The mean of the following distribution is 62.8 . . .

Question : The mean of the following distribution is 62.8 and sum of all frequencies is 50. Find the missing frequencies f₁ and f₂

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

Doubt by Pushkar 

Solution : 

Class Intervals (CI)	Frequency (fi)	Class Marks (xi)	di=xi-a	fidi
0-20	5	10	-40	-200
20-40	f1	30	-20	-20f1
40-60	10	50=a	0	0
60-80	f2	70	20	20f2
80-100	7	90	40	280
100-120	8	110	60	480
Total	Σfi=50			Σfidi= -20f1+20f2+560

Σf_i=50
30+f₁+f₂=50
f₁+f₂=50-30
f₁+f₂=20 — (1)

x̄=62.8 (Given)

Using Assumed Mean Method 
x̄ = a+[Σf_id_i/Σf_i]
62.8 = 50+[-20f₁+20f₂+560/50]
62.8-50 = [-20f₁+20f₂+560/50] 
12.8 × 50 = -20f₁+20f₂+560 
640-560 = -20f₁+20f₂ 
80 = 20[-f₁+f₂]
80/20 = -f₁+f₂
4 = -f₁+f₂
-f₁+f₂ = 4 — (1)

Adding equation (1) and (2)

f₁+f₂-f₁+f₂=20+4
2f₂=24
f₂ = 24/2
f₂ = 12
putting in equation (1)
f₁+1₂=20
f₁=20-12
f₁=8

Hence, f₁=8 and f₂ = 12