Chapter 18 : Statistics - Selina Solutions for Class 9 Maths ICSE

Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.

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Chapter 18 - Statistics Excercise Ex. 18(A)

Question 1

State, which of the following variables are continuous and which are discrete:

(a) Number of children in your class.

(b) Distance travelled by a car.

(c) Sizes of shoes.

(d) Time.

(e) Number of patients in a hospital.

Solution 1

(a)Discrete variable.

(b)Continuous variable.

(c)Discrete variable.

(d)Continuous variable.

(e)Discrete variable.

Question 2

Given below are the marks obtained by 30 students in an examination:

08

17

33

41

47

23

20

34

09

18

42

14

30

19

29

11

36

48

40

24

22

02

16

21

15

32

47

44

33

01

Taking class intervals 1 - 10, 11 - 20, ....., 41 - 50; make a frequency table for the above distribution.

Solution 2

The frequency table for the given distribution is

Marks

Tally Marks

Frequency

 

Question 3

The marks of 24 candidates in the subject mathematics are given below:

45

48

15

23

30

35

40

11

29

0

3

12

48

50

18

30

15

30

11

42

23

2

3

44

 

The maximum marks are 50. Make a frequency distribution taking class intervals 0 - 10, 10-20, ...... .

Solution 3

The frequency table for the given distribution is

Marks

Tally Marks

Frequency

 

In this frequency distribution, the marks 30 are in the class of interval 30 - 40 and not in 20 - 30. Similarly, marks 40 are in the class of interval 40 - 50 and not in 30 - 40.

Question 4

Fill in the blanks:

(a) A quantity which can very from one individual to another is called a .............

(b) Sizes of shoes are ........... variables.

(c) Daily temperatures is ........... variable.

(d) The range of the data 7, 13, 6, 25, 18, 20, 16 is ............

(e) In the class interval 35 - 46; the lower limit is .......... and upper limit is .........

(f) The class mark of class interval 22 - 29 is .......... .

Solution 4

(a)Variable.

(b)Discrete variables.

(c)Continuous variable.

(d)The range is

(e)Lower limit is and upper limit is

(f)The class mark is

Question 5

Find the actual lower class limits, upper class limits and the mid-values of the classes: 10 - 19, 20 - 29, 30 - 39 and 40 - 49.

Solution 5

In case of frequency 10 - 19 the lower class limit is 10, upper class limit is 20 and mid-value is fraction numerator 10 plus 19 over denominator 2 end fraction equals 14.5

In case of frequency 20 - 29 the lower class limit is 20, upper class limit is 29 and mid-value is fraction numerator 20 plus 29 over denominator 2 end fraction equals 24.5

In case of frequency 30 - 39 the lower class limit is 30, upper class limit is 39 and mid-value is fraction numerator 30 plus 39 over denominator 2 end fraction equals 34.5

In case of frequency 40 - 49 the lower class limit is 40, upper class limit is 49 and mid-value is fraction numerator 40 plus 49 over denominator 2 end fraction equals 44.5

Question 6

Find the actual lower and upper class limits and also the class marks of the classes:

1.1 - 2.0, 2.1 -3.0 and 3.1 - 4.0.

Solution 6

In case of frequency 1.1 - 2.0 the lower class limit is 1.1, upper class limit is 2.0 and class mark

is fraction numerator 1.1 plus 2.0 over denominator 2 end fraction equals 1.55

In case of frequency 2.1 - 3.0 the lower class limit is 2.1, upper class limit is 3.0 and class mark

is fraction numerator 2.1 plus 3.0 over denominator 2 end fraction equals 2.55

In case of frequency 3.1 - 4.0 the lower class limit is 3.1, upper class limit is 4.0 and class mark

is fraction numerator 3.1 plus 4.0 over denominator 2 end fraction equals 3.55

Question 7

Use the table given below to find:

(a) The actual class limits of the fourth class.

(b) The class boundaries of the sixth class.

(c) The class mark of the third class.

(d) The upper and lower limits of the fifth class.

(e) The size of the third class.

Class Interval

Frequency

30 - 34

7

35 - 39

10

40 - 44

12

45 - 49

13

50 - 54

8

55 - 59

4

 

 

Solution 7

(a)

The actual class limit of the fourth class will be:

44.5-49.5.

(b)

The class boundaries of the sixth class will be:

54.5-59.5

(c)

The class mark of the third class will be the average of the lower bound and the upper bound of the interval. Therefore class mark will be:

 

Syntax error from line 1 column 49 to line 1 column 73. Unexpected 'mathsize'.

(d)

The upper and lower limit of the fifth class is 54 and 50 respectively.

(e)

The size of the third class will be: 44 - 40 + 1 =5.

Question 8

Construct a cumulative frequency distribution table from the frequency table given below:

(i)

Class Interval

Frequency

0 - 8

9

8 - 16

13

16 - 24

12

24 - 32

7

32 - 40

15

(ii)

Class Interval

Frequency

1 - 10

12

11 - 20

18

21 - 30

23

31 - 40

15

41 - 50

10

 

Solution 8

(i)The cumulative frequency distribution table is

C.I

c.f

 

(ii)The cumulative frequency distribution table is

C.I

c.f

Question 9

Construct a frequency distribution table from the following cumulative frequency distribution:

(i)

Class Interval

Cumulative Frequency

10 - 19

8

20 - 29

19

30 - 39

23

40 - 49

30

(ii)

C.I.

C.F.

5 - 10

18

10 - 15

30

15 - 20

46

20 - 25

73

25 - 30

90

 

Solution 9

(i)The frequency distribution table is

C.I

c.f

 

(ii)The frequency distribution table is

C.I

c.f

Question 10

Construct a frequency table from the following data:

Marks

No. of students

less than 10

6

less than 20

15

less than 30

30

less than 40

39

less than 50

53

less than 60

70

Solution 10

The frequency table is

C.I

c.f

 

Question 11

Construct the frequency distribution table from the following cumulative frequency table:

Ages

No. of students

Below 4

0

Below 7

85

Below 10

140

Below 13

243

Below 16

300

(i) State the number of students in the age group 10 - 13.

(ii) State the age-group which has the least number of students.

Solution 11

The frequency distribution table is

C.I

c.f

(i)The number of students in the age group is

(ii)The age group which has the least number of students is

Question 12

Fill in the blanks in the following table:

Class Interval

Frequency

Cumulative Frequency

25 - 34

......

15

35 - 44

......

28

45 - 54

21

......

55 - 64

16

......

65 - 74

......

73

75 - 84

12

......

 

Solution 12

Class Interval

Frequency

Cumulative Frequency

 

65-74

Question 13

The value of   upto 50 decimal place is

3.14159265358979323846264338327950288419716939937510

(i) Make a frequency distribution table of digits from 0 to 9 after the decimal place.

(ii) Which are the most and least occurring digits?

Solution 13

X

0

1

2

3

4

5

6

7

8

9

F

2

5

5

8

4

5

4

4

5

8

 

Most occurring digits are 3 and 9. Least occurring digits are 0.

Chapter 18 - Statistics Excercise Ex. 18(B)

Question 1

Construct a frequency polygon for the following distribution:

Class-intervals

0 - 4

4 - 8

8 - 12

12 - 16

16 - 20

20 - 24

Frequency

4

7

10

15

11

6

 

Solution 1

The frequency polygon is shown in the following figure

Steps:

(i)Drawing a histogram for the given data.

(ii)Marking the mid-point at the top of each rectangle of the histogram drawn.

(iii)Also, marking mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.

(iv)Joining the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Question 2

Construct a combined histogram and frequency polygon for the following frequency distribution:

Class-Intervals

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

Frequency

3

5

6

4

2

Solution 2

Steps:

  1. Draw a histogram for the given data.
  2. Mark the mid-point at the top of each rectangle of the histogram drawn.
  3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
  4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

The required combined histogram and frequency polygon is shown in the following figure:

Question 3

Construct a frequency polygon for the following data:

Class-Intervals

10 - 14

15 - 19

20 - 24

25 - 29

30 - 34

Frequency

5

8

12

9

4

Solution 3

The class intervals are inclusive. We will first convert them into the exclusive form.

Class-Interval

Frequency

9.5 - 14.5

5

14.5 - 19.5

8

19.5 - 24.5

12

24.5 - 29.5

9

29.5 - 34.5

4

 

Steps:

  1. Draw a histogram for the given data.
  2. Mark the mid-point at the top of each rectangle of the histogram drawn.
  3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
  4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon. 

The required frequency polygon is as follows:

Question 4

The daily wages in a factory are distributed as follows:

Daily wages (in Rs.)

125 - 175

175 - 225

225 - 275

275 - 325

325 - 375

Number of workers

4

20

22

10

6

 

Draw a frequency polygon for this distribution.

Solution 4

Steps:

  1. Draw a histogram for the given data.
  2. Mark the mid-point at the top of each rectangle of the histogram drawn.
  3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
  4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

The required frequency polygon is as follows:

  

Question 5

Draw frequency polygons for each of the following frequency distribution:

(a) using histogram

(b) without using histogram

C.I

10 - 30

30 - 50

50 - 70

70 - 90

90 - 110

110 - 130

130 - 150

ƒ 

4

7

5

9

5

6

4

Solution 5

(a) Using Histogram:

C.I.

f

10 - 30

4

30 - 50

7

50 - 70

5

70 - 90

9

90 - 110

5

110 - 130

6

130 - 150

4

 

Steps:

  1. Draw a histogram for the given data.
  2. Mark the mid-point at the top of each rectangle of the histogram drawn.
  3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
  4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon. 

 

(b) Without using Histogram:

Steps: 

  1. Find the class-mark (mid-value) of each given class-interval.

    begin mathsize 12px style Class text-end text mark equals mid text -value end text equals fraction numerator text Upper limit end text plus Lower text    end text limit over denominator 2 end fraction end style
  2. On a graph paper, mark class-marks along X-axis and frequencies along Y-axis.
  3. On this graph paper, mark points taking values of class-marks along X-axis and the values of their corresponding frequencies along Y-axis.

  4. Draw line segments joining the consecutive points marked in step (3) above.

 

C.I.

Class-mark

f

-10 - 10

0

0

10 - 30

20

4

30 - 50

40

7

50 - 70

60

5

70 - 90

80

9

90 - 110

100

5

110 - 130

120

6

130 - 150

140

4

150 - 170

160

0

  

Question 6

Draw frequency polygons for each of the following frequency distribution:

(a) using histogram

(b) without using histogram

C.I

5 - 15

15 - 25

25 - 35

35 - 45

45 - 55

55 - 65

ƒ 

8

16

18

14

8

2

Solution 6

Using Histogram:

C.I.

f

5 - 15

8

15 - 25

16

25 - 35

18

35 - 45

14

45 - 55

8

55 - 65

2

 

Steps:

  1. Draw a histogram for the given data.
  2. Mark the mid-point at the top of each rectangle of the histogram drawn.
  3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
  4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

 

Without using Histogram:

Steps:

  1. Find the class-mark (mid-value) of each given class-interval.

     begin mathsize 12px style Class text-end text mark equals mid text -value end text equals fraction numerator text Upper limit end text plus Lower text    end text limit over denominator 2 end fraction end style

  2. On a graph paper, mark class-marks along X-axis and frequencies along Y-axis.
  3. On this graph paper, mark points taking values of class-marks along X-axis and the values of their corresponding frequencies along Y-axis.
  4. Draw line segments joining the consecutive points marked in step (3) above.

C.I.

Class-mark

f

-5 - 5

0

0

5 - 15

10

8

15 - 25

20

16

25 - 35

30

18

35 - 45

40

14

45 - 55

50

8

55 - 65

60

2

65 - 75

70

0

  

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