In statistics, variance is a measure of the spread or scatter in data.
Variance can also be a measure of the spread of the values of a random variable in a probability distribution.
Var(X) = Ïƒ_{x}^{2} = âˆ‘_{i=1}^{n}(x_{i}  Î¼)^{2}p(x_{i})
x_{1,}x_{2,}x_{3,.....,}x_{n} are the possible values of X and they occur with the probabilities p(x_{1}), p(x_{2}),p(x_{3}), ...p(x_{n}), respectively.
Î¼ = E(X) = Mean of X
Ïƒ_{x}^{2} = E(X  Î¼)^{2}
The square root of the variance of random variable X is called the standard deviation of X.
Ïƒ_{x = }âˆšVar(X) = âˆš(âˆ‘_{i=1}^{n}(x_{i}  Î¼)^{2}p(x_{i}))
Var(X) = Ïƒ_{x}^{2} = âˆ‘_{i=1}^{n}(x_{i}  Î¼)^{2}p(x_{i})
= âˆ‘_{i=1}^{n}(x_{i}^{2} + Î¼^{2}  2Î¼x_{i})p(x_{i})
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i}) + âˆ‘_{i=1}^{n} Î¼^{2} p(x_{i})  âˆ‘_{i=1}^{n} 2Î¼x_{i }p(x_{i})
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i}) + âˆ‘_{i=1}^{n} Î¼^{2} p(x_{i})  2Î¼ âˆ‘_{i=1}^{n} x_{i }p(x_{i})
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i}) + Î¼^{2 } 2Î¼^{2} ['.' âˆ‘_{i=1}^{n} p(x_{i}) = 1 and Î¼ = âˆ‘_{i=1}^{n} x_{i }p(x_{i}) ]
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i})  Î¼^{2}
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i})  [âˆ‘_{i=1}^{n} x_{i }p(x_{i})]^{2}
â‡’ Var(X) = E(X^{2})  [E(X)]^{2}
Ex:
The probability distributions of two experiments are
X 
0 
1 
2 
3 
Y 
â€‘ 1 
0 
3 
5 
6 
P(X) 
1/8 
3/8 
3/8 
1/8 
P(Y) 
2/8 
3/8 
1/8 
1/8 
1/8 
Calculate the mean of the random variables in both the cases.
Mean of X = E(X) = âˆ‘_{i=1}^{4} x_{i }p_{i}
= x_{1 }p_{1} + x_{2 }p_{2} + x_{3 }p_{3} + x_{4 }p_{4}
= (0 x 1/8) + (1 x 3/8) + (2 x 3/8) + (3 x 1/8)
= 0 + 3/8 + 6/8 + 3/8
= (3+6+3)/8
= 12/8 = 1.5
Mean of Y = E(Y) = âˆ‘_{i=1}^{5} x_{i }p_{i}
= x_{1 }p_{1} + x_{2 }p_{2} + x_{3 }p_{3} + x_{4 }p_{4 }+ x_{5 }p_{5}
= (1 x 2/8) + (0 x 3/8) + (3 x 1/8) + (5 x 1/8) + (6 x 1/8)
= 2/8 + 0 + 3/8 + 5/8 + 6/8
= (2+3+5+6)/8
= 12/8 = 1.5
Observe that both the distributions have the same mean.
It does not give us any information about the variability in the values of the random variables.
To understand the variability in the variables, draw two graphs using the distributions ... by taking the values of the random variables along the Xaxis and their probability along the Yaxis.


Var(X) = Ïƒ_{x}^{2} = âˆ‘_{i=1}^{4}(x_{i}  Î¼)^{2}p(x_{i})
= (01.5)^{2} x 1/8 + (1  1.5)^{2} x 3/8 + (2  1.5)^{2} x 3/8 + (3  1.5)^{2}x 1/8
= (1.5)^{2} x 1/8 + ( 0.5)^{2} x 3/8 + (0.5)^{2} x 3/8 + (1.5)^{2}x 1/8
= 2.25 x 1/8 + 0.25 x 3/8 + 0.25 x 3/8 + 2.25 x 1/8
= 0.281 + 0.094 + 0.094 + 0.281
= 0.469
Var(Y) = Ïƒ_{y}^{2} = âˆ‘_{i=1}^{4}(y_{i}  Î¼)^{2}p(y_{i})
= (1 1.5)^{2} x 2/8 + (0  1.5)^{2} x 3/8 + (3  1.5)^{2} x 1/8 + (5  1.5)^{2}x 1/8 + (6  1.5)^{2}x 1/8
= (2.5)^{2} x 2/8 + ( 1.5)^{2} x 3/8 + (1.5)^{2} x 1/8 + (3.5)^{2}x 1/8 + (4.5)^{2}x 1/8
= 6.25 x 2/8 + 2.25 x 3/8 + 2.25 x 1/8 + 12.25 x 1/8 + 20.25 x 1/8
= 1.562 + 0.844 + 0.781 + 2.531 + 3.781
= 6.749
â‡’ Var(Y) > Var(X)
The spread of values is more in Y than in X.
In statistics, variance is a measure of the spread or scatter in data.
Variance can also be a measure of the spread of the values of a random variable in a probability distribution.
Var(X) = Ïƒ_{x}^{2} = âˆ‘_{i=1}^{n}(x_{i}  Î¼)^{2}p(x_{i})
x_{1,}x_{2,}x_{3,.....,}x_{n} are the possible values of X and they occur with the probabilities p(x_{1}), p(x_{2}),p(x_{3}), ...p(x_{n}), respectively.
Î¼ = E(X) = Mean of X
Ïƒ_{x}^{2} = E(X  Î¼)^{2}
The square root of the variance of random variable X is called the standard deviation of X.
Ïƒ_{x = }âˆšVar(X) = âˆš(âˆ‘_{i=1}^{n}(x_{i}  Î¼)^{2}p(x_{i}))
Var(X) = Ïƒ_{x}^{2} = âˆ‘_{i=1}^{n}(x_{i}  Î¼)^{2}p(x_{i})
= âˆ‘_{i=1}^{n}(x_{i}^{2} + Î¼^{2}  2Î¼x_{i})p(x_{i})
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i}) + âˆ‘_{i=1}^{n} Î¼^{2} p(x_{i})  âˆ‘_{i=1}^{n} 2Î¼x_{i }p(x_{i})
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i}) + âˆ‘_{i=1}^{n} Î¼^{2} p(x_{i})  2Î¼ âˆ‘_{i=1}^{n} x_{i }p(x_{i})
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i}) + Î¼^{2 } 2Î¼^{2} ['.' âˆ‘_{i=1}^{n} p(x_{i}) = 1 and Î¼ = âˆ‘_{i=1}^{n} x_{i }p(x_{i}) ]
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i})  Î¼^{2}
= âˆ‘_{i=1}^{n} x_{i}^{2} p(x_{i})  [âˆ‘_{i=1}^{n} x_{i }p(x_{i})]^{2}
â‡’ Var(X) = E(X^{2})  [E(X)]^{2}
Ex:
The probability distributions of two experiments are
X 
0 
1 
2 
3 
Y 
â€‘ 1 
0 
3 
5 
6 
P(X) 
1/8 
3/8 
3/8 
1/8 
P(Y) 
2/8 
3/8 
1/8 
1/8 
1/8 
Calculate the mean of the random variables in both the cases.
Mean of X = E(X) = âˆ‘_{i=1}^{4} x_{i }p_{i}
= x_{1 }p_{1} + x_{2 }p_{2} + x_{3 }p_{3} + x_{4 }p_{4}
= (0 x 1/8) + (1 x 3/8) + (2 x 3/8) + (3 x 1/8)
= 0 + 3/8 + 6/8 + 3/8
= (3+6+3)/8
= 12/8 = 1.5
Mean of Y = E(Y) = âˆ‘_{i=1}^{5} x_{i }p_{i}
= x_{1 }p_{1} + x_{2 }p_{2} + x_{3 }p_{3} + x_{4 }p_{4 }+ x_{5 }p_{5}
= (1 x 2/8) + (0 x 3/8) + (3 x 1/8) + (5 x 1/8) + (6 x 1/8)
= 2/8 + 0 + 3/8 + 5/8 + 6/8
= (2+3+5+6)/8
= 12/8 = 1.5
Observe that both the distributions have the same mean.
It does not give us any information about the variability in the values of the random variables.
To understand the variability in the variables, draw two graphs using the distributions ... by taking the values of the random variables along the Xaxis and their probability along the Yaxis.


Var(X) = Ïƒ_{x}^{2} = âˆ‘_{i=1}^{4}(x_{i}  Î¼)^{2}p(x_{i})
= (01.5)^{2} x 1/8 + (1  1.5)^{2} x 3/8 + (2  1.5)^{2} x 3/8 + (3  1.5)^{2}x 1/8
= (1.5)^{2} x 1/8 + ( 0.5)^{2} x 3/8 + (0.5)^{2} x 3/8 + (1.5)^{2}x 1/8
= 2.25 x 1/8 + 0.25 x 3/8 + 0.25 x 3/8 + 2.25 x 1/8
= 0.281 + 0.094 + 0.094 + 0.281
= 0.469
Var(Y) = Ïƒ_{y}^{2} = âˆ‘_{i=1}^{4}(y_{i}  Î¼)^{2}p(y_{i})
= (1 1.5)^{2} x 2/8 + (0  1.5)^{2} x 3/8 + (3  1.5)^{2} x 1/8 + (5  1.5)^{2}x 1/8 + (6  1.5)^{2}x 1/8
= (2.5)^{2} x 2/8 + ( 1.5)^{2} x 3/8 + (1.5)^{2} x 1/8 + (3.5)^{2}x 1/8 + (4.5)^{2}x 1/8
= 6.25 x 2/8 + 2.25 x 3/8 + 2.25 x 1/8 + 12.25 x 1/8 + 20.25 x 1/8
= 1.562 + 0.844 + 0.781 + 2.531 + 3.781
= 6.749
â‡’ Var(Y) > Var(X)
The spread of values is more in Y than in X.