15[A Math CD]

1 © Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. The total number of fruit trees in the orchard= (3 + 1 + 4) × 5= 40

Answer: C

2. The mode = R

Answer: C

3. Total number of pupils

= 360°–––––90°

× 10

= 40

Number of pupils who passed

= 270°–––––360°

× 40

= 30

Answer: C

4. Mean score

= (2 × 0) + (6 × 1) + (4 × 2) + (8 × 3)–––––––––––––––––––––––––––––––

2 + 6 + 4 + 8

= 38–––20

= 1.9

Answer: A

5. Median is the 30 + 1––––––2

th = 15.5th value

Median mark = 3

Answer: C

Paper 2

1. Score Frequency10 − 14 215 − 19 320 − 24 825 − 29 730 − 34 535 − 39 5

2. Mass (kg) Frequency

30 – 32 133 – 35 436 – 38 939 – 41 1142 – 44 745 – 47 548 – 50 3

3. Temperature (°C)

Lower limit

Upper limit

Lower boundary

Upper boundary

11 – 15 11 15 10.5 15.516 – 20 16 20 15.5 20.521 – 25 21 25 20.5 25.526 – 30 26 30 25.5 30.531 – 35 31 35 30.5 35.5

Class interval = 15.5 − 10.5 = 5

4. Distance

(km)Lower limit

Upper limit

Lower boundary

Upper boundary

1.0 – 1.9 1.0 1.9 0.95 1.952.0 – 2.9 2.0 2.9 1.95 2.953.0 – 3.9 3.0 3.9 2.95 3.954.0 – 4.9 4.0 4.9 3.95 4.955.0 – 5.9 5.0 5.9 4.95 5.95

Class interval = 1.95 − 0.95 = 1

CHAPTER

15 Statistics IIICHAPTER

2

Mathematics SPM Chapter 15

© Penerbitan Pelangi Sdn. Bhd.

5. Class interval Frequency21 – 25 226 – 30 531 – 35 1036 – 40 941 – 45 346 – 50 1

6. Class interval Frequency24 – 26 127 – 29 330 – 32 833 – 35 1236 – 38 939 – 41 3

7. (a) Time (minutes) Frequency Midpoint11 – 13 3 1214 – 16 5 1517 – 19 11 1820 – 22 7 2123 – 25 4 24

Modal class = 17 – 19 minutes

(b) Score Frequency Midpoint10 – 14 2 1215 – 19 6 1720 – 24 13 2225 – 29 15 2730 – 34 6 3235 – 39 4 37

Modal class = 25 – 29

8. Mass (kg) Frequency Midpoint31 – 40 4 35.541 – 50 4 45.551 – 60 10 55.561 – 70 6 65.571 – 80 5 75.581 – 90 3 85.5

Modal class = 51 – 60 kg

9. (a) Distance (m) Frequency Midpoint1 – 3 2 24 – 6 6 57 – 9 12 8

10 – 12 5 1113 – 15 3 14

(b) Mean distance

=

(2 × 2) + (6 × 5) + (12 × 8) + (5 × 11) + (3 × 14)

––––––––––––––––––––––––2 + 6 + 12 + 5 + 3

= 227––––28

= 8.107 m

10. (a) Donation (RM) Frequency Midpoint

11 – 15 3 1316 – 20 4 1821 – 25 7 2326 – 30 10 2831 – 35 6 3336 – 40 6 38

(b) Mean donation

=

(3 × 13) + (4 × 18) + (7 × 23) + (10 × 28) + (6 × 33) + (6 × 38)

––––––––––––––––––––––––––––––––––– 3 + 4 + 7 + 10 + 6 + 6

= 978––––36

= RM27.17

11. (a) Total number of items = 2 + 4 + 7 + 9 + 5 + 3 + 2 = 32

(b) Modal class = 16 – 20 g

12. (a) Length (cm) Frequency Upper boundary

10 – 14 2 14.515 – 19 3 19.520 – 24 7 24.525 – 29 8 29.530 – 34 6 34.535 – 39 4 39.5

3



(b)

1

0

2

3

4

5

6

7

8

14.59.5 19.5 24.5 29.5

Freq

uenc

y

Length (cm)34.5 39.5

(c) Modal class = 25 – 29 cm

13. (a) Time (s) Frequency Upper boundary21 – 30 3 30.531 – 40 4 40.541 – 50 6 50.551 – 60 10 60.561 – 70 8 70.571 – 80 6 80.581 – 90 1 90.5

(b)

1

0

2

3

4

5

6

7

8

9

10

30.520.5 40.5 50.5 60.5

Freq

uenc

y

Time (s)70.5 80.5 90.5

(c) Modal class = 51 – 60 s

14. (a) Point Frequency Midpoint0 – 2 0 13 – 5 1 46 – 8 3 79 – 11 9 10

12 – 14 8 1315 – 17 4 1618 – 20 3 1921 – 23 0 22

(b)

1

0

2

3

4

5

6

7

8

9

41 7 10 13

Freq

uenc

y

Point16 19 22

15. Score Frequency Midpoint6 – 10 0 811 – 15 2 1316 – 20 3 1821 – 25 5 2326 – 30 10 2831 – 35 12 3336 – 40 5 3841 – 45 7 4346 – 50 0 48

4



1

0

2

3

4

5

6

7

8

9

10

11

12

138 18 23 28

Freq

uenc

y

Score33 38 43 48

16. (a) Number of

goals Frequency Cumulative frequency

0 3 31 5 3 + 5 = 82 6 8 + 6 = 143 3 14 + 3 = 174 2 17 + 2 = 195 1 19 + 1 = 20

(b) Size of shoes Frequency Cumulative

frequency2 3 33 7 3 + 7 = 104 10 10 + 10 = 205 8 20 + 8 = 286 2 28 + 2 = 30

17. Number of eggs Frequency Cumulative

frequency10 – 12 10 1013 – 15 14 10 + 14 = 2416 – 18 17 24 + 17 = 4119 – 21 20 41 + 20 = 6122 – 24 25 61 + 25 = 8625 – 27 14 86 + 14 = 100

18.

5

0

10

15

20

25

30

35

40

2.51.5 3.5 4.5 5.5

Cum

ulat

ive

frequ

ency

Size of shoes6.5 7.5 8.5

19.

5

0

10

15

20

25

30

35

25.520.5 30.5 35.5 40.5

Cum

ulat

ive

frequ

ency

Number of items45.5 50.5

5



20. (a) Range = 14 − 3 = 11

(b) Range = 25 − 8 = 17

21. Midpointofthefirstclass=9.5Midpoint of the last class = 89.5Range = 89.5 − 9.5 = 80 cm3

22.

Cum

ulat

ive

frequ

ency

Distance (metres)10 20

18 24 30

30 40 50

Q3

Median

Q1

10

20

30

40

50

60

0

(a) Median = 24 m

(b) Interquartile range = Q3 − Q1 = 30 − 18 = 12 m

(c) Number of children who can swim more than 30 m

= 64 − 48 = 16

Paper 1

1. The total number of DVDs

= 1 5 + 2 + 85

× 102 × 24

= 720

Answer: B

2. Angle of sector of the transport

= 34

× Angle of sector of the rental

= 34

× 72°

= 54°

54° + 90° + 72° + x° + 3x° = 360° 216° + 4x° = 360° 4x° = 360° – 216°

x° = 144°4

= 36°

Amount spent on transport

= 54°36°

× RM120

= RM180

Answer: A

3. Mass (g) Cumulative

frequency Frequency

20 – 29 x x

30 – 39 14 14 – x40 – 49 19 19 – 14 = 550 – 59 23 23 – 19 = 460 – 69 30 30 – 23 = 7

The modal class is 20 – 29.Hence x . 7 x = 8

Answer: D

4. Number of workbooks sold in October= (2 × 3) × 25= 150 Number of workbooks sold in December

= 1 54

× 82 × 25

= 250The total number of workbooks sold in October and December= 150 + 250= 400Answer: B

5. The heights in ascending order: 158, 159, 164, 170 , 171, 176

Median = 164 + 1702

= 167 cmAnswer: D

6



6. Angle of sector representing the number of pupils who read 3 books

= 60180

× 360°

= 120°

Answer: C

7. Number of students who like Mathematics in class Q= 64 − 37= 27

Number of students who dislike Mathematics

= 360° − 288°–––––––––––288°

× 64

= 16Number of students who dislike Mathematics in class Q= 16 − 5= 11

Hence, the total number of students in class Q= 27 + 11= 38

Answer: B

8. Number of jerseys sold from January to June= 6 + 10 + 14 + 12 + 8 + 10= 60150% of the number of jerseys sold from July to December = 60Number of jerseys sold from July to December

= 60 × 100150

= 40

Totalprofit= RM35 × (60 + 40)= RM3500

Answer: C

9. Number of pens sold in shops P and R= 12 × 10= 120Number of pens sold in shops Q and S= 240 – 120= 120

Number of pens sold in shop Q

= 23

× 120

= 80

Answer: C

10. Mode = 7 (Highest frequency)

Answer: B

11. Mean height

=

(26 × 2) + (29 × 5) + (32 × 8) + (35 × 6) + (38 × 3)

–––––––––––––––––––––––––2 + 5 + 8 + 6 + 3

= 777––––24

= 32.4

Answer: B

12. Mean mass

= (17 × 5) + (22 × 8) + (27 × 9) + (32 × 10) + (37 × 4)5 + 8 + 9 + 10 + 4

= 97236

= 27 kg

Answer: B

13. The minimum value of n is 11.

Answer: D

14. Number of students= 5 + 15 + 25 + 20 + 10= 75

Maximum difference in the sector angles

= 25 – 575

× 360°

= 96°

Answer: C

15. The median is the 20 + 1––––––2

th = 10.5th value

Median = 2 + 3–––––2

= 2.5

Answer: B

7



16. Let the number of cartons of oranges sold in the 3rd week = xThen, the number of cartons of oranges sold in the 4th week = 4x

Total number of cartons of oranges sold in the four weeks= (7 + 5) × 20 + x + 4x= 240 + 5x

Given total number of cartons = 440 240 + 5x = 440 5x = 440 − 240

x = 200––––5

= 40

Number of cartons of oranges sold in the 4th week= 4 × 40= 160

Answer: B

17. 6, 6, 6, k, k, 9

Median = 6 + k–––––2

7 = 6 + k–––––2

k = 14 − 6 = 8

Mean of the eight data

= 6 + 6 + 6 + 8 + 8 + 9 + 3 + 4–––––––––––––––––––––––––8

= 50–––8

= 6.25

Answer: A

Paper 2

1. (a) Height (cm) Frequency Cumulative

frequency15 – 19 0 020 – 24 2 225 – 29 3 530 – 34 6 1135 – 39 10 2140 – 44 9 3045 – 49 6 3650 – 54 4 40

(b)

5

0

10

15

20

25

30

35

40

24.519.5 29.5 34.5

34

39.5

Cum

ulat

ive

frequ

ency

Height (cm)44.5 49.5 54.5

(c) (i) 34 cm

(ii) 10 seedlings have heights less than 34 cm.

2. (a) Class interval Frequency Midpoint21 – 25 2 2326 – 30 3 2831 – 35 4 3336 – 40 8 3841 – 45 5 4346 – 50 2 48

(b) Modal class = 36 – 40 postcards

(c) Estimated mean

=

(23 × 2) + (28 × 3) + (33 × 4) + (38 × 8) + (43 × 5) + (48 × 2)2 + 3 + 4 + 8 + 5 + 2

= 87724

= 36.54

8



(d)

1

0

2

3

4

5

6

7

8

9

20.5 25.5 30.5 35.5 40.5 45.5 50.5

Fre

quen

cy

Number of postcards

(e) Number of students who collect less than 31 postcards

= 2 + 3 = 5

3. (a) Mass (kg) Frequency Midpoint

1 – 3 2 24 – 6 6 57 – 9 7 8

10 – 12 9 1113 – 15 5 1416 – 18 4 1719 – 21 2 2022 – 24 1 23

(b) Mean mass =

(2 × 2) + (6 × 5) + (7 × 8) + (9 × 11) + (5 × 14) + (4 × 17)

+ (2 × 20) + (1 × 23)––––––––––––––––––––––––––2 + 6 + 7 + 9 + 5 + 4 + 2 + 1

= 390––––36

= 10.83 kg

(c)

1

0

2

3

4

5

6

7

8

9

3.50.5 6.5 9.5 12.5

Freq

uenc

yMass (kg)

15.5 18.5 21.5 24.5

(d) Modal class = 10 − 12 kg

4. (a) Estimated mean mass

=

(53 × 2) + (58 × 5) + (63 × 15) + (68 × 18) + (73 × 32) + (78 × 9) + (83 × 3)

2 + 5 + 15 + 18 + 32 + 9 + 3

= 5852–––––84

= 69.67 g

(b) Cumulative frequency Upper boundary0 50.52 55.57 60.522 65.540 70.572 75.581 80.584 85.5

9



(c)

10

0

20

30

40

50

60

70

80

90

55.550.5 60.5 65.5 70.5

Cum

ulat

ive

freq

uenc

y

Mass (g)75.5 80.5 85.5

(d) 30 cakes

5. (a) Mark Midpoint Frequency

40 – 44 42 445 – 49 47 750 – 54 52 1055 – 59 57 860 – 64 62 565 – 69 67 470 – 74 72 2

(b) (i) Modal class = 50 − 54 marks

(ii) Mean mark

=

(42 × 4) + (47 × 7) + (52 × 10) + (57 × 8) + (62 × 5) + (67 × 4)

+ (72 × 2)––––––––––––––––––––––––––––

4 + 7 + 10 + 8 + 5 + 4 + 2

= 2195–––––40

= 54.88 marks

(c)

039.5 44.5 49.5

Freq

uenc

yMark

54.5 59.5 64.5 69.5 74.5

1

2

3

4

5

6

7

8

9

10

6. (a) Class interval (hours) Frequency Midpoint

6 – 10 0 811 – 15 3 1316 – 20 7 1821 – 25 9 2326 – 30 8 2831 – 35 5 3336 – 40 0 38

(b) Modal class = 21 – 25 hours

(c) Estimated mean

=

(13 × 3) + (18 × 7) + (23 × 9) + (28 × 8) + (33 × 5)

3 + 7 + 9 + 8 + 5

= 761––––32

= 23.78 hours

10



(d)

1

2

8 13 18 23 28 33 38

3

4

5

6

7

8

9

10

0

Fre

quen

cy

Time (hours)

(e) Number of students who spent less than 21 hours on Internet

= 3 + 7 = 10

7. (a) Mass (g) Midpoint Frequency

20 – 24 22 5

25 – 29 27 8

30 – 34 32 11

35 – 39 37 9

40 – 44 42 7

45 – 49 47 2

(b) Mean mass

=

(22 × 5) + (27 × 8) + (32 × 11) + (37 × 9) + (42 × 7) + (47 × 2)

––––––––––––––––––––––––––––––––––––5 + 8 + 11 + 9 + 7 + 2

= 1399–––––42

= 33.31 g

(c)

017 22 27

Freq

uenc

y

Mass (g)32 37 42 47 52

1

2

3

4

5

6

7

8

9

10

11

(d) Number of letters which have masses of less than 35 g

= 5 + 8 + 11 = 24

Paper 1

1. Number of students who obtained a score less than 8= (5 + 5) + (5 + 5 + 3)= 28

Answer: C

2. Each square grid represents 48–––3

= 16 members.

The difference between the number of members of Photography Club and Computer Club is= (6 − 2) × 16= 4 × 16= 64

Answer: B

11



3. Each square grid represents 1250–––––5 = 250 students.

Total number of students= (2 + 5 + 3 + 4 + 4) × 250= 18 × 250= 4500

Total donation= 4500 × RM1.50= RM6750

Answer: D

4. Angle of the pie chart which represents the number of students in the Physics tuition class

= 18––––––––––––18 + 12 + 15 × 360°

= 18–––45

× 360°

= 144°

Answer: C

5. Each square grid represents 240––––3

= 80 people

The difference between the numbers of youngsters and adults= (8 − 7) × 80= 80

Answer: B

6. Modal grade = Grade CNumber of candidates who achieved better grades than the modal grade= 25 + 35= 60

Answer: C

7. Number of people who chose durian

= 360° − 120° − 100°––––––––––––––––360°

× 900

= 140°–––––360° × 900

= 350

Answer: D

8.

54°90°

x

Average

Good

Excellent

x = 360° − 54° − 90° = 216°

Number of students who obtained excellent results

= 90°–––––216° × 120

= 50

Answer: B

9.

30°

x130°

III

III

x = 360° − 130° − 30° = 200°

Number of cakes of type I

= 200°–––––30° × 45

= 300

Number of cakes of type II

= 130°–––––30° × 45

= 195

The difference between the numbers of cakes of type I and type II sold= 300 − 195= 105

Answer: C

10. Total monthly expenditure

= 360°–––––75° × 250

= RM1200

Answer: B

11. Total frequency= 2 + 3 + 3 + 6 + 5 + 5 + 4 + 2= 30The median score is the 30 + 1––––––

2 th = 15.5th value

Median = 5

Answer: C

12. The modal mark is 2.Hence, the minimum value of y is 17.

Answer: B

12



13. Mean score

= (1 × 5) + (2 × 3) + (3 × 8) + (4 × 5) + (5 × 1)––––––––––––––––––––––––––––––––––––––

x + 5 + 3 + 8 + 5 + 1 = 60––––––

x + 22 Given mean = 2

60––––––x + 22

= 2

60 = 2(x + 22) 60 = 2x + 44 60 − 44 = 2x 2x = 16

x = 16–––2

= 8

Answer: C

14. Angle of the sector which represents the number of grade C durians

= 5–––16

× 360°

= 112.5°

Answer: B

15. Number of students whose distance from home to school is less than 6.5 km= 60 + 45 + 37= 142

Answer: D

16. Number of buns sold

= 130°–––––20° × 80

= 520

Answer: B

17. Each square grid represents 15–––3

= 5 students

Total number of students= (7 + 5 + 6) × 5= 90

Number of students who pass the test= (4 + 2 + 4) × 5= 50

% of students who pass the test

= 50–––90

× 100%

= 55.6%

Answer: D

18. Mean number of passengers in each vehicle

=

(1 × 6) + (2 × 19) + (3 × 10) + (4 × 8) + (5 × 3) + (6 × 4)

–––––––––––––––––––––––––––––––––6 + 19 + 10 + 8 + 3 + 4

= 145––––50

= 2.9

Answer: C

19. Mean grade

=

(5 × 1) + (0 × 2) + (10 × 3) + (25 × 4) + (15 × 5) + (10 × 6) + (15 × 7) + (20 × 8)––––––––––––––––––––––––––––––––––––

5 + 10 + 25 + 15 + 10 + 15 + 20

= 535––––100

= 5.4

Answer: C

Paper 2

1. (a) Mark Frequency Upper boundary

30 – 39 2 39.540 – 49 5 49.550 – 59 8 59.560 – 69 6 69.570 – 79 4 79.580 – 89 3 89.590 – 99 2 99.5

(b) (i) Modal class = 50 − 59 marks(c)

1

0

2

3

4

5

6

7

8

49.539.529.5 59.5 69.5 79.5 89.5 99.5

Freq

uenc

y

Mark

13



2. (a) Age (years) Frequency Upper boundary

1 – 5 1 5.56 – 10 4 10.511 – 15 10 15.516 – 20 8 20.521 – 25 5 25.526 – 30 2 30.5

(b)

1

0

2

3

4

5

6

7

8

9

10

10.55.50.5 15.5 20.5 25.5 30.5

Freq

uenc

y

Age (years)

(c) % of participants who are more than 10 years old

= 10 + 8 + 5 + 2–––––––––––––30

× 100%

= 25–––30

× 100%

= 83.3%

3. (a) Rainfall (cm) Frequency Midpoint11 – 15 0 1316 – 20 2 1821 – 25 3 2326 – 30 8 2831 – 35 6 3336 – 40 5 3841 – 45 1 4346 – 50 0 48

(b) (i) Modal class = 26 − 30 cm

(ii) Mean rainfall

=

(2 × 18) + (3 × 23) + (8 × 28) + (6 × 33) + (5 × 38) + (1 × 43)

––––––––––––––––––––––––––––2 + 3 + 8 + 6 + 5 + 1

= 760––––25

= 30.4 cm

(c)

1

0

2

3

4

5

6

7

8

1813 23 28 33

Freq

uenc

y

Rainfall (cm)38 43 48


1 – 10 0 5.511 – 20 1 15.521 – 30 4 25.531 – 40 6 35.541 – 50 11 45.551 – 60 7 55.561 – 70 5 65.571 – 80 4 75.581 – 90 0 85.5

14



(b)

1

0

2

3

4

5

6

7

8

9

10

11

15.55.5 25.5 35.5 45.5

Freq

uenc

y

Mass (kg)55.5 65.5 75.5 85.5


30 – 34 0 3235 – 39 1 3740 – 44 4 4245 – 49 8 4750 – 54 10 5255 – 59 6 5760 – 64 3 6265 – 69 0 67

(b) (i) Modal class = 50 − 54 kg

(ii) Mean mass

=

(1 × 37) + (4 × 42) + (8 × 47) + (10 × 52) + (6 × 57) + (3 × 62)

–––––––––––––––––––––––––––––32

= 1629–––––32

= 50.91 kg

(c)

1

0

2

3

4

5

6

7

8

9

10

3732 42 47 52 57Fr

eque

ncy

Mass (kg)

62 67

(d) There are 13 members whose masses are less than 50 kg.

6. (a) Height (cm) Frequency Midpoint

145 – 149 0 147150 – 154 6 152155 – 159 4 157160 – 164 6 162165 – 169 8 167170 – 174 9 172175 – 179 7 177180 – 184 0 182

(b) (i) Modal class = 170 − 174 cm

(ii) Mean height

=

(6 × 152) + (4 × 157) + (6 × 162) + (8 × 167) + (9 × 172) + (7 × 177)

–––––––––––––––––––––––––––––––6 + 4 + 6 + 8 + 9 + 7

= 6635–––––40

= 165.9 cm

15



(c)

1

0

2

3

4

5

6

7

8

9

152147 157 162 167 172

Freq

uenc

y

Height (cm)

177 182

(d) There are 16 students whose heights lie between 150 cm and 164 cm.

7. (a) Cumulative frequency Upper boundary

0 10.5

0 + 5 = 5 20.5

5 + 12 = 17 30.5

17 + 16 = 33 40.5

33 + 24 = 57 50.5

57 + 15 = 72 60.5

72 + 12 = 84 70.5

84 + 6 = 90 80.5

(b)

10

0

20

30

40

50

60

70

80

90

20.510.5 30.5 40.5 50.5

Cum

ulat

ive

frequ

ency

Number of letters60.5 70.5 80.5

8. (a) Modal class = 25 − 29 minutes Midpoint of the modal class

= 25 + 29–––––––2

= 27 minutes

(b) Time

(minutes)Cumulative frequency

Upper boundary

5 – 9 0 9.510 – 14 8 14.515 – 19 8 + 16 = 24 19.520 – 24 24 + 20 = 44 24.525 – 29 44 + 21 = 65 29.530 – 34 65 + 17 = 82 34.535 – 39 82 + 13 = 95 39.540 – 44 95 + 5 = 100 44.5

16



(c)

Time (minutes)

0

10

20

30

40

50

60

70

80

90

100

105 15 25 4020

Q1 =

20

Q2 =

26

Q3 =

32.

5

30 35 45

Cum

ulat

ive

frequ

ency

(d) (i) Median = 26 minutes

(ii) Interquartile range = 32.5 − 20 = 12.5 minutes

9. (a) Time

(minutes)Cumulative frequency

Upper boundary

16 – 17 0 17.5

18 – 19 1 19.5

20 – 21 1 + 8 = 9 21.5

22 – 23 9 + 21 = 30 23.5

24 – 25 30 + 14 = 44 25.5

26 – 27 44 + 7 = 51 27.5

28 – 29 51 + 5 = 56 29.5

30 – 31 56 + 4 = 60 31.5

(b)

0

5

10

15

20

25

30

Cum

ulat

ive

frequ

ency

35

40

45

50

55

60

16 18 20 22 24Time (minutes)

26 28 30 32

41

Q1 =

22.

2

Q3 =

25.

8

(c) (i) Interquartile range = 25.8 − 22.2 = 3.6 minutes

(ii) The number of participants who will receive prizes is 41.

10. (a) Mass of

parcels (kg)Cumulative frequency

Upper boundary

0.1 – 0.5 0 0.550.6 – 1.0 2 1.051.1 – 1.5 2 + 4 = 6 1.551.6 – 2.0 6 + 8 = 14 2.052.1 – 2.5 14 + 13 = 27 2.552.6 – 3.0 27 + 12 = 39 3.053.1 – 3.5 39 + 7 = 46 3.553.6 – 4.0 46 + 2 = 48 4.05

17



(b)

0

5

10

15

20

25

30

Cum

ulat

ive

frequ

ency 35

40

45

50

0.5 1.0 1.5 2.0Mass (kg)

2.5 3.0 3.5 4.0

Q1 =

1.9

5

33.5

(c) (i) First quartile = 1.95 kg

(ii) The number of parcels which are less than 2.8 kg is 33.5.

11. (a) Number of students who took part in the quiz = 5 + 15 + 30 + 40 + 35 + 25 + 15 = 165

(b) Modal class = 50 − 54 marks

(c) Mark Upper

boundaryCumulative frequency

30 – 34 34.5 035 – 39 39.5 540 – 44 44.5 2045 – 49 49.5 5050 – 54 54.5 9055 – 59 59.5 12560 – 64 64.5 15065 – 69 69.5 165

(d)

29.50

20

Mark

40

60

80

100

Cum

ulat

ive

frequ

ency

120

140

160

180

34.5 39.5 44.5 49.5

Q2 =

53.

5

54.5 59.5 64.5 69.5

144

(e) (i) Median = 53.5 marks

(ii) Number of students who will receive prizes = 165 − 144 = 21

12. (a) (i) Height (m) Midpoint Frequency1.0 – 1.4 1.2 21.5 – 1.9 1.7 92.0 – 2.4 2.2 262.5 – 2.9 2.7 243.0 – 3.4 3.2 193.5 – 3.9 3.7 84.0 – 4.4 4.2 34.5 – 4.9 4.7 1

(ii) Mean height

=

(1.2 × 2) + (1.7 × 9) + (2.2 × 26) + (2.7 × 24) + (3.2 × 19) + (3.7 × 8)

+ (4.2 × 3) + (4.7 × 1)––––––––––––––––––––––––––––––––

92

= 247.4–––––92

= 2.689 m

18



(b)

Upper boundary 0.95 1.45 1.95 2.45 2.95 3.45 3.95 4.45 4.95

Cumulative frequency 0 2 11 37 61 80 88 91 92

(c)

0.450

10

Heights (m)

20

30

40

50

Cum

ulat

ive

frequ

ency

60

70

80

90

0.95 1.45 1.95

Q1 =

2.2

0

Q2 =

2.6

5

Q3 =

3.1

5

2.45 2.95 3.45 3.95 4.45 4.95

(d) Median = 2.65 m Interquartile range = 3.15 − 2.20 = 0.95 m

13. (a) Mass (kg) Frequency Cumulative

frequencyUpper

boundary16 – 20 0 0 20.521 – 25 2 2 25.526 – 30 5 7 30.531 – 35 7 14 35.536 – 40 12 26 40.541 – 45 9 35 45.546 – 50 3 38 50.551 – 55 2 40 55.5

(b) Modal class = 36 − 40 kg

(c)

Cum

ulat

ive

frequ

ency

Mass (kg)15

0

5

10

15

20

25

30

35

40

20 25 30 35 40 45 50 55

Q2 =

38

(d) (i) Median = 38 kg

(ii) 50% of the students are less than 38 kg.

Documents

15[A Math CD]