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
9
Answers
In question last frequency is 8 ..
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….
=> 31 + f1 + f2 = 50
=> f1 + f2 = 50 - 31
=> f1 + f2 = 19 ----( 1 )
Mean = 62.8 ( given )
Σfixi/Σfi = 62.8
=> ( 2170 + 30f1 + 70f2 ) /50 = 62.8
=> 2170 + 30f1 + 70f2 = 62.8 × 50
=> 30f1 + 70f2 = 3140 - 2170
=> 30f1 + 70f2 = 970 -----( 2 )
Subtract equation 30 ×( 1 ) from (2 ) ,
we get
=> 40f2 = 400
=> f2 = 10
substitute f2 in equation ( 1 ), we get
f1 + 10 = 19
=> f1 = 19 - 10
=> f1 = 9
Therefore ,
f1 = 9
f2 = 10
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