Frank Solutions Class 9 Maths Chapter 22 Statistics

Frank Solutions Class 9 Maths Chapter 22 Statistics has accurate answers to every problem of the textbook. Students who find difficulty in solving problems can refer to Frank Solutions and grasp the concepts effortlessly. Practising textbook problems help to analyse the type of questions that would be asked in exams. For more conceptual knowledge, students can download the Frank Solutions Class 9 Maths Chapter 22 Statistics PDF, from the given links below.

Chapter 22 Statistics deals with the study of collection, analysis, interpretation and organization of data. Those who aspire to become an expert in Mathematics are advised to practise on a daily basis. By referring to solutions prepared by experts, students can understand the covered topics effectively. With the help of these solutions, students can easily solve the complex problems in a short duration.

Frank Solutions for Class 9 Maths Chapter 22 Statistics Download PDF

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Access Frank Solutions for Class 9 Maths Chapter 22 Statistics

1. Define primary data and secondary data.

Solution:

Data collected by investigator himself through personal observations with a definite plan or design in mind is known as primary data.

The data which has been collected previously by someone, other than the investigator but is used by the investigator for some specific purpose is known as secondary data

2. The marks obtained by the students in a class test are given below:

31, 12, 28, 45, 32, 16, 49, 12, 18, 26, 34, 39, 29, 28, 25, 46, 32, 13, 14, 26, 25, 34, 23, 23, 25, 45, 33, 22, 18, 37, 26, 19, 20, 30, 28, 38, 42, 21, 36, 19, 20, 40, 48, 15, 46, 26, 23, 33, 47, 40.

Arrange the above marks in classes each with a class size of 5 and answer the following:

(i) What is the highest score?

(ii) What is the lowest score?

(iii) What is the range?

(iv) If the pass mark is 20, how many students failed?

(v) How many students got 40 or more marks?

Solution:

Class

Tally Marks

Frequency

11 – 15

FRANK Solutions Class 9 Maths Chapter 22 - 1

5

16 – 20

FRANK Solutions Class 9 Maths Chapter 22 - 2||

7

21 – 25

FRANK Solutions Class 9 Maths Chapter 22 - 3|||

8

26 – 30

FRANK Solutions Class 9 Maths Chapter 22 - 4||||

9

31 – 35

FRANK Solutions Class 9 Maths Chapter 22 - 5||

7

36 – 40

FRANK Solutions Class 9 Maths Chapter 22 - 6|

6

41 – 45

|||

3

46 – 50

FRANK Solutions Class 9 Maths Chapter 22 - 7

5

Total

50

(i) The highest score is 49

(ii) The lowest score is 12

(iii) Range = 49 – 12 = 37

(iv) Given, pass marks is 20

So, all the students in the class 11 – 15 and 16 – 20 must have failed except for the students with score 20

Therefore,

Number of such students = 5 + 7 – 2 = 10

(v) Number of students scoring above 40 is the sum total of students in the classes 41 – 45 and 46 – 50

i.e 3 + 5 = 8

Number of students scoring exactly 40 = 2

Hence,

Number of students scoring 40 or more marks = 8 + 2 = 10

3. Find the class boundaries and class marks of the following classes:

55 – 59, 60 – 64, 65 – 69, 70 – 74, 75 – 79, 80 – 84, 85 – 89, 90 – 94 and 95 – 99

Solution:

For the class 55 – 59,

The actual lower limit = 55 – 0.5 = 54.5

The actual upper limit = 59 + 0.5 = 59.5

Therefore,

The class boundaries are 54.5 and 59.5

The class mark = (1/2) (54.5 + 59.5)

We get,

The class mark = 57

Similarly calculating for the other classes we get the following table:

Class

Class Boundaries

Class Mark

55 – 59

54.5 – 59.5

57

60 – 64

59.5 – 64.5

62

65 – 69

64.5 – 69.5

67

70 – 74

69.5 – 74.5

72

75 – 79

74.5 – 79.5

77

80 – 84

79.5 – 84.5

82

85 – 89

84.5 – 89.5

87

90 – 94

89.5 – 94.5

92

95 – 99

94.5 – 99.5

97

4. Find the actual (or true) lower and upper class limits and class – marks (or mid values) of the following classes: 2.1 – 4.0, 4.1 – 6.0 and 6.1 – 8.0

Solution:

Given classes are 2.1 – 4.0, 4.1 – 6.0 and 6.1 – 8.0

Since classes are inclusive,

We have,

Adjustment factor = {(4.1 – 4.0) / 2}

Adjustment factor = (0.1 / 2)

We get,

Adjustment factor = 0.05

True Lower Limit

True Upper Limit

Class – Mark

2.1 – 0.05 = 2.05

4.0 + 0.05 = 4.05

3.05

4.1 – 0.05 = 4.05

6.0 + 0.05 = 6.05

5.05

6.1 – 0.05 = 6.05

8.0 + 0.05 = 8.05

7.05

5. The number of goals scored by Arsenal Football Club in the English Premier League in the season 2007 were:

1, 2, 1, 3, 2, 5, 1, 6, 4, 4, 2, 3, 5, 6, 4, 2, 2, 3, 4, 1, 0, 5, 0, 5, 3, 2, 3, 4, 4, 1, 1, 2, 4, 3, 1, 4

Arrange these data in a discrete frequency distribution table and answer the following:

(i) What is the range of the number of goals scored by AFC?

(ii) How many times did AFC score 3 or more than 3 goals?

(iii) Which variate has the highest frequency?

Solution:

The discrete frequency distribution table is shown below

Number of goals

Tally Marks

Frequency

0

||

2

1

FRANK Solutions Class 9 Maths Chapter 22 - 8||

7

2

FRANK Solutions Class 9 Maths Chapter 22 - 9||

7

3

FRANK Solutions Class 9 Maths Chapter 22 - 10|

6

4

FRANK Solutions Class 9 Maths Chapter 22 - 11|||

8

5

||||

4

6

||

2

(i) Maximum goals scored = 6

Minimum goals scored = 0

Hence,

Range of the goals scored = 6 – 0 = 6

(ii) Number of times AFC scored 3 or more goals = 6 + 8 + 4 + 2 = 20

(iii) The variate which has highest frequency is 4

6. Prepare a cumulative frequency distribution table of the marks scored by 60 students in a test are given below:

Marks

Number of students

0 – 10

4

10 – 20

15

20 – 30

21

30 – 40

12

40 – 50

8

Solution:

The cumulative frequency distribution table of the marks scored by 60 students is shown below:

Marks

Number of students

Cumulative frequency

0 – 10

4

4

10 – 20

15

19

20 – 30

21

40

30 – 40

12

52

40 – 50

8

60

7. The table given below shows the ages of patients being treated in a hospital. Construct a cumulative frequency distribution table for the same:

Age

Number of patients

10 – 20

90

20 – 30

50

30 – 40

60

40 – 50

80

50 – 60

50

60 – 70

30

Solution:

The cumulative frequency distribution table is shown below

Age

Number of patients

Cumulative frequency

10 – 20

90

90

20 – 30

50

140

30 – 40

60

200

40 – 50

80

280

50 – 60

50

330

60 – 70

30

360

8. The electricity bills of 45 houses in a particular locality are given below. Tabulate the given data and present it as a cumulative frequency table with one of the classes being 300 – 450:

748, 567, 890, 231, 150, 458, 356, 762, 386, 824, 525, 663, 724, 841, 315, 641, 156, 712, 156, 317, 814, 547, 879, 456, 463, 664, 175, 584, 515, 487, 871, 511, 522, 454, 247, 819, 412, 326, 445, 311, 321, 545, 344, 266, 351.

Solution:

The table for the given data is as follows:

Class

Tally Marks

Frequency

Cumulative frequency

150 – 300

FRANK Solutions Class 9 Maths Chapter 22 - 12||

7

7

300 – 450

FRANK Solutions Class 9 Maths Chapter 22 - 13
FRANK Solutions Class 9 Maths Chapter 22 - 14|

11

18

450 – 600

FRANK Solutions Class 9 Maths Chapter 22 - 15
FRANK Solutions Class 9 Maths Chapter 22 - 16|||

13

31

600 – 750

FRANK Solutions Class 9 Maths Chapter 22 - 17||

7

38

750 – 900

FRANK Solutions Class 9 Maths Chapter 22 - 18||

7

45

9. From the cumulative frequency distribution given below, construct a frequency distribution table:

Marks

Cumulative frequency

Less than 10

10

Less than 20

18

Less than 30

32

Less than 40

45

Less than 50

50

Solution:

The frequency distribution table is shown below

Class

Cumulative frequency

Frequency

0 – 10

10

10

10 – 20

18

18 – 10 = 8

20 – 30

32

32 – 18 =14

30 – 40

45

45 – 32 = 13

40 – 50

50

50 – 45 = 5

10. Construct a frequency distribution table from the given cumulative frequency distribution showing the weights of 750 students in a school:

Weight (in kg)

Cumulative frequency

More Than 25

750

More Than 30

640

More Than 35

615

More Than 40

485

More Than 45

370

More Than 50

220

More Than 55

124

More Than 60

49

More Than 65

24

More Than 70

0

(a) Find the number of students whose weight lie in the interval 40 – 45

(b) Find the interval which has the most number of students

Solution:

Frequency distribution table is shown below

Weight (in kg)

Cumulative Frequency

25 – 30

(750 – 640) = 110

30 – 35

(640 – 615) = 25

35 – 40

(615 – 485) = 130

40 – 45

(485 – 370) = 115

45 – 50

(370 – 220) = 150

50 – 55

(220 – 124) = 96

55 – 60

(124 – 49) = 75

60 – 65

(49 – 24) = 25

65 – 70

(24 – 0) = 24

70 – 75

0

(a) The number of students whose weight lie in the interval 40 – 45 is 115

(b) The interval which has the most number of students is 45 – 50

11. For the set of numbers given below, find mean:

(i) 5, 7, 8, 4, 6

(ii) 3. 0, 5, 2, 6, 2

Solution:

Mean = (Σ x) / N

(i) Mean = {(5 + 7 + 8 + 4 + 6) / 5}

Mean = (30 / 5)

We get,

Mean = 6

(ii) Mean = {(3 + 0 + 5 + 2 + 6 + 2) / 6}

Mean = (18 / 6)

We get,

Mean = 3

12. Calculate mean of the following:

4, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 11, 3

Solution:

x

f

fx

3

1

3

4

1

4

6

3

18

7

4

28

8

2

16

9

2

18

11

1

11

Total

14

98

Hence,

Mean = (Σfx) / (Σf)

Mean = 98 / 14

We get,

Mean = 7

13. The weights of the seven members of a family, in kilograms are given below:

20, 52, 56, 72, 64, 13, 80.

Find mean weight.

Solution:

Mean = (Σx / N)

Mean = (20 + 52 + 56 + 72 + 64 + 13 + 80) / 7

Mean = (357 / 7)

We get,

Mean = 51 kg

14. A boy scored the following marks in various class tests during a terminal exam, each test being marked out of 20.

17, 15, 16, 7, 10, 14, 12, 19, 16, 12

Find his average mean marks.

Solution:

Mean = (Σx / N)

Average mean marks = {(17 + 15 + 16 + 7 + 10 + 14 + 12 + 19 + 16 + 12) / 10}

Average mean marks = (138 / 10)

We get,

Average mean marks = 13.8

15. Individual scores of a school cricket eleven in a match are given below:

10, 9, 31, 45, 0, 4, 8, 15, 12, 0, 6

Find the average score.

Solution:

Mean = (Σx / N)

Average score = {(10 + 9 + 31 + 45 + 0 + 4 + 8 + 15 + 12 + 0 + 6) / 11}

Average score = (140 / 11)

We get,

Average score = 12.7

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