RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.6 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 17.6.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 17 |

Chapter Name |
Measures of Central Tendency |

Exercise |
Exercise 17.6 |

Number of Questions Solved |
5 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.6

Question 1.

Find median of the following (RBSESolutions.com) variable values

25, 34, 33, 13, 20, 26, 36, 28, 19, 34

Solution :

Arranging given values is ascending order

13, 19, 20, 25, 26, 28, 33, 34, 34, 36

In given values. No. of terms n = 10 which is even No. Thus,

Thus, median = 27

Question 2.

Find the median of the data

19, 25, 59, 48, 35, 31, 30, 32, 51

If 25 is replaced by 52, then (RBSESolutions.com) find new median.

Solution :

Arranging given data in ascending order

19, 25, 30, 31, 32, 35, 48, 51, 59

Number of terms in given data, n =9 which is odd, thus

In second case, when 25 is replaced by 52 then (RBSESolutions.com) arranging data in ascending order

19, 30, 31, 32, 35, 48, 51, 52, 59

Thus, first median 32

New median = 35 (after replacement)

Question 3.

Marks obtained by students in a class are (RBSESolutions.com) given in following table. Find median.

Solution :

Writing the distribution in tabulation form

\(\frac { N }{ 2 }\)=\(\frac { 99 }{ 2 }\)=49.5

C.F. just above 49.5 is 52 whose corresponding variable value is 30.

∴ Required median = 30

Question 4.

Number of children in 100 families (RBSESolutions.com) are as follows :

Find their median.

Solution :

Here \(\frac { N }{ 2 }\)=\(\frac { 100 }{ 2 }\)=50

C.F. just above 50 is 72 whose corresponding variable value is 2.

∴ Required median = 2

Question 5.

Find median of following frequency (RBSESolutions.com) distribution :

Solution :

Writing the distribution in tabulation form :

Here \(\frac { N }{ 2 }\)=\(\frac { 160 }{ 2 }\)=80

C.F. just above 80 is 105 whose corresponding variable value is 35.

∴ Required median = 35

We hope the given RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.6 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.6, drop a comment below and we will get back to you at the earliest.

## Leave a Reply