Chapter 17 Measures of Central Tendency Ex 17.8

Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.8

Question 1.
Find mode of the following distribution :
(i) 2       5     7      5    3      1      5    8    7    5
(ii) 2      4     6      2    6      6      7    8
(iii) 2.5 2.5   2.1   2.5  2.7   2.8    2.5
Solution :
(i) Here 5 has maximum frequency (4 times) so mode of data will be 5.
(ii) Here 6 has maximum frequency (3 times) so mode of data will be 6.
(iii) Here 2.5 has maximum frequency (4 times) so 2.5 will be mode of data.

Question 2.
Find mode of the following frequency distribution :

Solution :
(i) 5 has maximum frequency (6), so mode = 5
(ii) 1.3 has maximum frequency (80), so mode = 1.3

Question 3.
The number of members in 30 families of a village are according to following table.
Find their mode.

Solution :
Here no. of members 6 has maximum frequency 10 so mode = 6.

Question 4.
Age in years of 20 students of a class are as follows:
15        16        13        14        14
13        15        14        13        13
14        12        15        14        16
13        14        14        13        15
Find mode by expressing them in frequency distribution :
Solution :

From data it is clear that students of age 14 yrs have maximum frequency, so mode = 14

Question 5.
Marks obtained by students are given below, find the mode of marks.

Solution :
From table it is clear that students who got 40 marks have maximum frequency (26) so mode = 40.

Find mode of following frequency distribution (Q. 6-8)
Question 6.

Solution :
Maximum frequency is 16 and its corresponding class is 20 – 25
So modal class = 20 – 25
Now, l = 20, f1 = 16, f0 = 7, f2 = 12 and h = 5

= 20 +  = 20 + 3.46 = 23.46
Thus, mode = 23.46 (approx)

Question 7.

Solution :
Here, Maximum frequency is 16 and its corresponding class is 20 – 30
Thus, modal class = 20 – 30
Now, l = 20, f1 = 14, f0 = 12, f2 = 10 and h = 10

Thus, required mode = 23.33 marks (approx)

Question 8.

Solution :
Here, Maximum frequency is 42 and its corresponding class is 40 – 50
Thus, modal class = 40 – 50
Now, l = 40, f1 = 42, f0 = 28, f2 = 20 and h = 10

= 40 +  = 40 + 3.88 = 43.88
Thus, required mode = 43.88 marks (approx)

Question 9.

Solution :
Here, Maximum frequency is 25 and its corresponding class is 58 – 61.
Thus, modal class = 58 – 61
Now, l = 58, f1 = 25, f0 = 20, f2 = 10 and h = 3

= 58 +  = 58 + 0.75 = 58.75
Thus, required mode = 58.75