Probability Distribution Mean (Expectation), Variance :: Problems

Solution

Since X is a random variable. Assuming the random variable to be discrete.. The given probability distribution would be a discrete probability distribution of "X" For a discrete probability distribution of a random variable "X", Σ p = 1 From the above distribution Σ p = 1 ⇒ 0 + 2K + 2K + K + 3K + K2 + 2K2 + (7K2 + K) = 1 ⇒ 9k + 10 k2 = 1 ⇒ 10 K2 + 9K − 1 = 0 ⇒ 10 K2 + 10K − k − 1 = 0 ⇒ 10 K (K + 1) − 1 (K + 1) = 0 ⇒ (10K − 1) (K + 1) = 0 ⇒ 10K − 1 = 0 (Or) K + 1 = 0 ⇒ 10K = 1 (Or) K = − 1 ⇒ K = 1 10 (Or) 0.1 Since probability cannot be negative, k = −1 is ignored. The probability distribution replacing the values of of "k" would be x P (X = x) In K terms Calculations Probability 0 0 0 1 2k 2 × (0.1) 0.2 2 2k 2 × (0.1) 0.2 3 k 1 × (0.1) 0.1 4 3k 3 × (0.1) 0.3 5 k2 (0.1)2 0.01 6 2k2 2 × (0.1) 0.02 7 7k2 + k 7 × (0.1) + (0.1) 0.17 Therefore, P(x < 6) = P(x=0) + P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) (Or) = 1 − P(x ≥ 6) = 1 − [P(x=6) + P(x=7)] = 1 − [0.02 + 0.17] = 1 − 0.19 = 0.81 P(x ≥ 6) = P(x=6) + P(x=7) = 0.02 + 0.17 = 0.19 P(0 < x < 5) = P(x=1) + P(x=2) + P(x=3) + P(x=4) = 0.2 + 0.2 + 0.1 + 0.3 = 0.8 Calculations for finding the mean and variance of the distribution x P (X = x) px [x × P (X = x)] x2 px2 [x2 × P (X = x)] 0 0.02 0 0 0 1 0.20 0.20 1 0.20 2 0.20 0.40 4 0.80 3 0.10 0.30 9 0.90 4 0.30 1.20 16 4.80 5 0.01 0.05 25 0.25 6 0.02 0.12 36 0.72 7 0.17 1.19 49 8.33 Total 1.00 3.46 16.00 Mean/Expectation of the distribution ⇒ E (x) (Or) x = Σ px = 3.46 Variance of the distribution ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 16 − (3.46)2 = 16 − 11.9716 = 4.0284 Standard Deviation of the distribution ⇒ SD (x) = + √ Var (x) ⇒ SD (x) = + √ 4.0284 ⇒ SD (x) = + 2.007