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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Variance and mean deviation for ungrouped data

Variance and mean deviation for ungrouped data

This article will give you the information about variance and mean deviation of grouped and ungrouped data, types of data , various formulas and techniques to find mean and median of grouped and ungrouped data, what is variance and variance formula.

Table of Content
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Introduction

The variance is the measure of variability. The more spread in your data the larger the variance is in relationship with mean. Mean deviation is used to compute how far the values in a data set are from the center. Data are of two types: grouped data and ungrouped data. Grouped data is a form in which information is arranged in tabular form.  Ungrouped data is a form of information in which the data is given in a raw form. 

Mean deviation of ungrouped data

Mean deviation: To find the measure to which extent the values in data from a fixed number ‘a’ we may take the mean of the absolute values of the deviations from the central value. This central value is called mean deviation. Thus, mean deviation about Central value is the mean of the absolute value of the deviations of the observations from ‘a’.

M.D. (a) = sum of absolute values of deviation from ‘a’/ number of observations 

Types of data 

We can categorize data in two ways: grouped data and ungrouped data.

Grouped data – As the name suggests it is the organized form of data i.e. raw data is categorized in different groups in the form of a table. Tables illustrate information of each group. There are many different ways to group a single data set.

Ungrouped data- it is a raw information of data that has not been organized into groups. Basically it is the data in its original form which we collect from our research. 

Calculating the mean deviation for ungrouped data

1.Calculating the central tendency i.e. mean and median 

To calculate the mean, first collect and count the data as it is given in raw format. For any set of data, the mean is the central value. So, we need to find the central tendency about which we are finding mean deviation. Let it be ‘a’.

M.D. =i=1n|xi-a| / n 

  1. Now find the deviation of each xi from a, i.e. x1 – a, x2 – a, x3– a …… xn – a 
  2. Find the absolute values i.e changing negative values to positive  

|x1-a|, |x2-a|, |x3-a|……..|xn-a|

  1. Now find the absolute values of the deviations. This is the mean deviation about ‘a’

M.D. =i=1n|xi-a| / n 

 M.D. (M) = 1/n i=1n|xi – M|, where M= median 

Mean deviation for grouped data 

We can grouped data in two ways :

  1. Discrete frequency distribution 
  2. Continuous frequency distribution 

Discrete frequency distribution: consider some data having n distinct values x1, x2, x3,……xn with frequency f1,f2,f2…..fn respectively. This can be represented in tabular form and known as discrete frequency distribution.

  1. Mean deviation about mean 

First find mean deviation of x of the given data by using the formula

X= i=1nxifi/i=1nfi=1Ni=1nxifi

Now, find the deviations of observations from mean , then find absolute value 

After this find the mean of the absolute value: 

M.D. (x) = i=1nfi|xi-x|/ i-1nfi

1/N i=1nfi | xi-x|

  1. Mean deviation about median 

Find the median of the given mean frequency distribution. So, arrange the observations in ascending order. This will give you the cumulative frequency . compare it with N/2 whether it is equal to or just greater .

M.D. (M) = 1/N i=1nfi |xi-M|

Continuous frequency distribution 

  1. Mean deviation about mean 

To find mean deviation in continuous frequency distribution we assume that frequency in each class is centered at its mid point. So, firstly we will find the midpoint of each class and then find deviations the same as in a discrete class.

  1. Mean deviation about median 

It is the same as we did earlier .

Median = l + N/2 – C / f *h 

After finding median , find absolute values of deviations of mid point xi of each class from the median |xi-M|

M.D. (M) = 1/N i=1nfi |xi-M|

Variance 

The mean of the squares of the deviations from mean is called variance. It is denoted by 2 .   therefore, the variance of n observations x1,x2,x3,…….xn. is given by 

2 = 1/n i=1n(xi-x) 2

Conclusion

Data is any piece of information collected for some specific purpose. It is of two types : grouped data and ungrouped data. In grouped data information is organized in  the tabular form whereas in ungrouped data raw information is provided. To find median of data Median for odd (n)= n+12the Observations .Median for even (n)= n2th + n2+1the observations / 2. To calculate mean : M.D. =i=1n|xi-a| / n 

To calculate median for ungrouped data M.D.(x) =1ni=1n|xi- x , where x = mean.

For grouped data, M.D. (x) = i=1nfi|xi-x|/ i-1nfi

1/N i=1nfi | xi-x|

Continuous frequency distribution 

M.D. (M) = 1/N i=1nfi |xi-M|

Mean deviation is the sum of absolute values of deviation from ‘a’ / number of observations.Variance is the mean of the square of the deviation. It is represented by  2

faq

Frequently asked questions

Get answers to the most common queries related to the IIT JEE Foundation Examination Preparation.

What is data?

Ans: Data is the information given in numerical figures by an individual obtained from various sour...Read full

Find the mean deviation about the mean of the following data: 6,7,10,12,13,4,8,12

Ans: Find mean of given data  X= 6+7+10+12+13+4+8+12 ...Read full

Find the mean deviation about the median for the following data: 3,9,5,3,12,10,18,4,19,21

Ans: Total number of observations = 11 which is odd, arrange data in ascending order, 3,3,4,5,7,9,1...Read full

Ans: Data is the information given in numerical figures by an individual obtained from various sources for some specific purpose.

Ans: Find mean of given data 

X= 6+7+10+12+13+4+8+12 / 8 = 72/8 = 9

  1. Find the deviation of the observations from mean x

|x1-a|, |x2-a|, |x3-a|……..|xn-a|

6-9 = -3 , 7-9=-2, 10-9=1, 12-9=3, 13-9=4, 4-9=-5, 8-9=-1, 12-9=3.

  1. The absolute values of deviations, |xi – x | 

3,2,1,3,4,5,1,3

  1. The required mean deviation about the mean is

M.D. (X) = i=18|xi–x| / 8

= 3+2+1+3+4+5+1+3/8 = 22/8 = 2.75 

 

 

Ans: Total number of observations = 11 which is odd, arrange data in ascending order, 3,3,4,5,7,9,10,12,18,19,21.

Median = 11+12th or 6th observation = 9

The absolute value of deviation |xi – M|

i=111|xi-M|=58

M.D. (M) = 1/11 i=111|xi-M | 

1/11 * 58 = 5.27

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