概率论与数理统计刘建亚习题解答第4章.doc

概率论与数理统计（第二版.刘建亚）习题解答第四章41解：E(X) =10.25+20.4+30.2+40.1+50.05= 2.342解： 由D(X) = E(X 2)-E(X)2 得E(X)E(X2)D(X) X15025011 X25025022 D(X1)0 0x0其中s0为常数，求E(X ), D(X) 。+ x -+- 分部积分 + -x/s=t解:E(X)= 022e 2xs22 dx=- 0xde 22sx20e 2xs22 dxs 2 p s2E(X 2) =0+ x 32e-2xs22 dx分部积分2 s2 D(X)=E(X 2)-E(X)2 = 4-ps2 s2418 解：E(X) = E1n k=n1 Xk= 1n k=n1 E(Xk) = 1n nm= mE(X) = D 1 n Xk = 12 Dn Xk = 12n D(Xk) = 12 n s2 = s2 n k=1nk=1n k=1nn X N(m, s2 )。n419 解：设进货量为 a，则利润为 500500aX+-300(100(Xa-Xa),10a XX30 aY(X) = 300600XX -+100200aa,),10a XX30 a期望利润为E(Y(X) = 1030 1 Y(x)dx= 1 10a(600x-100a)dx+ a30 (300x+200a)dx2020=-7.5a2 +350a+5250依题意有-7.5a2 +350a+52509280 -7.5a2 +350a4030020a26 故得利润期望不少于 9280 元的最少进货量为 21 单位。420 略。421 略422 解：（1） cov(X,Y) = EX -E(X)Y -E(Y)= EXY -XE(Y)-YE(X)+E(X)E(Y)= E(XY)-EXE(Y)-EYE(X)+EE(X)E(Y)= E(XY)-E(X)E(Y)（2） D(X Y) = E(X Y)-E(X Y)2= EX -E(X)Y -E(Y)2= EX -E(X)2 2X -E(X)Y -E(Y)+Y -E(Y)2= D(X)+D(Y)2cov(X,Y)423 解：由题意知 0 20x其它1, 0y1-xf (x, y) =+11-x1E(X) = - xf (x, y)dxdy =002xdxdy =0 2x(1-x)dx =2 + 211x212211E(X ) = - x f (x, y)dxdy =002x dxdy =0 2x (1-x)dx = 3- 2 = 6同理，E(Y) = 1,E(Y 2) = 13 6+11-x12 E(XY) = - xyf (x, y)dxdy =002xydxdy =0 x(1-x) dx =cov(X,Y) =E(XY)-E(X)E(Y) =D(X) =D(Y) = 1 -1 = 1 ，rXY = cov(X,Y) = 26918D(X)D(Y)11424 解：+1221227E(X) = - xf (x, y)dxdy = 800 x(x+ y)dxdy = 40 (x +x)dx = 62 + 222212325E(X ) = - - x f (x, y)dxdy = 0 0 x (x+ y)dxdy = 4 0 (x +x )dx = 3 D(X) = E(X 2)-E(X)2 = 同理 E(Y) = 1,D(Y) = 113 36+ 12212244E(XY) = - xyf (x, y)dxdy = 800 xy(x+ y)dxdy = 40 (x + 3 x)dx = 3cov(X,Y) = E(XY)-E(X)E(Y) =- 1 ，rXY =cov(X,Y)= 136D(X)D(Y)11425略。426略。427证明：E(Y) = E(aX +b) = aE(X)+bD(Y) = D(aX +b) = E(aX +b)-E(aX +b)2 = a2EX -E(X)2 = a2D(X) cov(X,Y) = EX -E(X)Y -E(Y)= EX -E(X)aX -E(X)= aEX -E(X)2 = aD(X) rXY = cov(X,Y) = aD(X) =1 D(X)D(Y) aD(X)428 解：由已知得：E(X1) = E(X2) =m, D(X1) = D(X2) =s2 ，则E(Y1) = E(aX1 +bX2) = aE(X1)+bE(X2) = (a+b)m， E(Y2) = (a-b)mD(Y1) = D(aX1 +bX2) = a2D(X1)+b2D(X2) = (a2 +b2)s2， E(Y2) = (a2 +b2)s2E(YY1 2) = E(aX1 +bX2)(aX1 -bX2)= E(a2X12 -b2X22) = a2E(X12)-b2E(X22)= a2(m2 +s2)-b2(m2 +s2) = (a2 -b2)(m2 +s2) cov(Y1,Y2) = E(YY1 2)-E(Y1)E(Y2) = (a2 -b2)(m2 +s2)-(a2 -b2)m2 = (a2 -b2)s2r YY1 2 =cov(Y1,Y2)= a22 -b22 D(Y1)D(Y2)a +b429 解：由题意知 E(X) = 7300,D(X) = 7002 x = 9400 X -E(X) = 2100,x = 5200 X -E(X) =-2100 5200 x9400X -E(X) 2100, e= 2100由切比雪夫不等式（4.2.6）得 P( X -E(X)3) = P X -500.05 3-500.05 = P X -2.5 0.5= P X -2.5 0.3160.05500.05502.52.5 2.5 =1-PX -2.5 2.5 0.316= 0.3745431 解：设某投保人出现意外为 Xk, (k =1, 2,L,10000), Xk B(1, 0.006)， m= E(Xk) = 0.006, s2 = D(Xk) = 0.006(1-0.006) = 0.005964，设遇到意外总人数为 X = 10000 Xk ，k=1保险公司亏损的条件为：2500x180000，解得x72， n=10000足够大，由中心极