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)
Ex 17.2 Class 10 RBSE Question 1.
Ex 17.2 Class 10 RBSE Measures of Central Tendency
Solution :
Table of A.M.
RBSE Solutions For Class 10 Maths Chapter 17.2 Measures of Central Tendency
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

RBSE Solutions For Class 10 Maths Chapter 17.2 Question 2.
Exercise 17.2 Class 10 RBSE Measures of Central Tendency
Solution :
Table for A.M.
Class 10 Maths RBSE Solution Chapter 17 Measures of Central Tendency
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

Exercise 17.2 Class 10 RBSE Question 3.
RBSE Class 10 Maths Chapter 17 Measures of Central Tendency
Solution:
Table for A.M.
RBSE Solution Class 10 Maths Chapter 17 Measures of Central Tendency
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

Class 10 Maths RBSE Solution Chapter 17 Question 4.
RBSE Solutions For Class 10 Maths Chapter 17 Measures of Central Tendency
Solution:
Table for A.M.
Ch 17 Maths Class 10 RBSE Measures of Central Tendency
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

RBSE Class 10 Maths Chapter 17 Question 5.
The number of children in 100 families ia as follows :
Class 10 Maths Chapter 17.2 RBSE Measures of Central Tendency
Find their arithmetic mean.
Solution:
calculation for (RBSESolutions.com) arithmetic mean
RBSE Class 10th Maths Chapter 17 Measures of Central Tendency
Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }{ x } }{ { \Sigma f } } =\frac { 200 }{ 100 } \) = 2
Thus, arithmetic mean = 2

RBSE Solution Class 10 Maths Chapter 17 Question 6.
Weight of students in a class are (RBSESolutions.com) given in following table :
Class 10 Maths 17.2 RBSE Measures of Central Tendency
Find their arithmetic mean :
Solution :
Table for A.M.
RBSE Solutions For Class 10 Maths Chapter 17.2 Measures of Central Tendency
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

RBSE Solutions For Class 10 Maths Chapter 17 Question 7.
If mean of following (RBSESolutions.com) distribution is 7.5 then find value of p
RBSE Class 10 Maths Chapter 17.2 Measures of Central Tendency
Solution :
Arithmetic mean = 7.5
Class 10 Maths Chapter 17 Measures of Central Tendency RBSE
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

Ch 17 Maths Class 10 RBSE Question 8.
If mean following frequency (RBSESolutions.com) distribution is 1.46, then find unknown frequencies.
Class 10 Math Chapter 17 RBSE Measures of Central Tendency
Solution:
Let unknown frequencies are f1 and f2 :
RBSE Solutions For Class 10 Maths Chapter 17.2 Measures of Central Tendency
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, (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),
Exercise 17.2 RBSE Measures of Central Tendency
Putting value of f2 in equation (i),
f1 + 38 = 114
⇒ f1 = 114 – 38 = 76
Thus, unknown (RBSESolutions.com) frequencies are 76 and 38

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