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

 Problem Back to Problems Page
The value of C for which P(X = K) = CK2 can serve as the probability function of a random variable x that takes values 0, 1, 2, 3, 4, is ...

 [Expectation: 3.33 ; Variance: 0.71 ; Standard Deviation : +0.843]

 Solution

"x" represent the value in the range of the random variable "X".
Therefore, f(x) = P(X=x) represents the probability mass function
 f(x) = P(X = k) = CK2 ⇒ where K = 1,   f(1) = C (1)2 ⇒ f(1) = C where K = 2,   f(2) = C (2)2 ⇒ f(2) = 4C where x = 3,   f(3) = C (3)2 ⇒ f(3) = 9C where x = 4,   f(4) = C (4)2 ⇒ f(4) = 16C

Since f(x) represents the probability mass function, the discrete probability distribution of "x" would be
 x P(X = x) 1 2 3 4 C 4C 9C 16C

Since f(x) is a probability mass function,
Σ p = 1
⇒ C + 4C + 9C + 16C = 1
⇒ 30C = 1
⇒ C =
 1 30

The discrete probability distribution of "x" with the values of "C" replaced

The probability distribution replacing the values of of "k" would be

x P (X = x)
In C terms Calculations Probability
1 C
1 × (
 1 30
)
 1 30
2 4C
4 ×(
 1 30
)
 4 30
3 9C
9 ×(
 1 30
)
 9 30
4 16C
16 ×(
 1 30
)
 16 30
Therefore, the distribution would be
x 1 2 3 4
P(X = x)
 1 30
 4 30
 9 30
 16 30

Calculations for Mean and Standard Deviations

x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
1
 1 30
 1 30
1
 1 30
2
 4 30
 8 30
4
 16 30
3
 9 30
 27 30
9
 81 30
4
 16 30
 64 30
16
 256 30
Total 1
 100 30
 354 30
= 3.33 = 11.8

Expectation/Mean of the distribution
 ⇒ E (x) (Or) x = Σ px = 3.33

Variance of the distribution
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = 11.8 − (3.33)2 = 11.8 − 11.09 = 0.71
Standard deviation of the distribution
 ⇒ SD (x) = + √ Var (x) = + √ 0.71 = + 0.843

 Credit : Vijayalakshmi Desu