.
(1 mark)
(b) (i)
Construct a 96% confidence interval for the mean weight of rice in a packet, giving
the limits to one decimal place.
(4 marks)
(ii) Hence comment on the manufacturer’s claim.
(2 marks)
The statement ‘250 grams’ is printed on each packet.
(c)
Explain, with reference to the values of x and s, why the consumer organisation may
consider this statement to be dubious.
(2 marks)
Emma visits her local supermarket every Thursday to do her weekly shopping.
5 (a)
The event that she buys orange juice is denoted by J , and the event that she buys
bottled water is denoted by W . At each visit, Emma may buy neither, or one, or
both of these items.
(i)
Complete the table of probabilities, printed below, for these events, where J ’ and W ’
denote the events ‘not J ’ and ‘not W ’ respectively.
(3 marks)
(ii) Hence, or otherwise, find the probability that, on any given Thursday, Emma buys
either orange juice or bottled water but not both.
(2 marks)
(iii) Show that:
(A) the events J and W are not mutually exclusive;
(B) the events J and W are not independent.
Turn over
s
(05)
(3 marks)
P38946/Jun11/MS/SS1B
6
Rhys visits the supermarket every Saturday to do his weekly shopping. Items that he
may buy are milk, cheese and yogurt.
(b)
The
The
The
The
probability, PðMÞ , that he buys milk on any given Saturday is 0.85 .
probability, PðCÞ , that he buys cheese on any given Saturday is 0.60 .
probability, PðY Þ , that he buys yogurt on any given Saturday is 0.55 .
events M, C and Y may be assumed to be independent.
Calculate the probability that, on any given Saturday, Rhys buys:
none of the 3 items;
(2 marks)
(ii) exactly 2 of the 3 items.
(3 marks)
(i)
J
J’
0.65
W
W’
Total
0.15
Total
0.30
1.00
An amateur tennis club purchases tennis balls that have been used previously in
professional tournaments.
6
The probability that each such ball fails a standard bounce test is 0.15 .
The club purchases boxes each containing 10 of these tennis balls. Assume that the
10 balls in any box represent a random sample.
Determine the probability that the number of balls in a box which fail the bounce
test is:
(a)
at most 2 ;
(1 mark)
(ii) at least 2 ;
(2 marks)
(iii) more than 1 but fewer than 5 .
(3 marks)
(i)
Determine the probability that, in 5 boxes, the total number of balls which fail the
bounce test is:
(b)
(i)
more than 5 ;
(ii) at least 5 but at most 10 .
(06)
(2 marks)
(3 marks)
P38946/Jun11/MS/SS1B
7
Three airport management trainees, Ryan, Sunil and Tim, were each instructed to
select a random sample of 12 suitcases from those waiting to be loaded onto aircraft.
7 (a)
Each trainee also had to measure the volume, x, and the weight, y, of each of the
12 suitcases in his sample, and then calculate the value of the product moment
correlation coefficient, r, between x and y.
*
*
Ryan obtained a value of 0.843 .
Sunil obtained a value of þ0.007 .
Explain why neither of these two values is likely to be correct.
(2 marks)
Peggy, a supervisor with many years’ experience, measured the volume, x cubic feet,
and the weight, y pounds, of each suitcase in a random sample of 6 suitcases, and
then obtained a value of 0.612 for r.
(b)
*
*
Ryan and Sunil each claimed that Peggy’s value was different from their values
because she had measured the volumes in cubic feet and the weights in pounds,
whereas they had measured the volumes in cubic metres and the weights in
kilograms.
Tim claimed that Peggy’s value was almost exactly half his calculated value
because she had used a sample of size 6 whereas he had used one of size 12 .
Explain why neither of these two claims is valid.
(2 marks)
Quentin, a manager, recorded the volumes, v, and the weights, w, of a random
sample of 8 suitcases as follows.
(c)
(i)
v
28.1
19.7
46.4
23.6
31.1
17.5
35.8
13.8
w
14.9
12.1
21.1
18.0
19.8
19.2
16.2
14.7
Calculate the value of r between v and w.
(ii) Interpret your value in the context of this question.
(3 marks)
(2 marks)
END OF QUESTIONS
Copyright ª 2011 AQA and its licensors. All rights reserved.
(07)
P38946/Jun11/MS/SS1B
MS/SS1B
Q
1
(a)(i)
Solution
(ii)
Marks
Total
Comments
Mode = 253
B1
1
Median = 252
B1
CAO
B1
CAO; either
May be implied by IQR = 3
CAO
Upper quartile = 253
Lower quartile = 250
Interquartile range = 3
B1
3
CAO; do not award if seen to be not
based on 253 and 250
Range = 271 – 227 = 44
B1
1
CAO; do not award if seen to be not
based on 271 and 227
(ii) Mean,
x = 251 to 251.4
Award B1 if divisor seen not to be 85 but answer
in range
B2
(b)(i)
AWFW
Note:
If B0 then can award
M1 for attempt at  fx  85 seen
Standard deviation,
 fx
= 21352
x = 251.2
Ignore notation and condone incorrect
midpoints (eg upper or lower limits
used)
s or  = 4.21 to 4.28
B2
4
AWFW
 = 4.217
 fx 2 = 5365134
s = 4.242
Award B1 if divisor seen not to be 84 or 85 but
answer in range
(c) Interquartile range (IQR)
B1
Not affected by unknown/large/small/extreme/
outlying/227 & 271 values
Bdep1
Named
2
Or equivalent
Dependent on previous B1
Only negative comments on other measures  Bdep0
More than one named  B0 Bdep0
Range  B0 Bdep0
OR
Standard deviation (s or  )
Named
(B1)
Uses all data values
Or equivalent
Dependent on previous (B1)
(Bdep1)
Only negative comments on other measures  Bdep0
Total
11
MS/SS1B (cont)
Q
2
Solution
Marks
Diameter, D ~ N(57.15, 0.042)
57.2  57.15 

P(D < 57.2) = P  Z 

0.04


(a)(i)
Total
Comments
M1
Standardising 57.2 with 57.15 and
0.04; allow (57.15 – 57.2)
= P(Z < 1.25)
A1
CAO; ignore inequality and sign
May be implied by a correct answer
= 0.894 to 0.895
A1
= p – (1 – p)
M1
= 2 × 0.89435 – 1 = 0.788 to 0.79(0)
A1
3
AWFW
(0.89435)
(ii) P(57.1 < D < 57.2)
Allow even if incorrect standardising
providing p – (1 – p) seen
May be implied by a correct answer
2
(b)(i)
P(16 balls < 57.2) = p16
0 < p < 1
M1
= [(a)(i)]16 = (0.89435)16 = 0.166 to 0.17(0)
A1
(ii)
with
2
= P(Z > 1) = 1 – P(Z < 1)
= 1 – 0.84134 = = 0.158 to 0.159
AWFW
B1
M1
Standardising 57.16 with 57.15 and
0.01 or equivalent;
allow (57.15 – 57.16)
m1
Area change
May be implied by a correct answer
or answer < 0.5
A1
4
Notes:
Ignore partial/incomplete attempts at (ii) in (i) if
followed by correct method
AWFW
(0.15866)
(1 – answer)  B1 M1 max
Mark two complete answers in (i) as
two attempts so (0 + 2)/2  1max
Answer to (i) or (ii) repeated
Mark as per scheme; thus
(2 max, 0) or (0, 4 max)
Total
(0.16754)
CAO
Stated or used (see Notes below)
CAO
If only seen in (b)(i), allow just B1
Sd of D16 = 0.04/16 = 0.01
57.16  57.15 

P  D16  57.16  = P  Z 

0.01


(0.78870)
Any probability to power 16 or 1 – p16;
do not allow multiplying factors
If only seen in (b)(ii), allow just M1
Variance of D16 = 0.042/16 = 0.0001
or
AWFW
11
MS/SS1B (cont)
Q
3
Solution
Marks
b (gradient) =
191
b (gradient) = 190 to 192
(a)
Total
B2
(B1)
Comments
CAO
AWFW
Treat rounding of correct answers as ISW
a (intercept) =
115
a (intercept) = 93 to 137
B2
(B1)
4
CAO
AWFW
OR
Attempt at
 x  x 2  y &  xy
or
Attempt at S xx & S xy
 y 
154 3452 30219 & 677042
(133170091)
(all 4 attempted)
2
(M1)
S 
12224 & 64 (2714668)
(both attempted)
yy
Attempt at correct formula for b (gradient)
b (gradient) = 191
a (intercept) = 115
(m1)
(A1)
(A1)
CAO
CAO
If a and b are not identified
anywhere in question, then:
190 to 192  B1
93 to 137  B1
Accept a & b interchanged only if identified
and used correctly in (ii)
(ii) y24 = 115 + 191 × 24
= £4699 or £4700
= £4650 to £4750
SC: (4290 + 5057)/2 = 4673 to 4674  B1
If B0 but clear evidence of correct use of c’s
equation with x = 24
(iii)
(Maximum) temperature (in February)
is likely to be/will be lower/different
B2
(B1)
2
B1
Rainfall amount/wind strength/sunshine hours/
daylight hours/opening times/day of week/
visitor numbers/public holidays/school holidays/
local attractions/etc
1
B1
1
Allow if at least 1 variable correctly identified
Total
(£4699)
(M1)
Must imply a temperature comparison with July
(iv)
Either; ignore units
AWFW
8
Or equivalent; must be clear
indication that (max) temperature is
less than/different
Extrapolation/not July/not
summer/winter/etc  B0
Or equivalent
Accept any sensible reason; do not
penalise for dubious ‘variable name’
so, for example, accept ‘rainfall’
Minimum/average temp/etc  B0
Quality or price of food/
staff/etc  B0
MS/SS1B (cont)
Q
3
Solution
Marks
Total
Comments
(b) Any line (straight, freehand, curve) from (0, –1)
on Figure 1 or from (0, 5) on Figure 2
B1
Accept clear marking of (0, –1)
or (0, 5) with no line
(i) Straight, not freehand, line from
(0, –1) to (40, 5) on F1 only; allow line
extensions and only very minor inaccuracies in
points plotted
B1
(10, 0.5)
(ii) Straight , not freehand, line from
(0, 5) to (10, 1) on F2 only; allow line
extensions and only very minor inaccuracies in
points plotted
B1
3
Notes:
Both lines on F1  B1 B1 B0 max
Both lines on F2  B1 B0 B1 max
>1 undeleted line on either F1 or F2  2 max
Total
3
(2, 4.2)
(20, 2)
(4, 3.4)
(30, 3.5)
(6, 2.6)
(8, 1.8)
MS/SS1B (cont)
Q
4
(a)
Solution
184.5
49
or
Marks
50
1.92 
49
B1
Total
1
= 1.94
(b)
(i)
96% (0.96)  z = 2.05 to 2.06
s
n
CI for  is
x  z 
Thus
251.1  2.0537 
Hence
or
1.94
50 or 49
251.1  0.6
Comments
Fully correct expression or equivalent
must be seen
Note: s  184.5 50  1.939  B0
AG
B1
AWFW
M1
Used with 251.1 and 1.94 correctly
Must have n with n > 1
AF1
F on z only
Adep1
4
(250.5, 251.7)
(2.0537)
CAO/AWRT
Dependent on AF1 but not on z so
can be gained using an incorrect z
AWRT
(ii) Claim is  > 250
Clear correct comparison of 250 with LCL
or CI
so
Claim is supported/reasonable/correct/true/etc
Must be consistent with c’s comparison
(c)
x  ns = 251.1  n  1.94 < 250
SC: Quoted values of 249.2, 247.2 or 245.3
(AWRT)  M1
BF1
Bdep1
F on CI
2
Dependent on BF1
Allow any multiple of 1.94
Must clearly indicate the value of a
numerical expression giving a result
less than 250
M1
so
Some individual packets are likely to/will
contain less than 250 grams
A1
Total
(250 < LCL or CI)
2
9
Or equivalent
MS/SS1B (cont)
Q
5
(a)(i)
Solution
W
W
Total
J
0.55
0.15
0.70
J
0.10
0.20
0.30
Marks
Total
0.65
0.35
1.00
Total
B1
0.35 and 0.7; CAO
B1
0.55; CAO
B1
3
Notes:
Use of Venn or tree diagrams without table
completion  B0 B0 B0
Printed table not completed but constructed and
completed on Page 12/13  B1 B1 B1 max
(ii) P(purchases exactly one)
= P W  J    0.15
(iii)
(A)
Only c’s equivalent to 0.10 shown
and added to 0.15
Can be implied by correct answer
M1
= 0.10 + 0.15
= 0.25 or 25/100 or 5/20 or 1/4
A1
P W  J  = 0.8 &/ P W   P  J  = 1.35
B1
2
&/ P W  = 0.65
or
Bdep1
P  J | W   0.55 0.65 = 0.85
Do not accept use of W  and/or J 
AWRT
3
&/ P  J  = 0.70
or
CAO
Any one of these three seen
Ignore contradictions, explanations &
justifications
B1
P W | J   0.55 0.70 = 0.79
0.1 and 0.2; CAO
Accept fractional answers
Do not accept percentages
or P W  J  = 0.55 (>0); accept if indicated
in a Venn diagram
or P W   P  J  = 1.35 >0 or impossible
(B)
Comments
P W   P  J  = 0.45 to 0.46
Any one of these three seen
Ignore contradictions, explanations &
justifications
AWFW
&/ P W  J  = 0.55
(b)
(i)
Do not allow multiplying factors in (b)
P(0) = 0.15 × 0.40 × 0.45
B1
= 0.027 or 27/1000
(ii) P(2) = 0.85 × 0.60 × 0.45 = 0.2295
+ 0.85 × 0.40 × 0.55 = 0.1870
+ 0.15 × 0.60 × 0.55 = 0.0495
or
B1
2
For either method:
At least two bold expressions correct
Only one bold expression correct
Can be implied by correct answer
For second method:
Must have ‘1 –’ for any marks
M2
(M1)
= 1 – (0.027 + 0.2265 + 0.2805)
= 0.466 or 466/1000 or 233/500
Total
A1
Can be implied by correct answer
or 1 – (0.2265 + 0.466 + 0.2805)
CAO
3
13
CAO; do not imply this from (i)
MS/SS1B (cont)
Q
6
(a)
(i)
Solution
X ~ B(10, 0.15)
P(X  2) = 0.82(0)
Marks
Total
B1
1
Comments
AWRT
(0.8202)
(ii) P(X ≥ 2) = 1 – P(X  1)
= 1 – (0.5443 or 0.8202)
M1
= 0.455 to 0.456
A1
(iii) P(1 < X < 5) = 0.9901 or 0.9986
(p1)
M1
minus 0.5443 or 0.1969
(p2)
M1
= 0.445 to 0.446
A1
OR
B(10, 0.15) expressions stated for at least 3
terms within 1  X  5 gives probability
= 0.445 to 0.446
(b)
Requires ‘1 –’
Accept 3/2 dp rounding or truncation
Can be implied by 0.455 to 0.456
but not by 0.179 to 0.18(0)
2
AWFW
(0.4557)
Accept 3 dp rounding or truncation
p2 – p1  M0 M0 A0
(1 – p2) – p1  M0 M0 A0
p1 – (1 – p2)  M1 M0 A0
only providing result > 0
Accept 3 dp rounding or truncation
3
AWFW
(0.4458)
(M1)
Can be implied by a correct answer
(A2)
AWFW
(0.4458)
Normal approximation  0 marks
Y ~ B(50, 0.15)
(i) P(Y > 5) = 1 – P(Y  5)
= 1 – (0.2194 or 0.1121)
M1
= 0.78(0) to 0.781
A1
(ii) P(5 ≤ Y ≤ 10) = 0.8801 or 0.7911
(p1)
M1
minus 0.1121 or 0.2194
(p2)
M1
= 0.768
A1
OR
B(50, 0.15) expressions stated for at least 3
terms within 4  Y  10 gives probability
= 0.768
Total
2
Requires ‘1 –’
Accept 3 dp rounding or truncation
Can be implied by 0.78(0) to 0.781
but not by 0.888 to 0.89
AWFW
(0.7806)
Accept 2/3 dp rounding or truncation
p2 – p1  M0 M0 A0
(1 – p2) – p1  M0 M0 A0
p1 – (1 – p2)  M1 M0 A0
only providing result > 0
Accept 3 dp rounding or truncation
3
AWRT
(0.7680)
(M1)
Can be implied by a correct answer
(A2)
AWRT
11
(0.7680)
MS/SS1B (cont)
Q
7
Solution
Marks
(a) Ryan:
Value indicates that as volume increases then
weight decreases
B1
Sunil:
Value indicates no correlation/relationship/
association/link between volume and weight
B1
SC:
If B0 B0:
Would expect weight to increase with volume
or
Would expect strong(er) positive correlation
between weight and volume
(b) Ryan & Sunil:
r is not affected by units/(linear) scaling
Tim:
r is not affected by sample size
or
2 × 0.612 > 1  impossibility
(c)
(i)
r = 0.541 to 0.543
r = 0.54 to 0.55
r = 0.5
to 0.6
Total
Comments
Or equivalent in context
2
(B1)
Or equivalent in context
Or equivalent in context
B1
Or equivalent
B1
2
Either; or equivalent
B3
(B2)
(B1)
3
AWFW
AWFW
AWFW
(0.54186)
OR
Attempt at
216 6633.16 136 2376.84
3795.5 (all 5 attempted)
Accept notation of x and y
801.16 64.84 & 123.5
(all 3 attempted)
 v  v2  w  w2 &  vw
(M1)
or
Attempt at Svv Sww & Svw
Attempt at substitution into correct
corresponding formula for r
r = 0.541 to 0.543
(m1)
(A1)
AWFW
Dependent on 0.5 ≤ r ≤ 0.6
Or equivalent; must qualify strength
and state positive
Bdep0 for very strong/strong/high/
good/average/medium/reasonable/
poor/very weak/little/etc
(ii)
(Quite or fairly) weak/some/moderate
positive (linear) correlation/relationship/
association/link (but not ‘ trend’)
between
volumes and weights of suitcases
Bdep1
B1
Total
TOTAL
&
2
9
75
Context; providing 0 < r < 1

Scaled mark unit grade boundaries - June 2011 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
62
68
62
88
69
67
68
68
68
63
58
67
62
59
64
55
62
54
54
76
64
59
59
62
61
57
51
60
55
52
56
48
55
47
46
66
55
51
52
54
53
51
46
52
48
46
49
41
48
41
38
57
46
43
46
46
46
45
41
44
41
40
42
34
41
35
30
48
38
36
40
39
39
39
36
37
34
34
35
28
34
29
23
39
30
29
34
32
32
33
31
30
28
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
70
74
55
63
61
68
47
54
50
56
40
45
39
45
33
36
28
34
26
28
18
23
MHEB1
MHEB2
GCE MODERN HEBREW UNIT 1
GCE MODERN HEBREW UNIT 2
100
100
80
61
71
54
62
47
54
40
46
34
38
MUSC1
MUS2A
MUS2B
MUS2C
GCE MUSIC UNIT 1
GCE MUSIC UNIT 2A
GCE MUSIC UNIT 2B
GCE MUSIC UNIT 2C
80
60
60
60
-
57
45
49
49
51
40
43
44
45
35
37
39
39
30
32
34
34
26
27
29