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Measures of Dispersion — Economics STD 11 Commerce — Question

CBSE BoardEnglish MediumSTD 11 CommerceEconomicsMeasures of Dispersion6 Marks
Question
Find out the standard deviation from the following data after finding mean:
Class lnterval: 0-10 10-20 20-30 30-40 40-50
Frequency: 8 13 16 8 5
✓

Answer

Class Interval Frequency M.V. (X) $dx$ $dx'$ $dx'^2$ $fdx'$ $fdx'^2$
0-10 8 5 -20 -2 4 -16 32
10-20 13 15 -10 -1 1 -13 13
20-30 16 25 0 0 0 0 0
30-40 8 35 10 1 1 8 8
40-50 5 45 20 2 4 10 20
  $\Sigma\text{f}=50$         $\Sigma\text{fdx}'=-11$ $\Sigma\text{fdx}'^2=73$
Mean $=\text{A}+\frac{\Sigma\text{fdx}'}{\Sigma\text{f}}\times\text{i}$
$=25-\frac{(-11)}{50}\times10$
$=25-\frac{(-11)}{5}$
$=25-2.2=22.8$
S.D. $=\sqrt{\frac{\Sigma\text{fdx}'^2}{\Sigma\text{f}}-\Big(\frac{\Sigma\text{fdx}'}{\Sigma\text{f}}\Big)^2}\times\text{C}$
$=\sqrt{\frac{73}{50}-\Big(\frac{11}{50}\Big)^2}\times10$
$=\sqrt{1.46-(0.22)^2}\times10$
$=\sqrt{1.46-.05}\times10=\sqrt{1.41}\times10$
$=1.19\times10=11.9$

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