RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.2

RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.2.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 17
Chapter Name	Measures of Central Tendency
Exercise	Exercise 17.2
Number of Questions Solved	8
Category	RBSE Solutions

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

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

Solution :
Table of A.M.

Thus, A.M. (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 99 }{ 14 } \) = 7.07
Thus, required A.M. = 7.07

Question 2.

Solution :
Table for A.M.

Thus, arithmetic (RBSESolutions.com) mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 151 }{ 20 } \) = 7.55
Thus, required A.M. = 7.55

Question 3.

Solution:
Table for A.M.

Thus, arithmetic (RBSESolutions.com) mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 72 }{ 210 } \) = 0.34
Thus, required A.M. = 0.34

Question 4.

Solution:
Table for A.M.

Thus, arithmetic (RBSESolutions.com) mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 27.5 }{ 50 } \) = 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 (RBSESolutions.com) arithmetic mean

Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }{ x } }{ { \Sigma f } } =\frac { 200 }{ 100 } \) = 2
Thus, arithmetic mean = 2

Question 6.
Weight of students in a class are (RBSESolutions.com) given in following table :

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

Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 717 }{ 30 } \) = 23.9
Thus, required A.M. = 23.9

Question 7.
If mean of following (RBSESolutions.com) distribution is 7.5 then find value of p

Solution :
Arithmetic mean = 7.5

Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }{ x } }{ { \Sigma f } } \)
7.5 = \(\frac { 303+9P }{ 41+P } \)
⇒ (41 + p) (7.5) = 303 + 9P
⇒ (41 × 7.5) + 7.5p = 303 + 9P
⇒ 307.5 – 303 = 9P – 7.5P
⇒ 1.5P = 4.5
⇒ P = \(\frac { 4.5 }{ 1.5 } \) = 3
Thus, value of P = 3

Question 8.
If mean following frequency (RBSESolutions.com) distribution is 1.46, then find unknown frequencies.

Solution:
Let unknown frequencies are f₁ and f₂ :

Given : Σf_i = 200
But from table Σf_i = 86 + f₁ + f₂
So, 86 + f₁ + f₂ = 200
⇒ f₁ + f₂ = 200 – 86 = 114
⇒ f₁ + f₂ = 114 ……(i)
According to question, (RBSESolutions.com) arithmetic mean = 1.46
or \(\overline { x }\) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } \)
⇒ 1.46 = \(\frac { { 140+f }_{ 1 }{ +2f }_{ 2 } }{ 200 } \)
⇒ 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 (RBSESolutions.com) frequencies are 76 and 38

We hope the given RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.2, drop a comment below and we will get back to you at the earliest.