Answer :

For equal class intervals, we will solve by finding mid points of these classes using direct method.


We have got


Σfi = 50 and Σfixi = 1500


mean is given by





Thus, mean is 30.


To find median, Assume


Σfi = N = Sum of frequencies,


h = length of median class,


l = lower boundary of the median class,


f = frequency of median class


and Cf = cumulative frequency


Lets form a table.



We have got


So, N = 50


N/2 = 50/2 = 25


The cumulative frequency just greater than (N/2 = ) 25 is 39, so the corresponding median class is 30 - 40 and accordingly we get Cf = 24(cumulative frequency before the median class).


Now, since median class is 30 - 40.


l = 30, h = 10, f = 15, N/2 = 25 and Cf = 24


Median is given by,




= 30 + 0.67


= 30.67


Thus, median is 30.67.


Since, we have got mean = 30 and median = 30.67


Applying the empirical formula,


Mode = 3(Median) – 2(Mean)


Mode = 3(30.67) – 2(30)


Mode = 92.01 – 60 = 32.01


Mean = 30, Median = 30.67 and Mode = 32.01


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