# Probability Distribution :: Mean and Variance

### Mean and variance

For a simple frequency distribution:

 Variable (x) Frequency (y) 5 8 10 15 50 125 512 1,000 3,375 8,000
Mean (x) =
 Σ fx Σ f
Variance (σ2) =
 Σ fx2 Σ f
(
 Σ fx Σ f
)2

In a frequency distribution the total frequency (Σf) indicates the total number of units in the data from which the simple frequency distribution has been constructed. From the above distribution we can say that the value 5 appears 125 times, 8 appers 512 times, 10 appears 1,000 times, 15 appears 3,375 times, 50 appears 8,000 times in the raw data from which the distribution has been constructed.

Therefore, we can say that a frequency distribution is a distribution where the total frequency is distributed over the different values of the variable in the distribution.

For a probability distribution

A probability distribution is analogous to a frequency distribution. The probability can be compared to the frequency in a frequency distribution.
x 0 1 2 3
P(X=x) or p
 1 8
 3 8
 3 8
 1 8

Applying similar logic to
Mean (x) =
 Σ px Σ p
[
 Σ fx Σ f
]
=
 Σ px 1
= Σ px
= E (x)
[Σ p = 1 for a probability distribution of a discrete random variable]

The mean of a probability distribution is nothing but its expectation.
Variance [Var (x)] =
 Σ px2 Σ p
(
 Σ px Σ p
)2       [
 Σ fx2 Σ f
(
 Σ fx Σ f
)2 ]
=
 Σ px2 1
(
 Σ px 1
)2
= (Σ px2) − (Σ px)2
= E (x2) − (E (x))2
[Σ p = 1 for a probability distribution of a discrete random variable]

We can say that a probability distribution is a distribution where the total probability (1) is distributed over the different values of the variable in the distribution.

### Illustration

If "x" represents a value within the range of a random variable "X" and
(i) If "a" is a constant then show that var(a) = 0 and var(ax) = a2 var(x)
(ii) If "a" and "b" are constants, then var (a + bx) = b2 var(x)
Sol. (i) We know, Var (x) = E(x2) − [ E(x)]2 (Or) Σ px2 − ( Σ px)2
 Therefore, Var (a) = Σ p a2 − (Σ p a) 2 = a2 Σ p − (a Σ p) 2 = a2 × 1 − (a × 1) 2 = a2 − (a) 2 = 0
 Var (ax) = Σ p (ax)2 − (Σ p ax) 2 = Σ p a2x2 − (a Σ px) 2 = a2 Σ p x2 − (a)2 (Σ px) 2 = a2 [ Σ p x2 − (Σ px) 2] = a2 [Var (x)]
 Σ pa = p1a + p2a + ... pna = a [p1 + p2 + ... pn] = c Σ p
(ii) We know, Var (x) = E(x2) − [ E(x)]2 (Or) Σ px2 − ( Σ px)2
 Therefore, Var (a + bx) = Σ p (a + bx)2 − ( Σ p (a + bx))2 = Σ p (a2 + 2abx + (bx)2) − ( Σ p a + pbx)2 = [Σ p a2 + 2 pabx + p(bx)2] − [ (Σ p a + Σ pbx)2] = [Σ p a2 + Σ 2 pabx + Σ p(bx)2] − [(a Σ p + b Σ px)2] = [a2 Σ p + 2ab Σ px + b2 Σ p x2] − [(a × 1 + b Σ px)2] = [a2 × 1 + 2ab Σ px + b2 Σ p x2] − [(a + b Σ px)2] = [a2 + 2ab Σ px + b2 Σ p x2] − [(a2 + 2a b Σ px + (b Σ px)2] = a2 + 2ab Σ px + b2 Σ p x2 − a2 − 2a b Σ px − (b Σ px)2 = b2 Σ p x2 − [ b2 (Σ px)2 = b2 { Σ p x2 − (Σ px)2} = b2 {Var (x)}
 Author Credit : The Edifier