The value of appropriate measure of dispersion for the following distribution of daily Wages Wages (`): Below 30 30-39 40-49 50-59 60-79 Above 80 No. of workers 5 7 18 32 28 10 is given by (a) `11.03 (b) ` 10.50 (c) 11.68 (d) `11.68.
Answers
Given :-
- Distribution = No. of workers.
- Below 30 = 5
- 30 - 39 = 7
- 40 - 49 = 18
- 50 - 59 = 32
- 60 - 79 = 28
- above 80 = 10
To Find :-
- which measure of dispersion is better suited to data ?
Formula used :-
- Mean = (m * f) / N
- Variance = [ (sum of m²f) - {(sum of m * f)² / N ] / N
- Standard Deviation = √(variance)
Solution :-
From image we get :-
- N = 100
- sum of (m * f) = 5682.5
- sum of (m² * f) = 342771.25
Putting all values in above told formulas now, we get :-
→ Mean wages of 100 workers = (5682.5/100) ≈ 56.83
and,
→ variance = [ (sum of m²f) - {(sum of m * f)² / N ] / N
→ variance = [ (342771.25) - {(5682.5)²/100} ] / 100
→ variance = (342771.25 - 322908.06) / 100
→ variance = (19863.18)/100
→ variance ≈ 198.63
and,
→ Standard Deviation = √(variance)
→ SD = √(198.63)
→ SD ≈ 14.09
Variance > Mean wages > SD.
Therefore,
if wages is given in variance than it is better .
AnswEr :-
Given:-
Distribution = No. of workers.
Below 30 = 5
30 - 39 = 7
40 - 49 = 18
50 - 59 = 32
60 - 79 = 28
above 80 = 10
To Find :-
measure of dispersion is better suited to data .
Formula used :-
Mean = (m * f) / N
Variance = [ (sum of m²f) - {(sum of m * f)² / N ] / N
Standard Deviation = √(variance)
Solution :-
N = 100
sum of (m * f) = 5682.5
sum of (m² * f) = 342771.25
Substituting all values in above told formulas,
→ Mean wages of 100 workers = (5682.5/100) ≈ 56.83
and,
→ variance = [ (sum of m²f) - {(sum of m * f)² / N ] / N
→ variance = [ (342771.25) - {(5682.5)²/100} ] / 100
→ variance = (342771.25 - 322908.06) / 100
→ variance = (19863.18)/100
→ variance ≈ 198.63
and,
→ Standard Deviation = √(variance)
→ SD = √(198.63)
→ SD ≈ 14.09
Variance > Mean wages > SD.
Therefore,
if wages is given in variance than it is better .
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