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

σ2 = 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:



σ2 = 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 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 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 yi 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, σ2 = i = 1 n f i ( x i - x _ ) 2 N ,

σ2 = i = 1 n f i ( A + hy i - A -  hy _ ) 2 N

σ2 = i = 1 n f i (  hy i -  h  y _ ) 2 N

σ2 = h 2 N i = 1 n f i ( y i - y _ ) 2

The variance deduced is in the term of variable yi.
 
Hence, this can also be written as: σx2 = h2σy2.
 
⇒ σx = hσy

σx  = N f i y i 2 - ( i = 1 n f i y i ) 2

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

Summary

Variance of a discrete frequency distribution:
 Observation (xi)   x1   x2   x3   ...   xn
 Frequency (fi)   f1   f2   f3   ...   fn

σ2 = 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:



σ2 = 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 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 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 yi 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, σ2 = i = 1 n f i ( x i - x _ ) 2 N ,

σ2 = i = 1 n f i ( A + hy i - A -  hy _ ) 2 N

σ2 = i = 1 n f i (  hy i -  h  y _ ) 2 N

σ2 = h 2 N i = 1 n f i ( y i - y _ ) 2

The variance deduced is in the term of variable yi.
 
Hence, this can also be written as: σx2 = h2σy2.
 
⇒ σx = hσy

σx  = N f i y i 2 - ( i = 1 n f i y i ) 2

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

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