Probability Distribution Mean (Expectation), Variance :: Problems

Solution

f(x) = ax ⇒ where x = 1,   f(1) = a × 1 ⇒   f(1) = a where x = 2,   f(2) = a × 2 ⇒   f(2) = 2a where x = 3,   f(3) = a × 3 ⇒   f(3) = 3a ... ... ... where x = n,   f(n) = a × n ⇒   f(n) = na The probability distribution of "x" which is a discrete probability distribution would be x 1 2 3 ... ... n P(X = x) a 2a 3a ... ... na Since f(x) is a probability mass function, Σ p = 1 ⇒ a + 2a + 3a + .... + na = 1 ⇒ a (1 + 2 + 3 + .... + n) = 1 ⇒ a n(n+1) 2 = 1 ⇒ a n(n + 1) = 2 ⇒ a = 2 n(n+1) Therefore, for a = 2 n(n+1) , f(x) represents a probability mass function 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 2 2a 22a 22 23a 3 3a 32a 32 33a ... ... ... ... ... ... ... ... ... ... n na n2a n2 n3a Total 1.00 4.36 22.54 Expectation/Mean of the distribution ⇒ E (x) (Or) x = Σ px = a + 22a + 32a + ... + n2a = a (1 + 22 + 32 + ... + n2) = a (12 + 22 + 32 + ... + n2) = a n(n+1)(2n+1) 6 = a n(n+1)(2n+1) 6 Variance of the distribution ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 22.54 − (4.36)2 = 22.54 − 19.0096 = 3.5304 Standard Deviation of the distribution ⇒ SD (x) = + √ Var (x) = + √ 3.5304 = + 1.879