Week 3 SolutionSeptember 8, 20171. Moment Generating functionE(X) = M0(0) =14+ 2 ∗18+ 3 ∗38+ 4 ∗14=218E(X2) = M00(0) =14+ 22∗18+ 32∗38+ 42∗14=658V ar(X) = E(X2) − [E(X)]2= 1.234375pmf: P (X = 1) =14, P (X = 2) =18, P (X = 3) =38, P (X = 4) =142. BinomialYu’s section version(a) X is B(20, 0.8)(b) E(X) = np = 20 ∗ 0.8 = 16; V ar(X) = np(1− p) = 20 ∗ 0.8∗ 0.2 = 3.2(c) P (X = 15) = C1520(0.8)15(0.2)5; P (X > 15) = P (X = 16) + P (X =17)+P (X = 18)+P (X = 19)+P (X = 20). P (X ≤ 15) = 1−P (X >15)Derivation of E(X) and V ar(X)E(X) = E(P20i=1Xi) =P20i=1E(X) = np = 20 ∗ (0.8 ∗ 1 + 0.2 ∗ 0) = 16V ar(X) = V ar(P20i=1Xi) =P20i=1E(Xi) = np(1 − p), using V ar(X1+X2) = V ar(X1) + V ar(X2) if X1,X2are independentThe second way of deriving V ar(X) is more complicated,1V ar(X) = E(X2) − [E(X)]2= E[(20Xi=1Xi)2] − E(X)2= E[20Xi=120Xj=1XiXj] − E(X)2=20Xi=120Xj=1E(XiXj) − E(X)2=20Xi=120Xj6=iE(Xi)E(Xj) +20Xi=1E(X2i) − E(X)2= n(n − 1)p2+ np − n2p2= np − np2= np(1 − p)Chris’ section version:(a) X is B(7, 0.15)(b) E(X) = np = 7 ∗ 0.15 = 1.05; V ar(X) = np(1 − p) = 0.8925(c) P (X ≥ 2) = 1 − P (X ≤ 1) = 0.2834; P (X = 1) = 0.3960. P (X ≤3) = 0.98793.(0.1)(1−0.955)(0.4)(1−0.975)+(0.5)(1−0.985)+(0.1)(1−0.955)≈ 0.1784. X is B(5, 0.05). The expected number of test is 1 ∗ P (X = 0) + 6P (X >0) = 1 ∗ (0.7738) + 6 ∗ (1 − 0.7738) =