# 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

Ex 17.7 Class 10 RBSE 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

RBSE Solutions For Class 10 Maths Chapter 17.7 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)
RBSE Solution Class 10 Maths Chapter 17 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

RBSE Solutions For Class 10 Maths Chapter 17 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

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