Q. 174.0( 4 Votes )

# Find the missing

Answer :

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

Mean = 50 and N = 120 (as given in the question)

Σfi = 68 + f1 + f2 and Σfixi = 3480 + 30f1 + 70f2

mean is given by  ( given: mean is 50)

3400 + 50f1 + 50f2 = 3480 + 30f1 + 70f2

50f1 – 30f1 + 50f2 – 70f2 = 3480 – 3400

20f1 – 20f2 = 80

f1 – f2 = 4 …(i)

As given in the question, frequency(Σfi) = 120

And as calculated by us, frequency (Σfi) = 68 + f1 + f2

Equalizing them, we get

68 + f1 + f2 = 120

f1 + f2 = 120 – 68 = 52

f1 + f2 = 52 …(ii)

We will now solve equations (i) and (ii), adding them we get

(f1 + f2) + (f1 – f2) = 52 + 4

2f1 = 56

f1 = 56/2

f1 = 28

Substitute f1 = 28 in equation (ii),

28 + f2 = 52

f2 = 52 – 28

f2 = 24

Thus, f1 = 28 and f2 = 24.

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