Mean (Expectation), Variance from a Probability Distribution of a Discrete Random Variable

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Find the mean and variance of the following probability distribution and hence find its standard deviation

− 10

− 20

P (X = x)

Net Answers :
[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

Probability Distribution Mean (Expectation), Variance :: Problems