STAT 418HW8 SolutionsSections 5.1-5.42, 3, 5, 6, 8, 10, 13, 15, 16, TE1, TE72. First, we must find C. By the substitution rule with y = −x/2; dy = −dx/2,Z∞0xe−x/2dx = −Z0−∞4yeydy.By integration by parts, with u = y; du = dy and dv = eydy; v = ey,−Z0−∞4yeydy = −4(yey|0−∞−Z0−∞eydy) = limx→∞[−2xe−x/2− 4e−x/2] + 4.By L’Hospital’s Rule,limx→∞− 2xe−x/2= limx→∞−2xex/2= limx→∞−2.5ex/2= 0.Hence,CZ∞0xe−x/2dx = 1 ⇒ C = 1/4P (X > 5) =14Z∞5xe−x/2dx=14[10e−5/2+ 4e−5/2]=144e−5/2= .2873.3. No. In both cases f52< 0.5. Must choose c such that 0.01 =R1c5(1 − x)4dx = (1 − c)5, therefore c = 1 − (.01)1/5.6.a. E(X) =14R∞0x2e−x/2dx. Let u = x/2; du = dx/2, then E(X) = 2R∞0u2e−udu =2Γ(3) = 2(3 − 1)! = 4b. By symmetry of f(x) about x = 0, E(X) = 0.c. E(X) =R∞55xdx = ∞8.EX =Z∞0x2e−xdx = Γ(3) = 2.10.a.P (X → a) = P (5 < X < 15 ∪ 20 < X < 30 ∪ 35 < X < 45 ∪ 50 < X < 60)=23, because X is U (0, 60)1b. Same as (a).13. X ∼ U(0, 30).P (X > 10) =23P (X > 25|X > 15) =P (X > 25)P (X > 15)=5/3015/30=13.15.a. P (X > 5) = PX−106>5−106= P (Z > −0.8333) = Φ(0.8333) = 0.7977b. P (4 < X < 16) = P4−106<X−106<16−106= Φ(1) − Φ(−1) = 0.6827c. P (X < 8) = PX−106<8−106= P (Z < −0.333) = 1 − Φ(0.333) = 0.3695d. P (X < 20) = PX−106<20−106= P (Z < 1.6667) = Φ(1.6667) = 0.9522e. P (X > 16) = PX−106>16−106= P (Z > 1) = 1 − Φ(1) = 0.158716. Let X = {annual rainfall}.P (X < 50) = P (X − 404<50 − 404) = Φ(2.5) = .9938.Assuming this mean and standard deviation for the next ten years and that each year’srainfall, denoted Xi, i = 1, . . . , 10, is independent of the others,P (∩10i=1[Xi< 50]) =10Yi=1P (Xi< 50) = .993810= .9397.TE 1. By integration by parts: u = −x/2b; du = −dx/2b and dv = −2bxe−bx2; v = e−bx2,Z∞0x2e−bx2=−xe−bx22b|∞0+12bZ∞0e−bx2dx=1(2b)3/2Z∞0e−y2/2dy=1(2b)3/2√2π2, because1√2πZ∞0e−y2/2dy =12=√π4b3/2,where the second equality follows by L’Hospital’s Rule and the substitution rule with y =√2bx. Therefore a =4b3/2√π.TE 7. SD(aX + b) =pV ar(aX + b) =√a2σ2=
View Full Document
Unlocking...