Notes On Shortcut Method To Find Variance and Standard Deviation - CBSE Class 11 Maths
Variance of a discrete frequency distribution:  Observation (xi)   x1   x2   x3   ...   xn  Frequency (fi)   f1   f2   f3   ...   fn Ïƒ2 = $\sqrt{\frac{\underset{\mathrm{\text{i = 1}}}{\overset{\text{n}}{âˆ‘}}\text{}{\text{f}}_{\text{i}}\mathrm{\text{}}\text{}{\left({\text{x}}_{\text{i}}\text{}\mathrm{\text{-}}\text{}\stackrel{_}{\text{x}}\text{}\right)}^{2}}{\text{N}}}$ Total of the frequencies: N = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}$   Variance of a continuous frequency distribution: Ïƒ2 = $\frac{\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}\text{}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}}{\text{N}}$ Total of the frequencies: N = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}$   Many a times, the values of xi in the discrete frequency distribution or the mid-values in the continuous distribution are large. In such cases, there are heavy calculations to find the mean, variance, standard deviation, and so on. To avoid such tedious and time-consuming calculations, step-deviation method by working with an assumed mean is used.   Let the assumed mean be A. The deviations of the mid-values can be reduced by $\frac{\text{1}}{\text{h}}$ times, where h is the common factor among the mid-values (we usually take the width of the class intervals). Let the step-deviations be yi $\frac{{\text{x}}_{\text{i}}\text{- A}}{\text{h}}$ â‡’ ${\text{x}}_{\text{i}}\text{= A + h}{\text{y}}_{\text{i}}$                    .... eq (1) $\stackrel{_}{\text{x}}\mathrm{\text{}}\mathrm{\text{=}}\frac{\underset{\mathrm{\text{i = 1}}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}\mathrm{\text{}}{\text{x}}_{\text{i}}}{\text{N}}$ $\mathrm{\text{=}}\frac{\underset{\mathrm{\text{i = 1}}}{\overset{\text{n}}{âˆ‘}}\mathrm{\text{}}{\text{f}}_{\text{i}}\mathrm{\text{}}\mathrm{\text{(A + h}}{\text{y}}_{\text{i}}\text{)}}{\text{N}}$   = $\text{}$$\frac{\text{A}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}}{\text{N}}\text{+}\frac{\text{h}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}{\text{N}}$$\text{}$   = $\frac{\text{AN}}{\text{N}}\text{+}\frac{\text{h}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{x}}_{\text{i}}\text{y}}_{\text{i}}}{\text{N}}$                             (âˆµ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}$  = N  ) Let $\text{}$  =  $\stackrel{_}{\text{y}}$ âˆ´ $\stackrel{_}{\text{x}}\text{= A + h}\stackrel{_}{\text{y}}$                          .... eq(2) Substituting eq (1) and eq (2) in the formula for the variance, Ïƒ2 = $\frac{\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}\text{}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}}{\text{N}}$, Ïƒ2 = Ïƒ2 = Ïƒ2 = $\underset{\text{i = 1}}{\overset{\text{n}}{\frac{{\text{h}}^{\text{2}}}{\text{N}}âˆ‘}}{{\text{f}}_{\text{i}}\text{(}{\text{y}}_{\text{i}}\text{-}\stackrel{_}{\text{y}}\text{)}}^{\text{2}}$ The variance deduced is in the term of variable yi.   Hence, this can also be written as: Ïƒx2 = h2Ïƒy2.   â‡’ Ïƒx = hÏƒy Ïƒx  = $\sqrt{\text{N}âˆ‘{{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}^{\text{2}}\text{-}{\text{(}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}\text{)}}^{\text{2}}}$   (âˆµ Ïƒy  = $\frac{\text{1}}{\text{N}}$ $\sqrt{\text{N}âˆ‘{{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}^{\text{2}}\text{-}{\text{(}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}\text{)}}^{\text{2}}}$)

#### Summary

Variance of a discrete frequency distribution:  Observation (xi)   x1   x2   x3   ...   xn  Frequency (fi)   f1   f2   f3   ...   fn Ïƒ2 = $\sqrt{\frac{\underset{\mathrm{\text{i = 1}}}{\overset{\text{n}}{âˆ‘}}\text{}{\text{f}}_{\text{i}}\mathrm{\text{}}\text{}{\left({\text{x}}_{\text{i}}\text{}\mathrm{\text{-}}\text{}\stackrel{_}{\text{x}}\text{}\right)}^{2}}{\text{N}}}$ Total of the frequencies: N = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}$   Variance of a continuous frequency distribution: Ïƒ2 = $\frac{\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}\text{}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}}{\text{N}}$ Total of the frequencies: N = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}$   Many a times, the values of xi in the discrete frequency distribution or the mid-values in the continuous distribution are large. In such cases, there are heavy calculations to find the mean, variance, standard deviation, and so on. To avoid such tedious and time-consuming calculations, step-deviation method by working with an assumed mean is used.   Let the assumed mean be A. The deviations of the mid-values can be reduced by $\frac{\text{1}}{\text{h}}$ times, where h is the common factor among the mid-values (we usually take the width of the class intervals). Let the step-deviations be yi $\frac{{\text{x}}_{\text{i}}\text{- A}}{\text{h}}$ â‡’ ${\text{x}}_{\text{i}}\text{= A + h}{\text{y}}_{\text{i}}$                    .... eq (1) $\stackrel{_}{\text{x}}\mathrm{\text{}}\mathrm{\text{=}}\frac{\underset{\mathrm{\text{i = 1}}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}\mathrm{\text{}}{\text{x}}_{\text{i}}}{\text{N}}$ $\mathrm{\text{=}}\frac{\underset{\mathrm{\text{i = 1}}}{\overset{\text{n}}{âˆ‘}}\mathrm{\text{}}{\text{f}}_{\text{i}}\mathrm{\text{}}\mathrm{\text{(A + h}}{\text{y}}_{\text{i}}\text{)}}{\text{N}}$   = $\text{}$$\frac{\text{A}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}}{\text{N}}\text{+}\frac{\text{h}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}{\text{N}}$$\text{}$   = $\frac{\text{AN}}{\text{N}}\text{+}\frac{\text{h}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{x}}_{\text{i}}\text{y}}_{\text{i}}}{\text{N}}$                             (âˆµ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}$  = N  ) Let $\text{}$  =  $\stackrel{_}{\text{y}}$ âˆ´ $\stackrel{_}{\text{x}}\text{= A + h}\stackrel{_}{\text{y}}$                          .... eq(2) Substituting eq (1) and eq (2) in the formula for the variance, Ïƒ2 = $\frac{\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{f}}_{\text{i}}\text{}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}}{\text{N}}$, Ïƒ2 = Ïƒ2 = Ïƒ2 = $\underset{\text{i = 1}}{\overset{\text{n}}{\frac{{\text{h}}^{\text{2}}}{\text{N}}âˆ‘}}{{\text{f}}_{\text{i}}\text{(}{\text{y}}_{\text{i}}\text{-}\stackrel{_}{\text{y}}\text{)}}^{\text{2}}$ The variance deduced is in the term of variable yi.   Hence, this can also be written as: Ïƒx2 = h2Ïƒy2.   â‡’ Ïƒx = hÏƒy Ïƒx  = $\sqrt{\text{N}âˆ‘{{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}^{\text{2}}\text{-}{\text{(}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}\text{)}}^{\text{2}}}$   (âˆµ Ïƒy  = $\frac{\text{1}}{\text{N}}$ $\sqrt{\text{N}âˆ‘{{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}^{\text{2}}\text{-}{\text{(}\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}\text{)}}^{\text{2}}}$)

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