NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1.
Board | CBSE |
Textbook | NCERT |
Class | Class 11 |
Subject | Maths |
Chapter | Chapter 15 |
Chapter Name | Statistics |
Exercise | Ex 15.1 |
Number of Questions Solved | 12 |
Category | NCERT Solutions |
NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1
Find the mean deviation about the mean for the data in Exercises 1 and 2.
Ex 15.1 Class 11 Maths Question 1.
4, 7, 8, 9, 10, 12, 13, 17
Solution:
Mean of the given data is
Ex 15.1 Class 11 Maths Question 2.
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
Mean of the given data is
Ex 15.1 Class 11 Maths Question 3.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Solution:
Arranging the data in ascending order, we have
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Here n = 12 (which is even)
So median is the average of 6th and 7th observations
Ex 15.1 Class 11 Maths Question 4.
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Solution:
Arranging the data in ascending order, we have 36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Here n = 10 (which is even)
So median is the average of 5th and 6th observations
Find the mean deviation about the mean for the data in Exercises 5 and 6.
Ex 15.1 Class 11 Maths Question 5.
Solution:
Ex 15.1 Class 11 Maths Question 6.
Solution:
Find the mean deviation about the median for the data in Exercises 7 and 8.
Ex 15.1 Class 11 Maths Question 7.
Solution:
Ex 15.1 Class 11 Maths Question 8.
Solution:
Find the mean deviation about the mean for the data in Exercises 9 and 10.
Ex 15.1 Class 11 Maths Question 9.
Solution:
Ex 15.1 Class 11 Maths Question 10.
Solution:
Ex 15.1 Class 11 Maths Question 11.
Find the mean deviation about median for the following data:
Solution:
Ex 15.1 Class 11 Maths Question 12.
Calculate the mean deviation about median age for the age distribution of 100 persons given below:
[Hint: Convert the given data into continuous frequency distribution by subtracting 0.5 from lower limit and adding 0.5 to the upper limit of each class interval]
Solution:
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