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.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 17 |

Chapter Name |
Measures of Central Tendency |

Exercise |
Exercise 17.4 |

Number of Questions Solved |
7 |

Category |
RBSE Solutions |

## 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)

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

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

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

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

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

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

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.

## Leave a Reply