# The daily expendi

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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