Let X be a Bernoulli rv with pmf as in Example 3.18. a.? ?Compute E(X2). b.?? how that V(X) = p(1 – p). c.? ?Compute E(X79). Reference example 3.18 Let X = 1 if a randomly selected vehicle passes an emissions test and X = 0 otherwise. Then X is a Bernoulli rv with pmf p(1) = p and p(0) = 1 - p , from which E(X) = 0. P(0) + 1. P(1) = 0(1 – p) + 1(p) = p. That is, the expected value of X is just the probability that X takes on the value 1. If we conceptualize a population consisting of 0s in proportion and 1 – p in proportion p, then the population average is µ = p.

Answer : Step 1 : Let X be a Bernoulli rv with pmf as Let X = 1 if a randomly selected vehicle passes an emissions test and X = 0 otherwise. a). 2 Now we have to calculate E(X ). SoE(X ) = X p(x) E(X ) = X p(x) 2 2 2 E(X ) = 1 × p + 0 × (1 p) E(X ) = p Therefore E(X ) = p. Step 2 : b). We have to show that V(X) = p(1 – p). Then, E(X ) = X p(x) 2 E(X ) = 1 × p + 0 × (1 p) E(X ) = p 2 2 V (X) = E(X ) E(X) V (X) = P (P) 2 V (X) = P(1 P) Therefore V (X) = P(1 P).