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 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 1 2 3 4 P(X = x) 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