# Probability Distribution Mean (Expectation), Variance :: Problems

 Problem Back to Problems Page
Find the mean and variance of the following probability distribution and hence find its standard deviation
x − 10 − 20 30 75 80
P (X = x)
 1 5
 3 20
 1 2
 1 10
 1 20

 [Expectation: 21.5; Variance: 932.25 ; Standard Deviation : +30.53]

 Solution

Calculations for Mean and Standard Deviations

x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
− 10
 1 5
=
 4 20
 − 40 20
100
 400 20
− 20
 3 20
=
 3 20
 − 60 20
400
 1,200 20
30
 1 2
=
 10 20
 300 20
900
 9,000 20
75
 1 10
=
 2 20
 150 20
5,625
 11,250 20
80
 1 20
=
 1 20
 80 20
6,400
 6,400 20
Total 1
 430 20
 27,890 20
= 21.5 = 1,394.5

Since the given distribution is the probability distribution of a discrete random variable "X", Σ p = 1.

For the probability distribution:

Mean of the distribution
⇒ Expectation of the variable
 ⇒ E (x) = Σ px = 21.5
Variance of the distribution
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 1,394.5 − (21.5)2 = 1,394.5 − 462.25 = 932.25
Standard Deviation of the distribution
 ⇒ SD (x) = + √ Var (x) ⇒ SD (x) = + √ 932.25 ⇒ SD (x) = + 30.53

 Credit : Vijayalakshmi Desu