Answer :

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

We have got


Σfi = 35 + f1 + f2 and Σfixi = 6150 + 190f1 + 210f2


mean is given by



( given: mean of pocket allowance is 188)


6580 + 188f1 + 188f2 = 6150 + 190f1 + 210f2


190f1 – 188f1 + 210f2 – 188f2 = 6580 – 6150


2f1 + 22f2 = 430 …(i)


As given in the question, frequency(Σfi) = 100


And as calculated by us, frequency (Σfi) = 35 + f1 + f2


Comparing them, we get


35 + f1 + f2 = 100


f1 + f2 = 65 …(ii)


We will now solve equations (i) and (ii), multiply eq.(ii) by 2 and then subtracting it from eq.(i), we get


(2f1 + 22f2) – [2(f1 + f2)] = 430 – 130


2f1 + 22f2 – 2f1 – 2f2 = 300


20 f2 = 300


f2 = 15


Substitute f2 = 15 in equation (ii),


f1 + 15 = 65


f1 = 65 – 15


f1 = 50


Thus, f1 = 50 and f2 = 15.


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