Variance of a discrete frequency distribution:
Observation (xi) |
x1 |
x2 |
x3 |
... |
xn |
Frequency (fi) |
f1 |
f2 |
f3 |
... |
fn |
σ
2 =
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
x2 = h
2σ
y2.
⇒ 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 )