Probability Distribution Mean (Expectation), Variance :: Problems

Solution

Let "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) = 3C3 ⇒ f(0) = 3C3 P(X = 1) = 4C − 10C2, ⇒ f(1) = 4C − 10C2 P(x = 2) = 5C − 1 ⇒ f(2) = 5C − 1 Since f(x) represents the probability mass function, the discrete probability distribution of "x" would be x 1 2 3 P(X = x) 3C3 4C − 10C2 5C − 1 Since f(x) is a probability mass function, Σ p = 1 ⇒ (3C3) + (4C − 10C2) + (5C − 1) = 1 ⇒ 3C3 + 4C − 10C2 + 5C − 1 = 1 ⇒ 3C3 − 10C2 + 9C − 1 − 1 = 0 ⇒ 3C3 − 10C2 + 9C − 2 = 0 ⇒ (C − 1) (3C2 − 7C + 2) = 0 ⇒ (C − 1) (3C2 − 6C − C + 2) = 0 ⇒ (C − 1) {3C(C − 2) − 1( C − 2)} = 0 ⇒ (C − 1) {(3C − 1) (C − 2)} = 0 ⇒ (C − 1) (3C − 1) (C − 2) = 0 ⇒ C − 1 = 0   (Or)   3C − 1 = 0   (Or)   C − 2 = 0 ⇒ C = 1   (Or)   3C = 1   (Or)   C = 2 ⇒ C = 1 3 C = 1 and C = 2 are ignored as they would give a value greater than 1 for 3C3. Since 3C3 represents probability it cannot be greater than 1. 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 3C3 3 × ( 1 3 )3 1 9 2 4C − 10C2 4 ×( 1 3 ) − 10×( 1 3 )2 2 9 3 5C − 1 5 ×( 1 3 ) − 1 2 3 (Or) 6 9 Therefore, the distribution would be x 1 2 3 P(X = x) 1 9 2 9 6 9 Calculations for Mean and Standard Deviations x P (X = x) px [x × P (X = x)] x2 px2 [x2 × P (X = x)] 1 1 9 1 9 1 1 9 2 2 9 4 9 4 8 9 3 6 9 18 9 9 54 9 Total 1 23 9 63 9 = 2.56 = 7 Expectation/Mean of the distribution ⇒ E (x) (Or) x = Σ px = 2.56 Variance of the distribution ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = 7 − (2.56)2 = 7 − 6.5536 = 0.4464 Standard deviation of the distribution ⇒ SD (x) = + √ Var (x) = + √ 0.4464 = + 0.668