Question

The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f₁ and f₂.

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

Answer 1

Arithmetic Mean `(barx)` = 62.8

Sum of all the frequencies `(sumf_"i")` = 50

Let the missing frequencies be f₁ and f₂ 

Class interval	frequency fi	mid value xi	fixi
0 − 20	5	10	0
20 − 40	f1	30	30 f1
40 − 60	10	50	500
60 − 80	f2	70	70 f2
80 − 100	7	90	630
100 − 120	8	110	880
	`sumf_"i" = 30 + f_1 + f_2`		`sumf_"i"x_"i" = 2060 + 30f_1 + 70f_2`

`30 + f_1 + f_2` = 50

`f_1 + f_2` = 20    ...(1)

Mean = `(sumf_"i"x_"i")/(sumf_"i")`

62.8 = `(2060 + 30f_1 + 70f_2)/50`

3140 = `2060 + 30f_1 + 70f_2`

∴ `30f_1 + 70f_2` = 3140 − 2060

`30f_1 + 70f_2` = 1080

(÷ by 10) ⇒ `3f_1 +  7f_2` = 108   ...(2)
(1) × 3    ⇒ `3f_1 +  3f_2` = 60      ...(3)
(2) × 3    ⇒ `3f_1 +  7f_2` = 108    ...(4)
                   (−)    (−)     (−)      
(3) − (4)  ⇒       `−4f_2` = − 48
                             `f_2 = 48/4 ⇒ f_2` = 12

Substitute of the value of f₂ in (1)

 f₁ + 12 = 20

⇒ f₁ = 20 – 12 = 8

The Missing frequency is 8 and 12.