Question

1 0.7 Find the Expected value and the variance of X. E (X)=EXP(X) Var(x)-o? And -E(x?)-me a) Note: = px E(X2)-DFP(%) b) Consider the following information for a binomial N- number of trials or experiments-5 distribution: x-number of success -3 Probability of uccess-sp- 04 and probability of filur 1p-0.6 Find the probability of 3 successes out of 5 trials: Note P(x): Nex px (1-p)NX Note: Nex NI / x! (Naji

Answer 1

The expected value of x E(x) = sigma x. p(r) = M_x = (0) (0.3) + (1) (0.7) E(x) = 0.7 variance of x Var (x) = sigma^2_x = E (r^2) - M^2_x E(x^2) = sigma x^2 P(x) = (0)^2 (0.3) + (1)^2 (0.7) = > 0.7 Var(x) = 0.7 - (0.7)^2 = > 0.7 - 0.49 Var(x) = 0.21 P(x) = N_c_gamma P^r (1 - p)^v - r probability of 3 success at of 5 = 5_C_3 (0.4)^3 (0.6)^2 \r\n = 5!/3! (5 - 3)! (0.4)^3 (0.6)^2 = 5!/3! 2! (0.064) (0.36) = (10) (0.064) (0.36) = 0.2304 P(x) = 0.2304