Chapter 17 Measures of Central Tendency Ex 17.2

Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.2

Find the mean (arithmetic) of following frequency distribution (Q. 1 – 4)
Question 1.

Solution :
Table of A.M.

Thus, A.M. () =  = 7.07
Thus, required A.M. = 7.07

Question 2.

Solution :
Table for A.M.

Thus, arithmetic mean () =  = 7.55
Thus, required A.M. = 7.55

Question 3.

Solution:
Table for A.M.

Thus, arithmetic mean () =  = 0.34
Thus, required A.M. = 0.34

Question 4.

Solution:
Table for A.M.

Thus, arithmetic mean () =  = 0.55
Thus, required A.M. = 0.55

Question 5.
The number of children in 100 families ia as follows :

Find their arithmetic mean.
Solution:
calculation for arithmetic mean

Thus, arithmetic mean () =  = 2
Thus, arithmetic mean = 2

Question 6.
Weight of students in a class are given in following table :

Find their arithmetic mean :
Solution :
Table for A.M.

Thus, arithmetic mean () =  = 23.9
Thus, required A.M. = 23.9

Question 7.
If mean of following distribution is 7.5 then find value of p

Solution :
Arithmetic mean = 7.5

Thus, arithmetic mean () = 
7.5 = 
⇒ (41 + p) (7.5) = 303 + 9P
⇒ (41 × 7.5) + 7.5p = 303 + 9P
⇒ 307.5 – 303 = 9P – 7.5P
⇒ 1.5P = 4.5
⇒ P =  = 3
Thus, value of P = 3

Question 8.
If mean following frequency distribution is 1.46, then find unknown frequencies.

Solution:
Let unknown frequencies are f1 and f2 :

Given : Σfi = 200
But from table Σfi = 86 + f1 + f2
So, 86 + f1 + f2 = 200
⇒ f1 + f2 = 200 – 86 = 114
⇒ f1 + f2 = 114 ……(i)
According to question, arithmetic mean = 1.46
or  = 
⇒ 1.46 = 
⇒ 140 + f1 + 2f2 = 292
f1 + f2 = 292 – 140
f1 + 2f2 = 152 ….(ii)
subtracting equation (ii)from(i),

Putting value of f2 in equation (i),
f1 + 38 = 114
⇒ f1 = 114 – 38 = 76
Thus, unknown frequencies are 76 and 38