Answer :

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

We have got


Σfi = 47 and Σfixi = 1516


mean is given by





Thus, mean is 32.26.


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 = 47


N/2 = 47/2 = 23.5


The cumulative frequency just greater than (N/2 = ) 23.5 is 29, so the corresponding median class is 30.5 - 35.5 and accordingly we get Cf = 15(cumulative frequency before the median class).


Now, since median class is 30.5 - 35.5.


l = 30.5, h = 5, f = 14, N/2 = 23.5 and Cf = 15


Median is given by,




= 30 + 3.03


= 33.03


Thus, median is 33.03.


Since, we have got mean = 32.26 and median = 33.03


Applying the empirical formula,


Mode = 3(Median) – 2(Mean)


Mode = 3(33.03) – 2(32.26)


Mode = 99.09 – 64.52 = 34.57


Mean = 32.26, Median = 33.03 and Mode = 34.57


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