RBSE Solutions for Class 9 Maths Chapter 15 Statistics Miscellaneous Exercise

RBSE Solutions for Class 9 Maths Chapter 15 Statistics Miscellaneous Exercise is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 15 Statistics Miscellaneous Exercise.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 9
Subject Maths
Chapter Chapter 15
Chapter Name Statistics
Exercise Miscellaneous Exercise
Number of Questions Solved 19
Category RBSE Solutions

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 15 Statistics Miscellaneous Exercise

Multiple Choice Questions

Question 1.
The frequency of the class interval 3 – 5 in the (RBSESolutions.com) following distribution is:
5, 5, 6, 4, 9, 5, 3, 2, 7, 6, 3, 8, 4
(A) 3
(B) 4
(C) 6
(D) 7

Question 2.
The range of the following frequency (RBSESolutions.com) distribution will be:
3.2, 2.8, 3.1, 2.1, 3.2, 2.4, 2.1, 2.8, 2.7, 2.7
(A) 2.7
(B) 3.1
(C) 2.4
(D) 1.1

RBSE Solutions

Question 3.
In the following frequency distribution, the number of students (RBSESolutions.com) whose age is less than 25 years is:
RBSE Solutions for Class 9 Maths Chapter 15 Statistics Miscellaneous Exercise - 1
(A) 8
(B) 16
(C) 9
(D) 25

Question 4.
In a histogram, the heights (RBSESolutions.com) of the rectangles are:
(A) inversely proportional to the frequencies of their classes.
(B) proportional to the frequencies of the classes.
(C) proportional to the class intervals.
(D) inversely proportional to the class intervals.

Question 5.
The comparative study of the result of (RBSESolutions.com) any class of a school can be done:
(A) by ogive curve
(B) by histogram
(C) by linear curve
(D) All the above

Question 6.
Range of the distribution 6, 1, 2, 3, 9, 8, 3, 4, 8, 2, 3 is:
(A) 4
(B) 8
(C) 7
(D) 6

Question 7.
If variate of the (RBSESolutions.com) distribution are 5, 5, 2, 3, 6, 5, 4, then frequency of variate 5 will be:
(A) 1
(B) 2
(C) 3
(D) 4

Question 8.
The median of 11, 2, 7, 8, 9, 3, 5 is:
(A) 7
(B) 9
(C) 5
(D) 10

Question 9.
The mean of 15, 0, 10, 5 will be:
(A) 15
(B) 10
(C) 5
(D) 7.5

Question 10.
The mode of the (RBSESolutions.com) distribution 4, 3, 4, 5, 4, 2, 4, 1 will be:
(A) 1
(B) 2
(C) 5
(D) 4

Answers
1. B
2. D
3. D
4. B
5. B
6. B
7. C
8. A
9. D
10. D

RBSE Solutions

Very Short Answer Type Questions

Question 11.
Write the frequency of the class interval 0-5 from the (RBSESolutions.com) distribution 3, 2, 0, 10, 8, 5, 13, 5, 6, 6, 0, 14.
Solution.
RBSE Solutions for Class 9 Maths Chapter 15 Statistics Miscellaneous Exercise

Question 12.
If the mean marks of 5, 8, 4, x, 6, 9, is 7, then find (RBSESolutions.com) the value of x.
Solution.
We have \(\bar { x }\) = 7
RBSE Solutions for Class 9 Maths Chapter 15 Statistics Miscellaneous Exercise

Question 13.
What is range?
Solution.
The difference of maximum and minimum values (RBSESolutions.com) of the variate is known as Range.
Range = Maximum value – Minimum value

Question 14.
What is histogram?
Solution.
The rectangular representation of classified and continuous frequency distribution is known as Histogram.

Question 15.
Prepare frequency distribution table (RBSESolutions.com) from the following data:
9, 7, 9, 8, 3, 9, 8, 3, 5, 7, 5, 3
Solution.
RBSE Solutions for Class 9 Maths Chapter 15 Statistics Miscellaneous Exercise

Question 16.
Mean of any frequency distribution is 15 and Σ f = 20, then find the value of Σ fx.
Solution.
We have
RBSE Solutions for Class 9 Maths Chapter 15 Statistics Miscellaneous Exercise

RBSE Solutions

Question 17.
Find the median of the following (RBSESolutions.com) distribution 5, 2, 3, 7, 5, 4, 3, 2, 1.
Solution.
Arranging the given data in ascending order, we get
1, 2, 2, 3, 3, 4, 5, 5, 7
Here, n = 9 (odd)
Median = \(\left( \frac { 9+1 }{ 2 } \right)\) th term = 5th term = 3

Question 18.
Find the median (RBSESolutions.com) of the following distribution 12, 1, 6, 4, 10, 8, 1, 4.
Solution.
Arranging the distribution in ascending order, we get
1, 1, 4, 4, 6, 8, 10, 12
Here, n = 8 (even). So, their will be two median terms
i.e. (\(\frac { n }{ 2 }\)) th, (\(\frac { 1 }{ 2 }\) + 1) th
Here, \(\frac { n }{ 2 }\) = \(\frac { 8 }{ 2 }\) = 4
Median = \(\frac { 4th term + 5th term }{ 2 }\) = \(\frac { 4 + 6 }{ 2 }\) = 5

Question 19.
Find the mode of the (RBSESolutions.com) distribution 4, 3, 4, 1, 2, 4, 7, 5, 3.
Solution.
The variate having the maximum frequency, i.e. 3 is 4.
Hence, mode of the distribution is 4.

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