07[Anal Add Math CD]

7/24/2019 07[Anal Add Math CD]

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1 Penerbitan Pelangi Sdn. Bhd.

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


2/15

2

Additional Mathematics SPM Chapter 7

Penerbitan Pelangi Sdn. Bhd.

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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3



(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



= 10 3

= 7

23. (a) The range = 3 0

= 3



= 8thterm 3rdterm

= 2 1

= 1

24. (a) The range = 5 0

= 5


=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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4



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


5/15

5



Interquartile range


= 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


= 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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6



(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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7



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



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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9



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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10



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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11



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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12



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


13/15

13



(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


14/15

14



(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


15/15

15



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

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07[Anal Add Math CD]