RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.2 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 17.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 17 |

Chapter Name |
Measures of Central Tendency |

Exercise |
Exercise 17.2 |

Number of Questions Solved |
8 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.2

Find the mean (arithmetic) of following (RBSESolutions.com) frequency distribution (Q. 1 – 4)

Question 1.

Solution :

Table of A.M.

Thus, A.M. (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 99 }{ 14 } \) = 7.07

Thus, required A.M. = 7.07

Question 2.

Solution :

Table for A.M.

Thus, arithmetic (RBSESolutions.com) mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 151 }{ 20 } \) = 7.55

Thus, required A.M. = 7.55

Question 3.

Solution:

Table for A.M.

Thus, arithmetic (RBSESolutions.com) mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 72 }{ 210 } \) = 0.34

Thus, required A.M. = 0.34

Question 4.

Solution:

Table for A.M.

Thus, arithmetic (RBSESolutions.com) mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 27.5 }{ 50 } \) = 0.55

Thus, required A.M. = 0.55

Question 5.

The number of children in 100 families ia as follows :

**Find their arithmetic mean.**

Solution:

calculation for (RBSESolutions.com) arithmetic mean

Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }{ x } }{ { \Sigma f } } =\frac { 200 }{ 100 } \) = 2

Thus, arithmetic mean = 2

Question 6.

Weight of students in a class are (RBSESolutions.com) given in following table :

Find their arithmetic mean :

Solution :

Table for A.M.

Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } =\frac { 717 }{ 30 } \) = 23.9

Thus, required A.M. = 23.9

Question 7.

If mean of following (RBSESolutions.com) distribution is 7.5 then find value of p

Solution :

Arithmetic mean = 7.5

Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }{ x } }{ { \Sigma f } } \)

7.5 = \(\frac { 303+9P }{ 41+P } \)

⇒ (41 + p) (7.5) = 303 + 9P

⇒ (41 × 7.5) + 7.5p = 303 + 9P

⇒ 307.5 – 303 = 9P – 7.5P

⇒ 1.5P = 4.5

⇒ P = \(\frac { 4.5 }{ 1.5 } \) = 3

Thus, value of P = 3

Question 8.

If mean following frequency (RBSESolutions.com) distribution is 1.46, then find unknown frequencies.

Solution:

Let unknown frequencies are f_{1} and f_{2} :

Given : Σf_{i} = 200

But from table Σf_{i} = 86 + f_{1} + f_{2}

So, 86 + f_{1} + f_{2} = 200

⇒ f_{1} + f_{2} = 200 – 86 = 114

⇒ f_{1} + f_{2} = 114 ……(i)

According to question, (RBSESolutions.com) arithmetic mean = 1.46

or \(\overline { x }\) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } } \)

⇒ 1.46 = \(\frac { { 140+f }_{ 1 }{ +2f }_{ 2 } }{ 200 } \)

⇒ 140 + f_{1} + 2f_{2} = 292

f_{1} + f_{2} = 292 – 140

f_{1} + 2f_{2} = 152 ….(ii)

subtracting equation (ii)from(i),

Putting value of f_{2} in equation (i),

f_{1} + 38 = 114

⇒ f_{1} = 114 – 38 = 76

Thus, unknown (RBSESolutions.com) frequencies are 76 and 38

We hope the given RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.2, drop a comment below and we will get back to you at the earliest.

## Leave a Reply