Notes On Mean Deviation about Mean for Grouped Data - CBSE Class 11 Maths
Data can be grouped in two ways - discrete frequency distribution and continuous frequency distribution.
 
A collection of data consisting of distinct values occurring with certain frequencies is called discrete frequency distribution.
 
Example of discrete frequency distribution:
The table displays the data of the performance of 50 students in a math test.
 


Classify the data into different class-intervals along with their respective frequencies.

When data are arranged in different class intervals without any gaps along with their frequencies, it is called continuous frequency distribution.
 
An example of continuous frequency distribution:
 
The table displays the marks scored by 50 students.
           Marks Scored          Number of Students
              90 - 100                     4
              75 - 90                     9
              60 - 75                    14
              40 - 60                    18
               0 - 40                      5
 
Mean deviation about the mean for discrete frequency distribution:
    Observations         Frequencies
              x1               f1
              x2               f2
              x3               f3
              ....               ....
              xn               fn
 
x _ = i = 1 n x i f i i = 1 n f i = 1 N i = 1 n x i f i

N = Sum of the frequencies (Number of observations)
Deviation from the mean, x _  = x i - x _ , where i = 1, 2, 3, ... , n
Absolute values of the deviations = |  x i - x _ |, where i = 1, 2, 3, ... , n
Mean deviation about the mean = MD ( x _ )

= i = 1 n f i | x i - x _ | i = 1 n f i

= 1 N i = 1 n f i | x i - x _ |

Ex: Consider the weights of 20 students.
  Weight in kg (xi)    Number of Students (fi)
          48            3
          49            4
          50            6
          51            4
          52            3

 
To find the mean deviation of the given data, mean of the data is found.
   Weight in kg (xi)     Number of Students (fi)        fixi
            48                       3        144
            49                       4        196
            50                       6        300
            51                      4        204
            52                      3        156
               i = 1 5 f i = 20   i = 1 5 f i x i   = 1000

 
Mean ( x _ ) = 1000 20 = 50 kg

Modulus of deviation of each observation from the mean and the sum of the products of the deviations of the observations and the frequency concerned:
   Weight in kg (xi)     Number of Students (fi)        fixi   | x i - x _ | = | x i - 50|          f i | x i - x _ |
            48                       3        144  |48 - 50| = 2            6
            49                       4        196  |49 - 50| = 1            4
            50                       6        300  |50 - 50| = 0            0
            51                      4        204  |51 - 50| = 1            4
            52                      3        156  |52 - 50| = 2            6
               i = 1 5 f i = 20   i = 1 5 f i x i   = 1000   i = 1 5 f i | x i   -  x _ |= 20
 
MD ( x _ ) = 1 n   i = 1 n f i | x i   -  x _ | = 20 20  = 1
 
Mean deviation about the mean for Continuous frequency distribution:
           Marks Scored               Number of Students (fi)        
              80 - 100                     3
              60 - 80                     15
              40 - 60                    19
              20 - 40                    10
               0 - 20                      3
 
Mean of the data:
           Marks Scored               Number of Students (fi)          Mid-values (xi)              fixi   
              80 - 100                     3         90  270
              60 - 80                     15         70 1050
              40 - 60                    19         50   950
              20 - 40                    10         30   300
               0 - 20                      3         10    30
                 ∑ fi = 50     i = 1 5 f i x i   = 2600
 
Mean ( x _ ) = x i f i f i   =  2600 50   = 52 
 
If the number of observations is large, then the mathematical calculation is tedious.
 
For data with a large number of observations, the assumed mean method or the step-deviation method can be used to find the mean.

Find the modulus of the deviations of the mid-values from the mean and find the product of the deviations with their respective frequencies.
           Marks Scored               Number of Students (fi)          Mid-values (xi)              fixi           | x i - x _ | = | x i - 52|                      f i | x i - x _ |
              80 - 100                     3         90  270    |90 - 52| = 38   114
              60 - 80                     15         70 1050    |70 - 52| = 18   270
              40 - 60                    19         50   950    |50 - 52| = 2   38
              20 - 40                    10         30   300    |30 - 52| = 22   220
               0 - 20                      3         10    30    |10 - 52| = 42    126
                 ∑ fi = 50     i = 1 5 f i x i   = 2600   i = 1 5 f i | x i   -  x _ |= 768
 
 
 Mean deviation about mean = MD ( x _ ) = 1 n   i = 1 n f i | x i   -  x _ | = 768 50  = 15.36
 
Therefore, the mean deviation about the mean is equal to 768 divided by 50, which is equal to 15.36.

Summary

Data can be grouped in two ways - discrete frequency distribution and continuous frequency distribution.
 
A collection of data consisting of distinct values occurring with certain frequencies is called discrete frequency distribution.
 
Example of discrete frequency distribution:
The table displays the data of the performance of 50 students in a math test.
 


Classify the data into different class-intervals along with their respective frequencies.

When data are arranged in different class intervals without any gaps along with their frequencies, it is called continuous frequency distribution.
 
An example of continuous frequency distribution:
 
The table displays the marks scored by 50 students.
           Marks Scored          Number of Students
              90 - 100                     4
              75 - 90                     9
              60 - 75                    14
              40 - 60                    18
               0 - 40                      5
 
Mean deviation about the mean for discrete frequency distribution:
    Observations         Frequencies
              x1               f1
              x2               f2
              x3               f3
              ....               ....
              xn               fn
 
x _ = i = 1 n x i f i i = 1 n f i = 1 N i = 1 n x i f i

N = Sum of the frequencies (Number of observations)
Deviation from the mean, x _  = x i - x _ , where i = 1, 2, 3, ... , n
Absolute values of the deviations = |  x i - x _ |, where i = 1, 2, 3, ... , n
Mean deviation about the mean = MD ( x _ )

= i = 1 n f i | x i - x _ | i = 1 n f i

= 1 N i = 1 n f i | x i - x _ |

Ex: Consider the weights of 20 students.
  Weight in kg (xi)    Number of Students (fi)
          48            3
          49            4
          50            6
          51            4
          52            3

 
To find the mean deviation of the given data, mean of the data is found.
   Weight in kg (xi)     Number of Students (fi)        fixi
            48                       3        144
            49                       4        196
            50                       6        300
            51                      4        204
            52                      3        156
               i = 1 5 f i = 20   i = 1 5 f i x i   = 1000

 
Mean ( x _ ) = 1000 20 = 50 kg

Modulus of deviation of each observation from the mean and the sum of the products of the deviations of the observations and the frequency concerned:
   Weight in kg (xi)     Number of Students (fi)        fixi   | x i - x _ | = | x i - 50|          f i | x i - x _ |
            48                       3        144  |48 - 50| = 2            6
            49                       4        196  |49 - 50| = 1            4
            50                       6        300  |50 - 50| = 0            0
            51                      4        204  |51 - 50| = 1            4
            52                      3        156  |52 - 50| = 2            6
               i = 1 5 f i = 20   i = 1 5 f i x i   = 1000   i = 1 5 f i | x i   -  x _ |= 20
 
MD ( x _ ) = 1 n   i = 1 n f i | x i   -  x _ | = 20 20  = 1
 
Mean deviation about the mean for Continuous frequency distribution:
           Marks Scored               Number of Students (fi)        
              80 - 100                     3
              60 - 80                     15
              40 - 60                    19
              20 - 40                    10
               0 - 20                      3
 
Mean of the data:
           Marks Scored               Number of Students (fi)          Mid-values (xi)              fixi   
              80 - 100                     3         90  270
              60 - 80                     15         70 1050
              40 - 60                    19         50   950
              20 - 40                    10         30   300
               0 - 20                      3         10    30
                 ∑ fi = 50     i = 1 5 f i x i   = 2600
 
Mean ( x _ ) = x i f i f i   =  2600 50   = 52 
 
If the number of observations is large, then the mathematical calculation is tedious.
 
For data with a large number of observations, the assumed mean method or the step-deviation method can be used to find the mean.

Find the modulus of the deviations of the mid-values from the mean and find the product of the deviations with their respective frequencies.
           Marks Scored               Number of Students (fi)          Mid-values (xi)              fixi           | x i - x _ | = | x i - 52|                      f i | x i - x _ |
              80 - 100                     3         90  270    |90 - 52| = 38   114
              60 - 80                     15         70 1050    |70 - 52| = 18   270
              40 - 60                    19         50   950    |50 - 52| = 2   38
              20 - 40                    10         30   300    |30 - 52| = 22   220
               0 - 20                      3         10    30    |10 - 52| = 42    126
                 ∑ fi = 50     i = 1 5 f i x i   = 2600   i = 1 5 f i | x i   -  x _ |= 768
 
 
 Mean deviation about mean = MD ( x _ ) = 1 n   i = 1 n f i | x i   -  x _ | = 768 50  = 15.36
 
Therefore, the mean deviation about the mean is equal to 768 divided by 50, which is equal to 15.36.

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