Question

The mean of each of the following frequency distributions is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f₁ and f₂.
Class:
0-20
20-40
40-60
60-80
80-100
100-120
Frequency:
5
f₁
10
f₂
7
8

Answer 1

Therefore

f1=8,

f2=12

i hope this will useful to u

Answer 2

STEP DEVIATION METHOD:

Step deviation method is used in the cases where the deviation from the assumed mean 'A' are multiples of a common number. If the values of ‘di’ for each class is a multiple of ‘h’ the calculation become simpler by taking ui= di/h = (xi - A )/h

Here, h is the class size of each class interval.

★★ Find the class marks of class interval. These class marks would serve as the representative of whole class and are represented by xi.  

★★ Class marks (xi)  = ( lower limit + upper limit) /2

★★ We may take Assumed mean 'A’ to be that xi which lies in the middle of x1 ,x2 …..xn

MEAN = A + h ×(Σfiui /Σfi) , where ui =  (xi - A )/h

[‘Σ’ Sigma means ‘summation’ ]

FREQUENCY DISTRIBUTION TABLE IS IN THE ATTACHMENT  

From the table : Σfiui = 28 - f1 + f2,  Σfi = 30 + f1 + f2

Let the assumed mean, A = 50,  h = 20

Given : Σfi = 50 , Mean = 62.8

Σfi = 30 + f1 + f2

50 = 30 + f1 + f2

f1 + f2 = 50 - 30

f1 + f2 = 20

f1 = 20 - f2 ………..(1)

MEAN = A + h ×(Σfiui /Σfi)

62.8 = 50 + 20(28 - f1 + f2) / (30 + f1 + f2)

62.8 -  50 =  20[(28 - f1 + f2) / (30 + f1 + f2)]

12.8 (30 + f1 + f2) = 20(28 - f1 + f2)

384 + 12.8f1 + 12.8f2 = 560 - 20f1 + 20f2

12.8f1 + 20f1 + 12.8f2 - 20f2 = 560 - 384

32.8f1 - 7.2f2 = 176

32.8 (20 - f2) - 7.2f2 = 176

[From eq 1]

656 - 32.8f2 - 7.2f2 = 176

- 32.8f2 - 7.2f2 = 176 - 656

- 40f2 = - 480

40f2 = 480

f2 = 480/40 = 48/4 = 12

f2 = 12

Putting the value of f2 in eq 1  

f1 = 20 - f2

f1 = 20 - 12

f1 = 8  

Hence, the missing frequency f1 = 8 & f2 = 12.

HOPE THIS ANSWER WILL HELP YOU….