Shortcut Method To Find Variance and Standard Deviation

Variance of a discrete frequency distribution:

Observation (xi)	x1	x2	x3	...	xn
Frequency (fi)	f1	f2	f3	...	fn

σ² = $$ \sqrt{\frac{\sum _{\mathrm{\text{i = 1}}}^{\text{n}}{\text{f}}_{\text{i}}{\left({\text{x}}_{\text{i}}\mathrm{\text{-}}\stackrel{\_}{\text{x}}\right)}^2}{\text{N}}}$$

Total of the frequencies: N = $\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}$
 
Variance of a continuous frequency distribution:



σ² = $\frac{\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{\_}{\text{x}}\text{)}}^{\text{2}}}{\text{N}}$

Total of the frequencies: N = $\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}$
 
Many a times, the values of x_i 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 y_i $\text{= }$$\frac{{\text{x}}_{\text{i}}\text{- A}}{\text{h}}$

⇒ ${\text{x}}_{\text{i}}\text{= A + h}{\text{y}}_{\text{i}}$                    .... eq (1)

$$ \stackrel{\_}{\text{x}}\mathrm{\text{=}}\frac{\sum _{\mathrm{\text{i = 1}}}^{\text{n}}{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}}{\text{N}}$$

$$ \stackrel{\_}{\text{x}}$$$$ \mathrm{\text{=}}\frac{\sum _{\mathrm{\text{i = 1}}}^{\text{n}}{\text{f}}_{\text{i}}\mathrm{\text{(A + h}}{\text{y}}_{\text{i}}\text{)}}{\text{N}}$$

  = $$$\frac{\text{A}\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}}{\text{N}}\text{+}\frac{\text{h}\sum _{\text{i = 1}}^{\text{n}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}{\text{N}}$$$

  = $\frac{\text{AN}}{\text{N}}\text{+}\frac{\text{h}\sum _{\text{i = 1}}^{\text{n}}{{\text{x}}_{\text{i}}\text{y}}_{\text{i}}}{\text{N}}$                             (∵ $\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}$  = N  )

Let $$$\frac{\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}}{\text{N}}$  =  $\stackrel{\_}{\text{y}}$

∴ $\stackrel{\_}{\text{x}}\text{= A + h}\stackrel{\_}{\text{y}}$                          .... eq(2)
Substituting eq (1) and eq (2) in the formula for the variance, σ² = $\frac{\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{\_}{\text{x}}\text{)}}^{\text{2}}}{\text{N}}$,

σ² = $\frac{\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}{\text{(}{\text{A + hy}}_{\text{i}}\text{- A - h}\stackrel{\_}{\text{y}}\text{)}}^{\text{2}}}{\text{N}}$

σ² = $\frac{\sum _{\text{i = 1}}^{\text{n}}{\text{f}}_{\text{i}}{\text{(} {\text{hy}}_{\text{i}}\text{- h }\stackrel{\_}{\text{y}}\text{)}}^{\text{2}}}{\text{N}}$

σ² = $\underset{\text{i = 1}}{\overset{\text{n}}{\frac{{\text{h}}^{\text{2}}}{\text{N}}\sum }}{{\text{f}}_{\text{i}}\text{(}{\text{y}}_{\text{i}}\text{-}\stackrel{\_}{\text{y}}\text{)}}^{\text{2}}$

The variance deduced is in the term of variable y_i.
 
Hence, this can also be written as: σ_x² = h²σ_y².
 
⇒ σ_x = hσ_y

σ_x  = $\sqrt{\text{N}\sum {{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}^{\text{2}}\text{-}{\text{(}\sum _{\text{i = 1}}^{\text{n}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}\text{)}}^{\text{2}}}$

  (∵ σ_y  = $\frac{\text{1}}{\text{N}}$ $\sqrt{\text{N}\sum {{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}}^{\text{2}}\text{-}{\text{(}\sum _{\text{i = 1}}^{\text{n}}{{\text{f}}_{\text{i}}\text{y}}_{\text{i}}\text{)}}^{\text{2}}}$)