Probability Distribution Mean (Expectation), Variance :: Problems

Solution

f(x) = k ⇒ where x = 1,   f(1) = k where x = 2,   f(2) = k where x = 3,   f(3) = k where x = 4,   f(4) = k where x = 5,   f(5) = k where x = 6,   f(6) = k The probability distribution of "x" which is a discrete probability distribution would be x 1 2 3 4 5 6 P(X = x) k k k k k k Since f(x) is a probability mass function, Σ p = 1 ⇒ k + k + k + k + k + k = 1 ⇒ 6 k = 1 ⇒ k 1 6 The discrete probability distribution of "x" with the values of "k" replaced x 1 2 3 4 5 6 P(X = x) 1 6 1 6 1 6 1 6 1 6 1 6 Calculations for Mean and Standard Deviations x P (X = x) px [x × P (X = x)] x2 px2 [x2 × P (X = x)] 1 1 6 1 6 1 1 6 2 1 6 2 6 4 4 6 3 1 6 3 6 9 9 6 4 1 6 4 6 16 16 6 5 1 6 5 6 25 25 6 6 1 6 6 6 36 36 6 Total 1 21 6 91 6 = 3.5 = 15.167 Expectation/Mean of the distribution ⇒ E (x) (Or) x = Σ px = 3.5 Variance of the distribution ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = 15.167 − (3.5)2 = 15.167 − 12.25 = 15.167 − 12.25 = 2.917 Standard deviation of the distribution ⇒ SD (x) = + √ Var (x) = + √ 2.917 = + 1.708