STAT 418HW13 SolutionsSections 6.6-6.734, 37, 42, 43; TE 1134. Let X be the number of men among the 200 who never eat breakfast. Then X ∼B(200, .252). Let Y be the number of women among the 200 who never eat breakfast. ThenY ∼ B(200, .236).a. Approximate P (X + Y ≥ 110). By the normal approximation to the Binomial, X ∼N(200·.252, 200 ·.252 ·(1 −.252)) and Y ∼ N(200·.236, 200 ·.236 ·(1 −.236)). ThereforeX + Y ∼ N(97.6, 73.8).P (X + Y ≥ 110) = P (X + Y ≥ 109.5)= PX + Y −97.6√73.8≥109.5 − 97.6√73.8≈ P (Z ≥ 1.39) = .0823.b. By the normal approximation to the Binomial, X ∼ N (50.4, 37.7) and Y ∼ N (47.2, 36.1).Therefore Y −X ∼ N(−3.2, 73.8).P (Y ≥ X) = P (Y −X ≥ −0.5)= PY −X + 3.2√73.8≥−0.5 + 3.2√73.8≈ P (Z ≥ 0.31) = .3783.37.a.P (Y1= 1|Y2= 1) =11111122=212= 1 − P (Y1= 0|Y2= 1).b.P (Y1= 1|Y2= 0) =11112123=312= 1 − P (Y1= 0|Y2= 0).142.a.fX|Y =y(x) =f(x, y)fY(y)fY(y) =Z∞0xe−x(y+1)dx = (y + 1)−2, y > 0fX|Y =y(x) =xe−x(y+1)(y + 1)−2, x > 0.fY |X=x(y) =f(x, y)fX(x)fX(x) =Z∞0xe−x(y+1)dy = e−x, x > 0fY |X=x(y) =xe−x(y+1)e−x= xe−xy, y > 0.b. Let Z = XY and V = X. Then X = V and Y = ZV−1. The Jacobian of thistransformation isabs0 1v−1∗=1v.Therefore, the joint p.d.f. of (Z, V ) isfZ,V(z, v) = ve−v(zv−1+1)v−1, v > 0, zv−1> 0= e−(z+v), v > 0, z > 0= e−ze−v, v > 0, z > 0.43.fY |X=x(y) =f(x, y)fX(x)fX(x) =Zx−xce−x(x2− y2)dy = ce−xx2y −y33|x−x= ce−x43x3, x > 0fY |X=x(y) =ce−x(x2− y2)ce−x43x3=34x3(x2− y2), −x ≤ y ≤ x.FY |X=x(y) =34x3Zy−x(x2− y2)dy=34x3x2y −y33|y−x=34x3x2y −y33+23x3, −x ≤ y ≤ x.2TE11.a. Since X1, . . . , X5are independent and all have the same p.d.f. f, then their joint p.d.fisf(x1)f(x2)f(x3)f(x4)f(x5).ThereforeP (X1< X2< X3< X4< X5) =Z···Z−∞<x1<···<x5<∞5Yi=1f(xi)dxi.Substitute u1= F (x1), u2= F (x2), . . . , u5= F (x5). Then du1= f(x1)dx1, . . . , du5=f(x5)dx5. Therefore,P (X1< ··· < X5) =Z···Z0<u1<···<u5<1du1du2···du5b.I = P (X1< ··· < X5) =Z10Zu50Zu40Zu30Zu20du1du2···du5=15!c. Because the Xiall have the same p.d.f. and are independent, thenP (X1< X2< X3< X4< X5) = P (X2< X3< X4< X5< X1).It does not matter how we pe rmute the Xi’s. Because there are 5! such permutations,and because all 5! such probabilities sum to 1, thenP (X1< X2< X3< X4< X5)
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