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# Shortcut Method To Find Variance and Standard Deviation

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#### Shortcut Method To Find Variance and Standard Deviation - Lesson Summary

Variance of a discrete frequency distribution:   Observation (xi)   x 1   x 2   x 3   ...   x n   Frequency (fi)   f 1   f 2   f 3   ...   f n
σ 2 = ∑ i = 1 n f i x i - x _ 2 N

Total of the frequencies: N = ∑ i = 1 n f i

Variance of a continuous frequency distribution:

σ 2 = ∑ i = 1 n f i ( x i - x _ ) 2 N

Total of the frequencies: N = ∑ i = 1 n f 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 1 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 =  x i - A h

⇒  x i = A + h y i                     .... eq (1)

x _ = ∑ i = 1 n f i x i N

x _   = ∑ i = 1 n f i (A + h y i ) N

= A ∑ i = 1 n f i N + h ∑ i = 1 n f i y i N

=  AN N + h ∑ i = 1 n x i y i N                             (∵ ∑ i = 1 n f i   = N  )

Let   ∑ i = 1 n f i N   =  y _

∴  x _ = A + h y _                          .... eq(2)
Substituting eq (1) and eq (2) in the formula for the variance, σ 2 = ∑ i = 1 n f i ( x i - x _ ) 2 N ,

σ 2 = ∑ i = 1 n f i ( A + hy i - A -  hy _ ) 2 N

σ 2 = ∑ i = 1 n f i (  hy i -  h  y _ ) 2 N

σ 2 = h 2 N ∑ i = 1 n f i ( y i - y _ ) 2

The variance deduced is in the term of variable y i.

Hence, this can also be written as: σ x 2 = h 2σ y 2.

⇒ σ x = hσ y

σ x  = N ∑ f i y i 2 - ( ∑ i = 1 n f i y i ) 2

(∵ σ y  =  1 N N ∑ f i y i 2 - ( ∑ i = 1 n f i y i ) 2 )