RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.7

RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.7 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 17.7.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 17
Chapter Name	Measures of Central Tendency
Exercise	Exercise 17.7
Number of Questions Solved	4
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.7

Question 1.
Marks obtained by 100 students are (RBSESolutions.com) given in the following table.

Find their median.
Solution :
Preparing cumulative (RBSESolutions.com) frequency table :

Here \(\frac { N }{ 2 }\) = 50
Cumulative frequency just above 50 is 70 whose (RBSESolutions.com) corresponding class-interval is 40 – 50.
Thus, median class =40 – 50
∵ l = 40, f = 44, C = 26, h = 10

Question 2.
Marks obtained by students of a class are given in the (RBSESolutions.com) following frequency distribution :

Find their median.
Solution :
Preparing cumulative frequency table :

Here \(\frac { N }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50
Cumulative frequency just above 50 is 74 whose (RBSESolutions.com) corresponding class-interval is 20 – 30.
Thus, median class =20 – 30
∵ l = 20, f = 42, C = 32, h = 10

= 20 + \(\frac { 180 }{ 42 }\) = 20 + 4.29 = 24.29
Thus, median class = 24.29

Find median from the following frequency (RBSESolutions.com) distribution (Q.3 and 4)
Question 3.

Solution :

Here \(\frac { N }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50
Cumulative frequency just above 50 is 65 whose (RBSESolutions.com) corresponding class-interval is 40 – 50.
Thus, median class =40 – 50
∵ l = 40, f = 30, C = 35, h = 10

= 40 + \(\frac { 150 }{ 30 }\) = 40 + 5 = 45
Thus, required median = 45

Question 4.

Solution :

Here \(\frac { N }{ 2 }\) = \(\frac { 157 }{ 2 }\) = 78.5
Cumulative frequency just above 78.5 is 122 whose (RBSESolutions.com) corresponding class-interval is 16 – 24.
Thus, median class =16 – 24
∵ l = 16, f = 50, C = 72, h = 8

= 16 + \(\frac { 52 }{ 50 }\) = 16 + 1.04 = 17.04
Thus, median (RBSESolutions.com) class = 17.04

We hope the given RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.7 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.7, drop a comment below and we will get back to you at the earliest.