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1. (a) Mean =2 + 3 + 5 + 1 + 7
5
=18
5
(b) Mean =2 + 3 + 7 + 6 + 9 + 10
6
=336
(c) Mean =1 + 2 + 0 + 3 + 8
5
=12
5
2. (a) Mode = 2
(b) Modes = 0, 2
(c) None
3. (a) 1, 1, 2, 3, 4, 5, 7
Median = 3
(b) 2, 0, 1, 4, 5, 7
Median =1 + 4
2
=52
4. Mode = 1
5. Modal class = 30 39
6. Mode = 2
7. Modal class = (6 10) kg
Mode =5.5 + 10.5
2
= 8 kg
8. Mean =(7 5) + (2 10) + (3 15) + (4 20)
7 + 2 + 3 + 4
= 11.25
9. Mean =
(2 4.5) + (3 14.5) + (8 24.5)
+ (7
34.5) + (10
44.5)2 + 3 + 8 + 7 + 10
= 31.17
10. (a) Median = 21 + 12 th
term
= 11thterm
Median = 2
(b) Median =10thterm + 11thterm
2
=1 + 2
2
= 1.5
11. (a) Median = L+ 12N F
fm
C Median class = 10 14
Median = 9.5 + 12
(30) 10
12
5 = 9.5 + 5
125
= 11.58 cm
(b) Median class = 15 19
Median = 14.5 + 12
(21) 10
8
5 = 14.5 + 0.5
85
= 14.81 cm
CHAPTER
7 Statistics
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12. (a)
29.50
Numberofstudents
Weight (kg)
2
4
6
8
10
12
14
16
34.5 39.5 44.5 49.5
(b) Median = 39.5 kg
13. (a)Height less than Number of students
99.5 0
109.5 3
119.5 8
129.5 15
139.5 21
149.5 25
(b)
Numberofstudents
Height (cm)99.50
5
10
15
20
25
109.5 119.5 129.5 139.5 149.5
126
(c) Median = 126 cm
14. (a) Mean = 4 + 5
= 9
Mode = 2 + 5
= 7
Median = 3 + 5
= 8
(b) Mean = 4 1 = 3
Mode = 2 1
= 1
Median = 3 1
= 2
(c) Mean = 4 4
= 16
Mode = 2 4
= 8
Median = 3 4
= 12
(d) Mean = 42
= 2
Mode =22
= 1
Median =32
= 112
(e) Mean = 4 2 + 3
= 11
Mode = 2 2 + 3
= 7
Median = 3 2 + 3
= 9
(f) Mean =4 1
2
=32
Mode =2 1
2
=12
Median =3 1
2
= 1
15. (a) (i) Mean =3 + 5 + 6 + 7 + 7
5
= 5.6
Mode = 7
Median = 6
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(ii) Mean =30 + 5 + 6 + 7 + 7
5
= 11
Mode = 7
Median = 7
(b) The mean
16. (a) (i) Mean =20 + 25 + 30 + 30
4
= 26.25
Mode = 30
Median =25 + 30
2
= 27.5
(ii) Mean =1 + 25 + 30 + 30
4
= 21.5
Mode = 30
Median =
25 + 30
2
= 27.5
(b) The mean
17. (a) Sum of 4 numbers = 4 8
= 32
The new mean =32 + 3
5
=35
5
= 7
(b) The new sum = 32 5
= 27
The new mean =27
3
= 9
18. (a) 1, 3, 4, 5, 8
The median = 4
(b) (i) 1, 4, 5, 8
The median =4 + 5
2
=92
= 4.5
(ii) 1, 3, 3, 4, 5, 8
The median =3 + 4
2
= 3.5
19. (a) Mode = 9
(b) Mode = 9
(c) Modes = 3, 9
(d) None
20. (a) Mode
(b) Median, because there is a extreme value, 50
(c) Median or mean
21. (a) 1, 2, 3, 5, 7, 10
The range = 10 1
= 9
(b) The interquartile range
= Upper quartile Lower quartile
= 7 2
= 5
22. (a) 1, 3, 4, 6, 9, 10, 12
The range = 12 (1)
= 13
(b) The interquartile range
= Upper quartile Lower quartile
= 10 3
= 7
23. (a) The range = 3 0
= 3
(b) The interquartile range
= Upper quartile Lower quartile
= 8thterm 3rdterm
= 2 1
= 1
24. (a) The range = 5 0
= 5
(b) The interquartile range
=13th+ 14th
2
4th+ 5th
2
=3 + 3
2
1 + 1
2
= 3 1
= 2
25. (a) The range
= The midpoint of last class The midpoint of
rst class
=70 + 79
2 10 + 19
2
= 74.5 14.5
= 60
(b) Lower quartile, Q1 =
14
thof 20
= 5thterm
= 19.5 + 14
(20) 4
5
10 = 19.5 + 2
= 21.5
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Upper quartile, Q3 =
34
thof 20
= 15thterm
= 39.5 + 34
(20) 13
3
10 = 39.5 + 6.67
= 46.17
Therefore, the interquartile range
= 46.17 21.5
= 24.67
26. (a)Height, less than Number of poles
9.5 0
19.5 4
29.5 9
39.5 13
49.5 16
59.5 18
69.5 19
79.5 20
Numberofpoles
Height (cm)9.5 19.5 29.5
21.5 44.5
39.5 49.5 59.5 69.5 79.50
5
10
15
20
(b) (i) Lower quartile = 21.5 cm
(ii) Upper quartile = 44.5 cm
(iii) The range of interquartile = 44.5 21.5
= 23 cm
27. x2= 22+ 32+ 12+ 52+ 42+ 92
= 136
x= 2 + 3 + 1 + 5 + 4 + 9 = 24
(a) The variance, 2=x2
N
x
N
2
=136
6
2462
= 22.67 16
= 6.67
(b) The standard deviation, = 6.67
= 2.583
28. (a) fx= (2 0) + (3 1) + (4 2) + (1 3)
= 14
fx2= (2 02) + (3 12) + (4 22) + (1 32)
= 28
f = 2 + 3 + 4 + 1
= 10
The variance, 2= fx2
f
fx f2
=28
10
14102
= 2.8 (1.4)2
= 0.84
(b) The standard deviation = 0.84
= 0.9165
29. (a)
Midpoint oflength, x Number ofinsects, f fx fx2
0.2 5 1 0.2
0.5 3 1.5 0.75
0.8 2 1.6 1.28
1.1 3 3.3 3.63
1.4 1 1.4 1.96
f= 14 fx = 8.8fx2= 7.82
The variance =fx2
f
fx f
2
=7.82
14
8.8142
= 0.163 cm2
(b) The standard deviation = 0.163 = 0.4037 cm
30.(a) (b) (c) (d) (e) (f)
Range6 6 6 5
= 30
65
6 5
= 30
6 5
= 30
Interquartile
range
3 3 3 5
= 15
35
3 5
= 15
3 5
= 15
Variance4 4 4 52
= 100
425
4 52
= 100
4 52
= 100
Standard
deviation
2 2 2 5
= 10
25
2 5
= 10
2 5
= 10
31. 1, 2, 3, 4, 5, 6, 7
Range = 7 1
= 6
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Interquartile range
= Upper quartile Lower quartile
= 6 2
= 4
Variance
=x2
N
x
N
2
=12+ 22+ 32+ 42+ 52+ 62+ 72
7
1 + 2 + 3 + 4 + 5 + 6 + 77
2
= 20 16
= 2
Standard deviation = 4 = 2
32. (a) 1, 2, 3, 4, 5, 6, 50
Range = 50 1
= 49 Interquartile range
= Upper quartile Lower quartile
= 6 2
= 4
Variance
=x2
N
x
N
2
=12+ 22+ 32+ 42+ 5 + 62+ 502
7
1 + 2 + 3 + 4 + 5 + 6 + 50
7
2
= 370.1 102.9
= 267.2
Standard deviation = 267.2= 16.35
(b) The range, variance and standard deviation have
the signicant difference compared to Question
31.
33. (a) 1, 2, 3, 4, 5, 18
The range = 18 1
= 17
Interquartile range = 5 2
= 3
Variance, 2
=x2
N
x
N
2
=12+ 22+ 32+ 42+ 52+ 182
6
1 + 2 + 3 + 4 + 5 + 186
2
= 32.92
Standard deviation, = 32.92 = 5.738
(b) 1, 2, 3, 4
Range = 4 1
= 3
Interquartile range =3 + 4
2
1 + 2
2
= 2
Variance
=x2
N
x
N
2
=12+ 22+ 32+ 42
4
1 + 2 + 3 + 44
2
=30
4
104
2
= 1.25
Standard deviation = 1.25 = 1.118
34. (a) Mean of Siti =85 + 87 + 90 + 93 + 95
5 = 90
Standard deviation
= x2
N
(x)2
= 852+ 872+ 902+ 932+ 9525 902 = 3.688
Mean of Fatimah =91 + 90 + 88 + 90 + 91
5
= 90 Standard deviation
= 912+ 902+ 882+ 902+ 9125 902 = 1.095
(b) Fatimah. Even though both have the same mean
marks but Fatimah is more consistent in the
performance with smaller standard deviation.
1. (a) (x x)2= 360
Standard deviation = (x x)2
10
= 36010
= 36 = 6
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(b)x2
N
xN
2
= 62
x2
10 40010
2
= 36
x2
10 = 36 + 1 600
= 1 636 x2 = 16 360
2. (a) x =x
N
6 =30
N
N =30
6
= 5
(b) 2=x2
N (x)2
23 = x2
5 62
x2
5= 23 + 36
= 59
x2= 59 5
= 295
3. (a)
Marks 0 19 20 39 40 59 60 79 80 99
Number
ofstudents
5 10 25 16 4
(b) Lower quartile = 19.5 + 14
60 5
10
20 = 19.5 + 20
= 39.5
Upper quartile = 59.5 + 34
60 40
16
20 = 59.5 + 6.25
= 65.75
Semi interquartile range =12
(65.75 39.5)
= 13.125
4. x= 15
N= 5
x2= 55
(a) Mean,x =x
N
=15
5
= 3
Standard deviation =
x2
N
(x)2
= 555 32 = 2
(b) (i) The value of the number removed is 3 (the
mean itself).
(ii) N= 4
x= 15 3
= 12
x2= 55 32
= 46
Variance, 2=x2
N
x
N
2
=46
4
1242
= 2.5
5. (a) Total workers = 30
Median is the 302th
= 15thterm
The median class is 40 49.
The median = L+ 12N F
fm C
= 39.5 + 12
(30) 10
10
10 = 39.5 + 5
= 44.5
(b) The mean,x
=fx
f
=
(4 24.5) + (6 34.5) + (10 44.5)
+ (8 54.5) + (2 64.5)
30 = 43.83
(c) 2=fx2
N
(x)2
=
(4 24.52) + (6 34.52) + (10 44.52)+ (8 54.52) + (2 64.52)
30
(43.83)2
= 126.5
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6. (a) Given total number of students = 30
1 + 8 + h+ k+ 3 = 30
h+ k+ 12 = 30
h+ k = 18 ............................
Median = L+
12N F
fm
C
45.5 = 39.5 + 12
30 9
h
10 45.5 39.5 =
60h
6 =60
h
6h = 60
h = 10
From , h+ k= 18
10 + k= 18 k= 8
(b) Since h = 10, the modal class is 40 49, and
k= 8, therefore the mode is the middle value of
39.5 and 49.5.
Mode =12
(39.5 + 49.5)
= 44.5 kg
(c) New mode = 44.5 3
= 41.5 kg
1. (a) Mean = 4
2 + x + 5 + (x 4) + 3 + 4
6
= 4
10 + 2x = 4 6
2x = 24 10
x =14
2
= 7
(b) The numbers are 2, 7, 5, 3, 3, 4 Therefore, the mode is 3.
2. (a) Mean = 9
7 + x+ 11 + 5x+ 9
5
= 9
27 + 6x= 5 9
6x= 45 27
= 18
x= 3
(b) The numbers are 7, 3, 11, 15, 9
Rearrange in ascending order: 3, 7, 9, 11, 15
Therefore, the median is 9.
3. Mean = 7
8 + 7 + x+ 7 + y+ 10 + 11
7
= 7
43 + x+ y = 7 7
x+ y= 49 43
x+ y = 6
y= 6 x
4. Mean = 20
x
5 = 20
x= 100
New mean = 10
100 + x
6
= 10
x= 60 100
x= 40
5. Mean = 6
x+ 3 + 4 + 5 + y+ 9 + 10
7= 6
x+ y+ 31 = 7 6
x+ y= 42 31
x+ y= 11
Since the mode is 5, x + y = 11 and yx, then
x= 5, y= 6.
6. Givenx = 7
x
5= 7
x= 35
{5, 5, 6, a, b}
5 + 5 + 6 + a+ b= 35
a+ b= 35 16
= 19
a= 7, b= 12
or a= 8, b= 11
or a= 9, b= 10
Therefore, the possible 5 positive integers are
{5, 5, 6, 7, 12} or {5, 5, 6, 8, 11} or {5, 5, 6, 9, 10}.
7. Givenx = 5
x
8= 5
x= 40
40 + a+ 3a
10= 6
4a= 60 40
= 20
a= 5
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8. (a) 3, 4, 5, 8, 11
Mean =3 + 4 + 5 + 8 + 11
5
= 6.2
Mode = None
Median = 5
(b) 2, 2, 3, 4, 5, 6, 7
Mean =2 + 2 + 3 + 4 + 5 + 6 + 7
7
=29
7
Mode = 2
Median = 4
9. (a) 1, 2, 3, 4, 6
Range = 6 1
= 5
First quartile =1 + 2
2 = 1.5
Third quartile =4 + 6
2
= 5
(b) 1, 2, 5, 6, 7, 8
Range = 8 1
= 7
First quartile = 2
Third quartile = 7
10. Mean = 5, mode = 3, median = 4
(a) New mean = 5 + 2
= 7
New mode = 3 + 2
= 5
New median = 4 + 2
= 6
(b) New mean = 5 6 2
= 28
New mode = 3 6 2
= 16
New median = 4 6 2
= 22
(c) New mean =5 2
4
=34
New mode =3 2
4
=14
New median =4 2
4
=12
11. (a) x = 2
fx
f
= 2
(5 0) + (4 1) + (6 2) + 3x
15 + x
= 2
4 + 12 + 3x= 2(15 + x)
16 + 3x= 30 + 2x x= 30 16
x= 14
(b) 0 x6
12. (a) Mode = 2
(b) Mean =fx
f
=
(3 0) + (5 1) + (7 2) + (4 3)+ (1 4)
20 = 1.75
(c) Median =10thterm + 11thterm
2
=2 + 2
2
= 2
13. (a) There are 10 numbers.
The median is between 5thand 6thnumbers.
Median =6 + (x+ 2)
2
=8 + x
2
= 4 +x2
(b) 1 x 1 6 and 6 x+ 2 10
2 x 7 4 x 8
2 4
2 < x< 7
4 < x< 8
7 8
The possible values of xare 5 and 6.
14. (a) Team A, because both teams A and Chave the
same mean score but the standard deviation of
team Ais smaller.
(b) The combined mean score
=(32 76.8) + (36 70.3) + (32 76.8)
32 + 36 + 32
= 74.46
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15. (a) Mean,x = 5, variance, 2= 2, N= 30
x =x
N
5 =x
30
x= 150
Sum of the 30 numbers = 150
(b) 2=x2
N
x
N
2
2 =x2
30
52
x2
30
= 27
x2= 810
Sum of the square of the numbers = 810
16. (a) Mode = 3, mean = 6
2, 3, 3, 5, 9, 11, 11, p, q
Mean = 6
2 + 3 + 3 + 5 + 9 + 11 + 11 + p+ q
9
= 6
44 + p+ q= 9 6
p+ q= 54 44
p+ q= 10
Since 3 is the mode and pq,
then q= 3 and p= 7.
(b)x2
N =
22+ 32+ 32+ 52+ 92+ 112+ 112+ 72+ 32
9
=428
9
2=x2
N
(x)2
=428
9
62
=428
9
36
=104
9
17. 2 =134
9
x2
N
x
N
2
=134
9
y2+ 9y2+ 100
3
4y+ 103
2
=134
9
y2+ 9y2+ 100
3
16y2+ 80y+ 100
9 =
134
9
9, 3(10y2+ 100) 16y2 80y 100 = 134
30y2+ 300 16y2 80y 100 134 = 0
14y2 80y+ 66 = 0
2, 7y2 40y+ 33 = 0
(y 1)(7y 33) = 0
y= 1,33
7
18. (a) (i) Mean = 7
5 + 13 + 5 + n+ 5 + 10 + 11 + 10+ n2+ 9
10
= 7
68 + n+ n2= 70
n2+ n 2 = 0
(n+ 2)(n 1) = 0
n= 2, 1
(ii) Since n0, then n= 1
When n= 1,
5, 13, 5, 1, 5, 10, 11, 10, 1, 9
That is, 1, 1, 5, 5, 5, 9, 10, 10, 11, 13
Mode = 5
Median = 5 + 92
= 7
(b) Standard deviation =
Variance = 2
New standard deviation = 3
New variance = 32(2)
= 92
19. x = 5, 2= 4, N= 10
(a) (i) x =x
N
5 =x
10
x= 50
(ii) 2=x2
N
(x )2
4 =x2
10
52
x2
10
= 29
x2
= 290
(b) n= 10,y = 10, y
2= 9
Mean =x+ y
N+ n
=50 + 10(10)
10 + 10
=150
20
= 7.5
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2=x2+ y2
N+ n
(7.5)2
=290 + [2
y+ (y)2]10
10 + 10
(7.5)2
=290 + [9 + 100]10
20
7.52
= 69 56.25= 12.75
20. (a) x = 2, 2= 6, x2= 100
2=x2
n
(x)2
6 =100
n 2
2
100
n = 10
10n= 100
n= 10
(b) (i) x = 12
4 + 8 + 4 + 4 + 10 + x+ 10 + y+ 20
9
= 12
60 + x+ y = 108
x+ y = 108 60
x+ y = 48
(ii) When y= 3x,
x+ 3x= 48
4x= 48
x= 12 Since x+ y= 48,
then 12 + y= 48
y= 36
Arrange in ascending order:
4, 4, 4, 8, 10, 10, 12, 20, 36
The median is 10.
21.Weight
(kg)
Midpoint,
x
Number
of water-
melons, f
Weight,
less
than
Number
of water-
melons
0.0 0.9 0.45 0 0.95 0
1.0 1.2 1.1 6 1.25 6
1.3 1.5 1.4 10 1.55 16
1.6 1.8 1.7 12 1.85 28
1.9 2.1 2.0 13 2.15 41
2.2 2.4 2.3 9 2.45 50
(a) The range = 2.3 1.1
= 1.2 kg
(b) Mean =fx
f
=
6(1.1) + 10(1.4) + 12(1.7)+ 13(2.0) + 9(2.3)
50
= 1.754 kg
(c) Median = 12 50th
= 25thposition
The median class is 1.6 1.8.
Median = L+ 12N F
fm
C
= 1.55 + 12
(50) 16
12
(0.3) = 1.55 +
9
12
0.3
= 1.775 kg
22. (a)
Numberoffish
Weight (g)34.5 44.5 54.5 64.5
61
74.5 84.50
10
20
30
40
From the histogram, mode = 61 g
(b) Lower quartile = 14 120th
= 30thposition
Upper quartile = 34 120th
= 90thposition
Lower quartile = 44.5 + 1
4 120 18
20 10 = 50.5 g
Upper quartile = 64.5 + 34
120 78
30
10 = 68.5 g
Interquartile range = 68.5 50.5
= 18 g
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23.
MarksNumber of
students
Marks,
less than
Number of
students
0 0 0.5 0
1 20 15 20.5 15
21 40 20 40.5 35
41 60 32 60.5 67
61 80 23 80.5 90
Numbe
rofstudents
Marks
10.50.5 20.5 30.5
28.5 47.5 61
40.5 50.5 60.5 70.5 80.50
10
20
30
40
50
60 66
70
80
90
(a) Median = 47.5
(b) Interquartile range = 61 28.5
= 32.5
(c) Number of students who scored more than 60marks = 24
Percentage =24
90
100
= 26.67%
24.
MarksNumber of
students,f
Midpoint,
x fx fx2
5 9 5 7 35 245
10 14 10 12 120 1440
15 19 15 17 255 4335
20 24 12 22 264 5808
25 29 8 27 216 5832
f= 50 fx= 890 fx2= 17 660
(a) The mean score =fx
f
=890
50
= 17.8
(b) The variance =fx2
f
fx f
2
=17 660
50
89050
2
= 36.36
25. (a) Median class at 12 40th
position,
that is, 30 39.
Median = L+ 12N F
fm
C
= 29.5 + 12
40 14
8
10 = 37 cm
(b) 15 14 + x25
1 x 11
26. (a)Age Number of workers
25 29 16
30 34 20
35 39 24
40 44 2045 49 14
50 54 6
(b)
Numberofworkers
Age24.5 29.5 39.5
37
34.5 44.5 49.5 54.50
5
10
15
20
25
The mode is 37.
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27. (a) The median class = 12 80th
= 40thposition
Therefore, the median class is 71 x80.
Score Number of students
41 x50 12
51
x
60 861 x70 10
71 x80 15
81 x90 25
91 x100 10
The median marks
= 70.5 + 12
80 30
15
10 = 77.17
(b) The modal class is 81x90.
(c)15 + 25 + 10
80
100% = 62.5%
28. (a) 3 n+ 1 8 and 10 2n+ 3 15
7 2n 12
72
n 6
3 1 n 8 1
2 n 7
2 72
7
2
6 7
< n< 6
2 < n< 7
Therefore, 4 n6.
Hence, n= 4, 5.
(b) (i) When n= 5,
The mean,x =
(3 3) + (2 6) + (1 8)
+ (1 10) + (2 13)+ (1 15)
10
= 8
(ii) Variance =fx2
f
(x)2
=
(3 32) + (2 62)
+ (1 82) + (1 102)
+ (2 132) + (1 152)
10 82
= 18.6
Standard deviation = 18.6 = 4.313
29. n= 5,x = 10, x2= 558
(a) 2=x2
n
(x)2
=558
5
102
=58
5
Standard deviation = 585 = 3.406
(b) (i) x= Original mean
= 10
(ii) 2 =y2
6
(x)2
=x2+ 102
6
102
=558 + 102
6
100
= 9.67
(iii) (I) The new mean = 10 + 2
= 12
(II) The new variance = 9.67
30. (a) x=150
360
3600
= 1500
y=30
360
3600
= 300
z =180
360
3600
= 1800
(b) The mean =
(1500 14.5) + (300 24.5)+ (1800 34.5)
3600 = 25.33 g
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(c)
Numberofeggs
300
0
600
900
1200
1500
1800
34
9.5 19.5 29.5 39.5Weight (g)
The mode = 34 g
(d)300 + 1800
3600
100%
=2100
3600
100%
= 58.33%
1. x= 4, x
2= 2, nx= 5
y= 40, y2= 400, ny= 5
z= m, z2= n, N= 10z=
x+ y
10
=4 5 + 40
10
= 6
Therefore, m= 6
x
2=x2
5
x 2
2 =x2
5
42
x2
5
= 18
x2= 90
z
2=x2+ y2
10
m2
=90 + 400
10
62
= 49 36
= 13
Therefore, n= 13
2. (a) x= 10, nx= 5
y= 10, ny= 6
x2= 558
x=x
5
10 =x
5
x= 50
y=50 + p
6
= 10
p= 10
(b) y
2 =y2
6
(y)2
=558 + 102
6
102
= 9.667
3. (a)13 + 5 + 5 + n+ 5 + 10 + 10 + 11 + 9 + n2
10
= 7
68 + n+ n2= 70
n2+ n 2 = 0
(n+ 2)(n 1) = 0
n= 2 or
n= 1
(b) (i) When n= 1,
1, 1, 5, 5, 5, 9, 10, 10, 11, 13
Median =5 + 9
2
= 7
(ii) When n= 2,
2, 4, 5, 5, 5, 9, 10, 10, 11, 13
Median =5 + 9
2
= 7
4. (a) 5, 6, 7, x, 12, y, 17, 23
Given median = 11
x+ 12
2
= 11
x+ 12 = 22
x = 10
Given mean = 12
5 + 6 + 7 + 10 + 12 + y+ 17 + 23
8
= 12
80 + y
8
= 12
80 + y = 96
y = 16
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(b) 2=x2
8
x82
=52+ 62+ 72+ 102+ 122+ 162+ 172+ 232
8
(12)2
= 34.5
= 34.5
= 5.874
5. (a)5 + 12 + y+ (y+ 1) + 6 + 9 + 10 + 7
8
= 8
50 + 2y = 64
2y = 14
y = 7
(b) 2=52+ 122+ 72+ 82+ 62+ 92+ 102+ 72
8
82
= 4.5
= 4.5 = 2.121
(c) (i) Mean = 8 2
= 6
(ii) Standard deviation = 2.121
6. (a) + 2 =54
54 = 2
14= 2
= 8
Therefore, mean = 8
3 + 6 + 5 + 10 + 8 + 12 + x+ 3x
8
= 8
44 + 4x = 64
4x = 20
x = 5
Variance, 2
=32+ 62+ 52+ 102+ 82+ 122+ 52+ 152
8
82
= 14.5
(b) The number that is added is 8 since mean does
not change.
Variance, 2
=32+ 62+ 52+ 102+ 82+ 122+ 52+ 152+ 82
9
82
= 12.89
Standard deviation, = 12.89 = 3.59
3.6
7. (a) Mean = 6
3 + 5 + 8 + 6 + 8 + 10 + 5 + 3 + x+ y
10
= 6
48 + x+ y= 60
x+ y= 60 48
x+ y= 12
(b) (i) When x= y, therefore x= 6, y= 6. Hence, the mode is 6.
(ii) When x y,
x= 1, y= 11
The modes are 3, 5 and 8.
x= 2, y= 10
The modes are 3, 5, 8 and 10.
x= 3, y= 9
The mode is 3.
x= 4, y= 8
The mode is 8.
x = 5, y= 7
The mode is 5.
Therefore, the possible modes are 3, 5, 8 or
10.
(c) = 315 2 =
31
5
32+ 52+ 82+ 62+ 82+ 102+ 52+ 32+ x2+ (12 x)2
10 (6)2
=31
5
332 + x2+ (12 x)2
10 =
31
5 + 36
332 + x2+ (12 x)2 =211
5
10
332 + x2+ 144 24x+ x2= 422
2x2 24x+ 476 = 422
2x2 24x+ 54 = 0
2, x2 12x+ 27 = 0
(x 3)(x 9) = 0
x= 3, 9
8. Giveny
10
= 7
y = 70
New mean =70 + (7 x) + (7 + x)
12
=84
12
= 7
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9. Mean = 9
2 + (x y) + (x+ y) + 16
4
= 9
18 + 2x= 36
2x= 18
x= 9
2= 25
22+ (9 y)2+ (9 + y)2+ 162
4
92= 25
260 + 92 18y+ y2+ 92+ 18y+ y2
4
= 25 + 81
2y2+ 422 = 4(106)
2y2= 4(106) 422
= 2
y2= 1
y= 1
10. x2= 100, = 4
2
= 16
x2
4
x4
2
= 16
100
4
x4
2
= 16
x4
2
= 25 16
= 9
x
4
= 3
Therefore, the mean is 3.
11. (a) If the mode is 1, the minimum value of xis 7.
(b) If the median is 2,
3 + x+ 3 = 8
x = 2
and 3 + x = 11
x = 8
The range of xis 1
x
9.
(c) Mean = 1.95
(0)(3) + (1)(x) + (2)(4) + (3)(6) + (4)(2)
15 + x= 1.95
x+ 8 + 18 + 8 = 1.95(15 + x)
x+ 34 = 29.25 + 1.95x
0.95x= 4.75
x= 5