NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1 Statistics  PDF Download
JEE Mains & AdvancedNCERT solutions for class 11 maths chapter 13 exercise 13.1 Statistics are based on the mean deviation. The mean deviation is a measure of dispersion. There are three measures, namely the range, standard deviation, and quartile deviation. While the range of data provides a general indication of variability or dispersion, it does not provide an indication of dispersion from a central tendency measure. For this, we use average and standard deviation. These measurements are done on both ungrouped and grouped data. Grouped data can also be subdivided into discrete and continuous frequency distributions. Students will also learn how to calculate the mean deviation about the median.
Class 11 maths chapter 13 exercise 13.1 NCERT solutions consists of 12 questions based on mean deviation formulas. Ex 13.1 class 11 maths solutions are prepared by the academic team of maths at eSaral. These NCERT solutions provide students with the opportunity to refine their knowledge of concepts and solve complex questions without any doubt. Students can download the NCERT solution PDF of ex 13.1 class 11 maths chapter 13 from eSaral website and practice the questions offline to improve their maths skills. Download the free PDF of these solutions from the link mentioned below.
Topics Covered in Exercise 13.1 Class 11 Mathematics Questions
Ex 13.1 class 11 maths chapter 13 includes topics based on measures of dispersion, range, mean deviation, mean deviation for ungrouped data, mean deviation for grouped data, discrete frequency distribution and continuous frequency distribution. Here, you can get a precise explanation of these topics which are provided by subject experts of eSaral.
1. 
Measures of Dispersion 

2. 
Range 

3. 
Mean Deviation 
(a) Discrete frequency distribution (b) Continuous frequency distribution 

Measures of Dispersion
The dispersion or scatter in a data is measured on the basis of the observations and the types of the measure of central tendency, used there. There are following measures of dispersion
(i) Range
(ii) Quartile deviation
(iii) Mean deviation
(iv) Standard deviation.
In this chapter, you will study at all of these dispersion measures with the exception of the quartile deviation.

Range  The difference between the maximum and minimum value is called the range.
Range = Maximum value – Minimum value

Mean Deviation  Mean deviation is a measure of the deviation from the value. The value is usually a mean or median value. To find the mean deviation, we first take the mean deviation for observations from the value d = x –a. Here, x is the observation and a is the constant value.
The mean deviation of observations from a central value of 'a' is the mean of the absolute value of the deviations of observations from 'a'. The mean deviation from ‘a’ is denoted as M.D. (a).
The formula to find out the mean deviation is
M.D. (a) $=\frac{\text { Sum of absolute values of deviations from ' } a \text { ' }}{\text { Number of observations }}$

Mean deviation for ungrouped data
In order to calculate the average deviation about the mean or median, the following steps must be taken:
Let’s take ‘n’ observations be x_{1}, x_{2}, x_{3},....., x_{n}.
Step 1 Calculate the measure of central tendency about which we are to find the mean deviation. Let it be ‘a’.
Step 2 Find the deviation of each xi from a, i.e., x_{1}a, x_{2}a, x_{3}a, …., x_{n} a.
Step 3 Find the absolute values of the deviations, i.e., drop the minus sign (–), if it is there, i.e., $\leftx_1a\right,\leftx_2a\right,\leftx_3a\right, \ldots,\leftx_na\right$
Step 4 Find the mean of the absolute values of the deviations. This mean is the mean deviation about a, i.e.,
M.D.(a $)=\frac{\sum_{i=1}^n\leftx_ia\right}{n}$
M.D. $(\bar{x})=\frac{1}{n} \sum_{i=1}^n\leftx_i\bar{x}\right$, where $\bar{x}=$ Mean
M.D. $(M)=\frac{1}{n} \sum_{i=1}^n\leftx_iM\right$, where $M=$ Median

Mean deviation for grouped data
The data can be grouped into two categories:
(a) Discrete frequency distribution
(b) Continuous frequency distribution.
Let's explore the methods for determining the mean deviation for both data types.
(a) Discrete frequency distribution  Let the given data consist of n distinct values x_{1} , x_{2 }, ..., x_{n} occurring with frequencies f_{1} , f_{2} , ..., f_{n} respectively. This data can be represented in the tabular form which is called discrete frequency distribution.
(i) Mean deviation about mean
(ii) Mean deviation about median
In this topic, you will also learn mean deviation about mean and median which are discussed in detail format in NCERT maths textbook for ex 13.1 class 11.
(b) Continuous frequency distribution  A continuous frequency distribution is a series in which the data are classified into different classintervals without gaps along with their respective frequencies.
(i) Mean deviation about mean  While calculating the mean of a continuous frequency distribution, we had made the assumption that the frequency in each class is centred at its midpoint. Here also, we write the midpoint of each given class and proceed further as for a discrete frequency distribution to find the mean deviation.
(ii) Mean deviation about median  The process of finding the mean deviation about median for a continuous frequency distribution is similar as we did for mean deviation about the mean. The only difference lies in the replacement of the mean by median while taking deviations.
Tips for Solving Exercise 13.1 Class 11 Chapter 13 Statistics
Exercise 13.1 class 11 maths ch 13 NCERT solutions includes important formulas that students need to remember so they can solve problems. Here, our experienced faculty of maths have combined some important tips and tricks to remember these formulas and solve questions. You can check them below.

When students are trying to solve maths problems in class 11 maths ex 13.1, they need to know what kind of data is in the question so they can use the right formula. This way, they won't make mistakes when trying to solve problems.

Ex 13.1 class 11 maths chapter 13 NCERT solutions has questions where students have to change a frequency distribution from a discrete one to a continuous one. You have to be able to mix and match the data according to the total.

As questions of ex 13.1 class 11 maths are quite lengthy, you must pay special attention to making accurate calculations and avoiding errors in addition or subtraction.
Importance of Solving Ex 13.1 Class 11 Maths Chapter 13 Statistics
Ex 13.1 class 11 maths chapter 13 mainly deals with mean deviation about mean and median. Therefore NCERT solution for chapter 13 ex 13.1 class 11 maths can be considered as scoring exercise. Here, we have provided some significant benefits of solving ex 13.1 class 11 maths.

NCERT solutions help students to gain a deep understanding of the intricate concepts given in the maths ex 13.1 class 11 chapter 13. This helps students to score higher in the exams.

NCERT solutions for class 11 maths chapter 13 ex 13.1 contains important topics that are highly descriptive and are explained stepwise by eSaral’s subject experts to help students solve difficult questions easily.

By practicing questions from NCERT solutions class 11 maths chapter 13 ex 13.1, you will grasp the basic knowledge and understand the concepts to solve any question quickly.

NCERT solutions for class 11 maths ex 13.1 prepared by eSaral’s expert faculties . They have provided accurate and precise answers for all questions that you can trust without any doubt.
Frequently Asked Questions
Question 1. How many ways are there to define mean deviation for grouped data?
Answer 1. Mean deviation for grouped data can be defined in two ways.

Discrete frequency distribution

Continuous frequency distribution.
Question 2. What is the range in NCERT solutions class 11 maths chapter 13 ex 13.1?
Answer 2. The difference between the maximum and minimum value is called the range.
Thus, Range = Maximum value – Minimum value.
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