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The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency is 50. Compute the missing frequency.

 Class:  0 - 20  20 - 40  40 - 60  60 - 80  80 - 100  100 - 120
 Frequency:  5  f1  10  f2  7  8

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 Class interval  Mid value (xi)  Frequency (fi)  fiui
 0 - 20 10 5 50
 20 - 40 30 f1 30f1
 40 - 60 50 10 500
 60 - 80 70 f2 70f2
 80 - 100  90 7 630
 100 - 120  110  8  880
 N = 50 \(\sum\)fiui = 30f1 + 70f2 + 2060

Given, 

Sum of frequency = 50

5+ f1 + 10 + f2 + 7 + 8 = 50

f1 + f2 = 50 – 5 – 10 – 7 – 8

f1 + f2 = 20

3f1 + 3f2 = 60 (i) [Multiply by 3]

And mean = 62.8

\(\frac{\sum f_iu_i}{N}=62.8\)

\(\frac{30f1+70f2+2060}{50}=62.8\)

30f1 + 70f2 = 3140 – 2060 

30f1 + 70f2 = 1080 

3f1 + 7f2 = 108 (ii) [Divide by 10] 

Subtract (i) from (ii), we get 

3f1 + 7f2 – 3f1 – 3f2 = 108 – 60 

4f2 = 48 

f2 = 12 

Put value of f2 in (i), we get 

3f1 + 3 x 12 = 60 

3f1 + 36 = 60 

3f1 = 24 

f1 = 8 

So, 

f1 = 8 and f2 = 12

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