西安交大数理统计作业

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1、第一章1.1X〜N(Pq2)22则—),所以片一“〜N(O,UnnP{X-//<1}=P{X-/z

2、V«ovl}=0.95X-jLly[n(J〜N(0,1),而①(0.975)=1.96所以n最小要®J[1.962xct2]+l1.2(1)至800小时，没有一个元件失效这个事件等价于?{%,%2%3%4^5%6>800}的概率由己知X服从指数分布，可求得P{>800}=e~72(2)至3000小时，所有六个元件都失效的概率等价与P{X1X2X3X4X5X6<3000}的概率可求得P{X1X2X3X4X5X6<3000}=(1一严)61.5£(尤一°)2=

3、£[(尤一文)+(戈-q)F/=!/=!=Z-^)2+2(J-X)+£(片一°)2/=!/=!/=!因为-来)=0J=1所以£亿-°)2=£亿-劝2+£(戈_"/=!/=!/=!=nS2+》(X-a)?»=1所以当a=X时，^(Xi-a)2有最小值且等于刃S?/=!_1n1.6（1）^x=-yxs“Z=1有等式的左边=£X；-2“£X,+n/Ll2/=!/=!等式的右边=£X;-2片乞/+nX2+nX2—2nX/j+n）r/=!Z=1=£X；_2nX2+nX2+nX2-2立Xp+np_1_1X1_1_百纠H齐(心-初;=1/=!=£疋-2“£尤+必2/=

4、1/=1左边等于右边，结论得证。(2)等式的左边=-IX^X^nX2=£x；-nX2/=!/=1/=!等式的右边=£/；-川片2左边等于右边,/=1结论得证。_1刃17⑴由i”及■驴刃_1有左边=^+1n+11n1ny讪y产荷（驴+x亠市y+7+]n+1/=!〃+1nXX-1_二—土+」■二云+―丁(£+厂孟戸右边7?+1n+〃+1左边等于右边，结论得证。(2)由左边=S：+i1n+l]〃+i_2__1_—1_1_市卩-和+荷（粘-孝-科（兀Yr]-flSn.n（yyx2「市+丙严"）二七£+47（尤+厂KJ”右边77+1刃+1左边等于右边，结论得

5、证。1.9因为yi-axi+b1n1n1n所以p=—=—V（ax.b）=—axib=axbn/=1n匸in匸iifj]〃iS七0厂刃2=詔卫⑷=再（叶词再令必=79.98,……，儿=79.96再令a=l,b=80由yi-ax：+b=兀+80得：兀为:-0.02,0.04,0.02,0.04,0.03,0.03,0.04,-0.03,0.05,0.03,0.02,0.00,0.02,-0.04]147=—=0.016414台z_1”1.9由X=~Y^i—1n1"故施)吧驴)宁(pH)1n1nE(少)=一£[£(«-劝2]=—可£(疋_2£?+若)ni=

6、ni=(1)二项分布E(X)=mpD(X)=mp(-p)E(X)—E(X)=mpD(心丄D(X)=徑SnnE&)=心D(X)=口mp([-p)nn(2)泊松分布E(X)=AD(X)=AE(X)=E(X)=A一12D(X)=-D(X)=-nn£(S2)=^!-Z)(X)=^—!-ann(3)均匀分布E(X)=a+b"T-D(X)=(i)212E(来)=E(X)=仝D(X)=-D(X)=^^-n2nE(S2)=(1)指数分布E(X)=*D(X)=-1E(X)=E(X)=-A一11D(X)=-D(X)=—nnAre(sj4d(x)TnnA~(2)正态分布

7、E(X)=jUD(X)=o-E(X)=E(X)=“D(X)=-D(X)=—nnE^)=^D(X)=^-a2nn1.9统计量有：(1),(3),(4),(5),(6),(7)顺序统计量有：(5)1.12顺序统计量为：-4,・2.1,・2.1,-0.1,・0.1,0,0,1.2,1.2,2.01,2.22,3.2,3.21所以阻二叫空广禺=0厂=尤3-尤=3.21-(-4)=7.21添加2.7后：顺序统计量为：・4,・2.1,・2.1,-0.1,・0.1,0,0,1.2,1.2,2.01,2.22,2.7,3.2,3.21所以叫二丄(/+血)=0.621.1

8、6因为X服从正态分布故Z=口N(0,l)故由定理1.2.1知：Y=tz"=Z(^J£)2=-tE(^-a)2dz2wj=1/=!/=11.20Y已知X-/(/?),即有Y〜N(0,l),Z〜力2(刃)使得X=-^=yjz/n1.22Y2尸〜力⑴所以屮〜F(l’)结论得证。已知X〜N(2.5,36),—心),乔(y_〃)〜幷(0,1)(J(1)^{30

9、{玉竺sh}P{—0.2忌厉吐2.5)<盘6cr966=0.19294*0.63