SELINA Solutions for Class 9 Maths Chapter 18 - Statistics

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Chapter 18 - Statistics Exercise Ex. 18(A)

Solution 1

(a)Discrete variable.

(b)Continuous variable.

(c)Discrete variable.

(d)Continuous variable.

(e)Discrete variable.

Solution 2

The frequency table for the given distribution is

Marks

Tally Marks

Frequency

 

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.

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

Solution 5

In case of frequency 10 - 19 the lower class limit is 10, upper class limit is 19 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

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

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:

 

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(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.

Solution 8

(i)The cumulative frequency distribution table is

C.I

c.f

 

(ii)The cumulative frequency distribution table is

C.I

c.f

Solution 9

(i)The frequency distribution table is

C.I

c.f

 

(ii)The frequency distribution table is

C.I

c.f

Solution 10

The frequency table is

C.I

c.f

 

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

Solution 12

Class Interval

Frequency

Cumulative Frequency

 

65-74

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 Exercise Ex. 18(B)

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.

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:

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:

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:

  

Solution 5(i)

(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

  

Solution 5(ii)

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