RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.4

RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.4 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 17.4.

Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.4

Find the mean (arithmetic) of following (RBSESolutions.com) frequency distribution (Q. 1 – 4)
Ex 17.4 Class 10 RBSE Question 1.

Solution :
Let A = 900, h = 20
Table of find arithmetic mean :

Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma u }_{ i }{ f }_{ i } }{ { \Sigma f }_{ i } } \) × h = 900 + \(\frac { -44 }{ 100 }\) × 20 = 900 – 8.8 = 891.2
Thus, required arithmetic mean = 891.2

Exercise 17.4 Class 10 RBSE Question 2.

Solution :
Let A = 63
Table to find arithmetic (RBSESolutions.com) mean :

Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma f }_{ i }{ d }_{ i } }{ { \Sigma f }_{ i } } \) = 63 – \(\frac { 21 }{ 60 }\) = 63 – 0.35 = 62.65
Arithmetic mean = 62.59

RBSE Solutions For Class 10 Maths Chapter 17.4 Question 3.

Solution:
Let A = 275, h = 50

Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma u }_{ i }{ f }_{ i } }{ { \Sigma f }_{ i } } \) × h = 275 – \(\frac { 35 }{ 200 }\) × 50 = 275 – 8.75 = 266.25
Thus, required arithmetic (RBSESolutions.com) mean = 266.25

RBSE Solutions For Class 10 Maths Chapter 17 Question 4.

Solution:
Let A = 42.5, h = 5

Arithmetic mean = A + \(\frac { { \Sigma f }_{ i }{ u }_{ i } }{ { \Sigma f }_{ i } } \) × h = 42.5 + \(\frac { -41 }{ 70 }\) × 5 = 42.5 – 2.93 = 39.57
Thus, required A.M. = 39.57

RBSE Class 10 Maths Chapter 17 Question 5.
Find mean of the following frequency (RBSESolutions.com) distribution taking assumed mean 25.

Solution:
Let Assumed Mean (A) = 25
Table for calculation of arithmetic mean :

Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma f }_{ i }{ d }_{ i } }{ { \Sigma f }_{ i } } \) = 25 + \(\frac { -70 }{ 40 }\) = 25 – 1.75 = 23.25
Thus, required arithmetic mean = 23.25

RBSE Solution Class 10 Maths Chapter 17 Question 6.
In the following table, age distribution of patients (RBSESolutions.com) suffering from a disease in a specify class in a city are given. Find average age (in yrs) per patient.

Solution :
Let assumed mean (A) = 29.5

Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma f }_{ i }{ d }_{ i } }{ { \Sigma f }_{ i } } \) = 29.5 + \(\frac { 430 }{ 80 }\) = 29.5 + 5.37 = 34.87
Thus, arithmetic mean = 34.87 yrs

Class 10 Maths RBSE Solution Chapter 17 Question 7.
Find mean from the following (RBSESolutions.com) frequency distribution :

Solution :
Let A = 65, h = 10

Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma u }_{ i }{ f }_{ i } }{ { \Sigma f }_{ i } } \) × h = 65 + \(\frac { 32 }{ 100 }\) × 10
= 65 + 3.2 = 68.2
Thus, required A.M. = 68.2

We hope the given RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.4 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.4, drop a comment below and we will get back to you at the earliest.