Probability Distribution Mean (Expectation), Variance :: Problems

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 − 1 0 1 P(X = x) 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 − 1 0 1 P(X = x) 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