(1) For class interval 20 – 25 write the lower class limit and the upper class limit.

Solution:

The lower class limit is 20 and the upper class limit is 25.

(2) Find the class-mark of the class 35 – 40.

Solution:

 Class Mark = \(\displaystyle \frac{\text{Lower Class Limit + Upper Class Limit}}{2}\)

∴ Class Mark = \(\displaystyle \frac{35 + 40}{2}\)

∴ Class Mark = \(\displaystyle \frac{75}{2}\)

∴ Class Mark = 37.5

(3) If class mark is 10 and class width is 6 then find the class.

Solution:

Let, the lower class limit be L and the upper class limit be H.

Class Mark = \(\displaystyle \frac{\text{Lower Class Limit + Upper Class Limit}}{2}\)

Class Mark = 10 ... (Given)

∴ \(\displaystyle 10 = \frac{\text{L + H}}{2}\)

∴ \(\displaystyle 10 \times 2 = \text{L + H}\)

∴ 20 = L + H

i.e. L + H = 20 ... (i)

Also, Class Width = 6 ... (Given)

∴ H − L = 6

∴ H = 6 + L

i.e. H = L + 6 ... (ii)

Substituting the value of H in (i),

 L + H = 20 ... (i)

∴ L + L + 6 = 20

∴ 2L + 6 = 20

∴ 2L = 20 − 6

∴ 2L = 14

∴ L = \(\displaystyle \frac{14}{2}\)

∴ L = 7

Substituting the value of L in (ii),

 H = L + 6 ... (ii)

∴ H = 7 + 6

∴ H = 13

Therefore, the class is 7 – 13.

(4) Complete the following table:

Classes (age)	Tally Marks	Frequency (No. of students)
12 – 13	\(\enclose{downdiagonalstrike}{||||}\)
13 – 14	\(\enclose{downdiagonalstrike}{||||}\enclose{downdiagonalstrike}{||||} ||||\)
14 – 15
15 – 16	\(||||\)
\(\displaystyle N = \sum{f} = 35\)

Solution:

The completed table is given below:

Classes (age)	Tally Marks	Frequency (No. of students)
12 – 13	\(\enclose{downdiagonalstrike}{||||}\)	5
13 – 14	\(\enclose{downdiagonalstrike}{||||}\enclose{downdiagonalstrike}{||||} ||||\)	14
14 – 15	\(\enclose{downdiagonalstrike}{||||}\enclose{downdiagonalstrike}{||||} ||\)	12
15 – 16	\(||||\)	4
\(\displaystyle N = \sum{f} = 35\)

(5) In a ‘tree plantation’ project of a certain school there are 45 students of ‘Harit Sena’. The record of trees planted by each student is given below:
3, 5, 7, 6, 4, 3, 5, 4, 3, 5, 4, 7, 5 , 3, 6, 6, 5, 3, 4, 5, 7, 3, 5, 6, 4, 4, 3, 5, 6, 6, 4, 3, 5, 7, 3, 4, 5, 7, 6, 4, 3, 5, 4, 4, 7.
Prepare a frequency distribution table of the data.

Solution:

The completed table is given below:

Class (No. of saplings)	Tally Marks	Frequency (f)
3	\(\enclose{downdiagonalstrike}{||||} \enclose{downdiagonalstrike}{||||}\)	10
4	\(\enclose{downdiagonalstrike}{||||}\enclose{downdiagonalstrike}{||||} |\)	11
5	\(\enclose{downdiagonalstrike}{||||}\enclose{downdiagonalstrike}{||||} |\)	11
6	\(\enclose{downdiagonalstrike}{||||} ||\)	7
7	\(\enclose{downdiagonalstrike}{||||} |\)	6
\(\displaystyle N = \sum{f} = 45\)

(6) The value of π upto 50 decimal places is given below:
3 . 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0 2 8 8 4 1 9 7 1 6 9 3 9 9 3 7 5 1 0
From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.

Solution:

The completed table is given below:

Digit	Tally Marks	Frequency (f)
0	\(||\)	2
1	\(\enclose{downdiagonalstrike}{||||}\)	5
2	\(\enclose{downdiagonalstrike}{||||}\)	5
3	\(\enclose{downdiagonalstrike}{||||} |||\)	8
4	\(||||\)	4
5	\(\enclose{downdiagonalstrike}{||||}\)	5
6	\(||||\)	4
7	\(||||\)	4
8	\(\enclose{downdiagonalstrike}{||||}\)	5
9	\(\enclose{downdiagonalstrike}{||||} |||\)	8
\(\displaystyle N = \sum{f} = 50\)

(7)(i) In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.

Class Mark	Frequency
5	3
15	9
25	15
35	13

Solution:

For the class having class mark 5, Let L be the lower class limit and H be the upper class limit.

Class Mark = \(\displaystyle \frac{\text{Upper Class Limit + Lower Class Limit}}{2}\)

\(\displaystyle \therefore 5 = \frac{\text{H} + \text{L}}{2}\)

\(\displaystyle \therefore 5 \times 2 = \text{H} + \text{L}\)

\(\displaystyle \therefore 10 = \text{H} + \text{L}\)

i.e. H + L = 10 ... (i)

And Class Width = 35 − 25 = 25 − 15 = 15 − 5 = 10

∴ H − L = 10 ... (ii)

Adding (i) and (ii):

	H	+	L	=	10	... (i)
+	H	−	L	=	10	... (ii)
	2H			=	20

\(\therefore \displaystyle \text {H} = \frac{20}{2}\)

∴ H = 10 ... (iii)

Substituting the value of H in (i):

H + L = 10

∴ 10 + L = 10

∴ L = 10 − 10

∴ L = 0 ... (iv)

∴ The first class is 0 – 10

∴ The other classes are: 10 – 20, 20 – 30, 30 – 40

∴ The exclusive classes are as follows:

Class	Class Mark	Frequency
0 – 10	5	3
10 – 20	15	9
20 – 30	25	15
30 – 40	35	13

And the inclusive classes are:

Class	Class Mark	Frequency
0   9	4.5	3
10 – 19	14.5	9
20 – 29	24.5	15
30 – 39	34.5	13
40 – 49	44.5	...

(7)(ii) In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.

Class Mark	Frequency
22	6
24	7
26	13
28	4

Solution:

For the class having class mark 22, Let L be the lower class limit and H be the upper class limit.

Class Mark = \(\displaystyle \frac{\text{Upper Class Limit + Lower Class Limit}}{2}\)

\(\displaystyle \therefore 22 = \frac{\text {H} + \text{L}}{2}\)

\(\displaystyle \therefore 22 \times 2 = \text {H} + \text{L}\)

\(\displaystyle \therefore 44 = \text {H} + \text{L}\)

i.e. H + L = 44 ... (i)

And Class Width = 28 − 26 = 26 − 24 = 24 − 22 = 2

∴ H − L = 2 ... (ii)

Adding (i) and (ii):

	H	+	L	=	44	... (i)
+	H	−	L	=	2	... (ii)
	2H			=	46

\(\therefore \displaystyle \text H = \frac{46}{2}\)

∴ H = 23 ... (iii)

Substituting the value of H in (i):

 H + L = 44

∴ 23 + L = 44

∴ L = 44 − 23

∴ L = 21 ... (iv)

∴ The first class is 21 – 23

∴ The other classes are: 23 – 25, 25 – 27, 27 – 29

∴ The exclusive classes are as follows:

Class	Class Mark	Frequency
21 – 23	22	6
23 – 25	24	7
25 – 27	26	13
27 – 29	28	4

And the inclusive classes are:

Class	Class Mark	Frequency
21.5 – 22.5	22	6
23.5 – 24.5	24	7
25.5 – 26.5	26	13
27.5 – 28.5	28	4
29.5 – 30.5	30	...