# The mean of the f

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

We have got

Σfi = 32 + f1 + f2 and Σfixi = 1940 + 30f1 + 70f2

mean is given by

( given: mean of pocket allowance is 57.6)

1843.2 + 57.6f1 + 57.6f2 = 1940 + 30f1 + 70f2

57.6f1 – 30f1 + 57.6f2 – 70f2 = 1940 – 1843.2

27.6f1 – 12.4f2 = 96.8

69f1 – 31f2 = 242 …(i)

As given in the question, frequency(Σfi) = 50

And as calculated by us, frequency (Σfi) = 32 + f1 + f2

Comparing them, we get

32 + f1 + f2 = 50

f1 + f2 = 18 …(ii)

We will now solve equations (i) and (ii), multiply eq.(ii) by 31 and then adding to eq.(i), we get

(69f1 – 31f2) + [31(f1 + f2)] = 242 + 558

69f1 – 31f2 + 31f1 + 31f2 = 800

100f1 = 800

f1 = 8

Substitute f1 = 8 in equation (ii),

8 + f2 = 18

f2 = 18 – 8

f2 = 10

Thus, f1 = 8 and f2 = 10.

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