The mean of the following frequency distribution is 132 and the sum of observations is 50. Find the missing frequency f₁ and f₂.

Class

0-40

40-80

80-120

120-160

160-200

200-240

Freq.

4

7

f₁

12

f₂

9

Here, we have

sum from blank to blank of f subscript i space equals space 50 comma space sum from blank to blank of f subscript i u subscript i space equals space 2 f subscript 2 space plus space 24 comma space

A (assumed mean) = 100

top enclose straight x space equals space 132 space a n d space h space equals space 40

It is given that

sum from blank to blank of f subscript i space equals space 50

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f₁ + f₂ + 32 = 50

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f₁+ f₃ = 18

Now,

top enclose straight x space equals space A plus fraction numerator begin display style sum from blank to blank of end style f subscript i u subscript i over denominator begin display style sum from blank to blank of end style f subscript i end fraction cross times h

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132 = 100 +

fraction numerator 24 plus 2 straight f subscript 2 over denominator 50 end fraction cross times 40

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132 = 100 +

fraction numerator 4 left parenthesis 2 straight f subscript 2 plus 24 right parenthesis over denominator 5 end fraction

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132 =

fraction numerator 500 plus 8 straight f subscript 2 plus 96 over denominator 5 end fraction

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132 x 5 = 8f₂+ 596

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660 = 8f₂+ 596

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8f₂ = 64

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f₂= 8
Putting this value in (i), we get

straight f subscript 1 plus 8 space equals space 18

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f₁ = 10
Hence. f₁= 10, f₂ = 8

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