The mean of the following frequency distribution of $62.8$ and the sum of all frequencies is $50$ Compute the sum of missing frequencies $f_{1}$ and $f_{2}$
Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120
Frequency $5$ $f_{1}$ $10$ $f_{2}$ $7$ $8$

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Solution

Let A = 30, h = 20
Class $x_{i}$ $f_{i}$ $u_{i} = \frac{x_{i} - A}{h}$ $f_{i} u_{i}$
0 - 20 10 5 -1 -5
20 - 40 30 $f_{i}$ 0 0
40 - 60 50 10 +1 10
60 - 80 70 $f_{2}$ +2 $2 f_{2}$
80 - 100 90 7 +3 21
100 - 120 110 8 +4 32
$\sum f_{i} = 30 + f_{1} + f_{2}$ $\sum f_{i} u_{i} = 58 + 2 f_{2}$
Given $30 + f_{1} + f_{2} = 50 \Rightarrow f_{1} + f_{2} = 20$ .........(i)
$¯ ¯ ¯ x = A + (\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h) \Rightarrow 62.8 = 30 + (\frac{58 + 2 f_{2}}{50} \times 20)$
$32.8 = (58 + 2 f_{2}) \times \frac{2}{5} \Rightarrow 32.8 \times 5 = 116 + 4 f_{2} \Rightarrow 164 = 116 + 4 f_{2}$
$4 f_{2} = 164 - 116 \Rightarrow 4 f_{2} = 48 \Rightarrow f_{2} = 12$
Now $f_{1} + f_{2} = 20 \Rightarrow f_{1} + 12 = 20 \Rightarrow f_{1} = 8$
So the missing frequencies are $f_{1} = 8 a n d f_{2} = 12$

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Class	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency	$5$	$f_{1}$	$10$	$f_{2}$	$7$	$8$

Class	$x_{i}$	$f_{i}$	$u_{i} = \frac{x_{i} - A}{h}$	$f_{i} u_{i}$
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