应用数理统计,施雨,课后答案.doc

该【应用数理统计,施雨,课后答案 】是由【泰山小桥流水】上传分享，beplayapp体育下载一共【48】页，该beplayapp体育下载可以免费在线阅读，需要了解更多关于【应用数理统计,施雨,课后答案 】的内容，可以使用beplayapp体育下载的站内搜索功能，选择自己适合的beplayapp体育下载，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此beplayapp体育下载到您的设备，方便您编辑和打印。::x upnnx un~N0,1这可通过查N(0,1)分布表,():(1)至800小时,没有一个元件失效,则说明所有元件的寿命>800小时。|,(2)至300小时,所有元件失效,则说明所有元件的寿命<|,:(1){(x1,x2,x3)|xk0,1,2,L,k1,2,3}因为Xi~P(),所以P{X1x1,X2x2,X3x3}P{X1x1}P{X2x2}P{X3x3x1x2x3e3}x1!x2!x3!其中,xk0,1,2,L,k1,2,3(2){(x1,x2,x3)|xk0;k1,2,3}ex,x 0因为Xi~Exp(),其概率密度为 f(x)0, x 0所以,f(x1,x2,x3)3e(x1,x2,x3),其中xk0;k1,2,3(3){(x1,x2,x3)|axkb;k1,2,3}1xb因为Xi~U(a,b),其概率密度为f(x)b,aa0,xa|xb所以,f(x1,x2,x3)1,其中axkb;k1,2,3(ba)3(4){(x1,x2,x3)|xk;k1,2,3}1(x因为Xi~N(,1),其概率密度为f(x)e2,(x)2131(xk所以,f(x1,x2,x3)32e2k1,其中xk;k1,2,3)(2解:由题意可得: f(xi)则f(xi,...xn)lnxu21i22,0xiexi20,其它1nu)2e22(lnxinni1,0xi,i1,...f(xi)=(2)2nni1ixi10,其它n(Xia)2证:令F(a)i1则F'(a)na),F''(a)2n02(Xii1令F'(a)n2(Xia)0,则可解得a1nXiXi1ni1由于这是唯一解,又因为F''(a)2n0,1n因此,当aXiX时,F(a)取得最小值ni1nn证:(1)等式左边(Xi(XiXXi1i1nnnn(XiX2(X)(XiX)(X(XiXn(X)2i1i1i1i1左边=右边,(2) 等式左边 (Xi 1nXi2 2nX2 nXi1nniXXi22XXinX2i1i1n2Xi2nX2i1左边=右边,:(1)1nxnxini11n1xn1xin1i1_1_那么xn(xn1xn)n1=1n111ni1xin1xn1n1?nnxii11n11n_xixi=xn1=xn1=n1i1n1n1i1原命题得证21n22(2)xnsnnixi11n12sn211ixi2xn1n1那么nsn21(xn1xn)2n1n1=1nnxn2+n2nn2xi2-(n2xn21-2xn1xn+2xnn1i1n11)(n1)(n1)=1nxi2-n22xn2+1xn21-12xn21-2n2xn1xnn1i1(n1)n1(n1)(n1)=1nxi2-(1xn1+nxn)21i11nnn1由(1)可得:1xn1+nxn=xn1n1n11n2则上式=1ixi2-xn1=sn21n1原命题得证解:因为X1nXi,S21n(XiX)21nXi2X2ni1ni1ni1所以(1)二项分布B(m,p)E(X)E(1nXi)E(Xi)mpni1D(X)D(1nXi)1nXi)mp(1p)nin2D(n1i121nX)2)1n2)E(X2)n1E(S)E(ni(XinE(Ximp(1p)1i1n(2)泊松分布P()E(X),D(X),E(S2)n1nn(3)均匀分布U(a,b)E(X)ba,D(X)ba)2,E(S2)n1(ba)2212n12n(4)指数分布Exp()E(X)1,D(X)1,E(S2)n1nn(5)正态分布N(,2)E(X)1,D(X)n2,E(S2)n12n解:(1)是统计量不是统计量,因为u未知统计量统计量统计量,顺序统计量统计量统计量不是统计量,:因为Xi独立同分布,并且Xi~(a,1n,XXini1n所以Xi~(na,;i1n1令YXi,则XY,由求解随机变量函数的概率密度公式可得i1nnafX(x)(nx)na1enxn,x0na):(1)()的概率密度为:xmf(m)(x)n!F(x)m11F(x)nmf(x)(m1)!(nm)!又F(x)=x2且f(x)=2x,011可得该概率p1==个样品的均值大于9分钟,即x9可得该概率为 p2==100 个样品的均值大于分钟 即x P3==综上所述,第一种情况更有可能发生。:=2=36n=5(1)30s244s2n2555?2(,)69而s22n~2(n1)即5s22(4)36通过查表可得 P=样本方差落在30~40的概率为样品均值x落在~的概率即:P{T=x~t(n)Yn2E(nxn)=0D(nxn)=n2n2n1n则xi服从N(0,1)分布。ni1xi~N(0,2)E(nxn)=0D(xi)=2则xi服从N(0,1)分布nm(xi)2服从 2(m)分布i n11nxini1则服从t(m)分布m(xi)2in1m1nnxiCxin令i1=i1nm(xi)2nm(xi)2in1in1m这样可得 C=,X~2(n1),Y~2(n2)=>F=X/m~F(n1,n2)Y/n2xi~N(0, 2)则 xi~N(0,1)nxi22()nmxi22()这样有()~n)~(mi1in1n(xi)2nmxi)2/m)~F(n,m)可得/((i1in1n2nm2令其=dxi/xii1in1则d=mn证:Xi~N(1,1)2Yi~N(2,2)2则Xi1~N(0,1)1Yi2~N(0,1)2=>n1(Xi21)~2(n1)i11n2Yiu22)~2(n2)i1(2n1Xiu12n2Yiu22)/n1)/()/n2)~F(n1,n2)=>(((i11i12n1n22(Xiu1)22=>i1~F(n1,n2)n2n12(Yi2)21i1****题2解:(1)x~Exp( )