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Question

Calculate the mean deviation about the median for the following data:

Class16202125263031353640414546505155Frequency5612142612169

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Solution

Converting the given series into an exclusive series, we prepare the table, given below:

ClassFrequency (fi)cfMidpoint (xi)15.520.5551820.525.56112325.530.512232830.535.514373335.540.526633840.545.512754345.550.516914850.555.5910053 N=100

Thus N = 100 and therefore, N2=50

median class is 35.5 - 40.5

L = 35.5, f = 26, h = 5 and c = 37.

median=l+(N2c)f×h

={35.5+(5037)26×5}=(35.5+2.5)=38

Thus, M = 38.

Now, we prepare the table given below.

fixi|xiM|fi×|xiM|5182010062315901228101201433570263800124356016481016095315135N=100 735

Thus, fi×|xiM|=735 and N = 100

MD(M)=fi×|xiM|N=735100=7.35

Hence, the mean deviation about the median is 7.35


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