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

 Problem Back to Problems Page
A random variable x takes the values of &m inus;1, 0, 1. Its mean is 0.6.
If P(X = 0) = 0.2, then P(X = 1) = ?

 [Expectation: 0.6 ; Variance: 0.44 ; Standard Deviation : +0.663; P(X=1): 0.7]

 Solution

"x" represent the value in the range of the random variable "X".
Therefore, f(x) = P(X=x) represents the probability mass function

P(X = 0) = 0.2   ⇒ f(0) = 0.2

Let P(X = − 1) = "a" and P(X = 1) = "b"

Since f(x) represents the probability mass function, the discrete probability distribution of "x" would be
 x P(X = x) − 1 0 1 a 0.2 b

Since f(x) is a probability mass function,
Σ p = 1
⇒ a + 0.2 + b = 1
⇒ a + b = 1 − 0.2
⇒ a + b = 0.8     → (1)

Calculations for Mean and Standard Deviations

x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
− 1 a − a 1 a
0 0.2 0 0 0
+ 1 b b 1 b
Total 1 − a + b a + b

Expectation/Mean of the distribution
 ⇒ E (x) (Or) x = Σ px ⇒ 0.6 = − a + b → (2)

Solving (1) and (2) we get,

 (1) → a + b = 0.8 (2) → − a + b = 0.6 (1) + (2) 2b = 1.4 ⇒ b = 0.7

Substituting the value of "b" in (1) we get,

a + b = 0.8   ⇒   a + 0.7 = 0.8   ⇒   a = 0.8 − 0.7   ⇒   a = 0.1

The distribution with the values of "a" and "b" replaced would be
 x P(X = x) − 1 0 1 0.1 0.2 0.7

Therefore, P(x=1) = b   ⇒   P(x=1) = 0.7

Variance of the distribution
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = (a + b) − (− a + b)2 = 0.8 − (0.6)2 = 0.8 − 0.36 = 0.44
Standard deviation of the distribution
 ⇒ SD (x) = + √ Var (x) = + √ 0.44 = + 0.663

 Credit : Vijayalakshmi Desu