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1、應(yīng)用數(shù)理統(tǒng)計(jì)課后答案第一章概率論基礎(chǔ)3.解：10?62?6（1）P{2?X?10}?P{X?10}?P{X?2}??()??()55?2?(0.8)?1?0.56725?6P{X?5}?1?P{X?5}?1-?()??(0.2)?0.57935(2)P{X?d}?0.5?1?P{X?d}?0.5?P{X?d}?0.5d?6??()?0.5?d?65即d至多為6.5.解：∵X~E(?)∴??e??x，x?0，?1?e??x，x?0，f(x)??F(x)???0，x?0.?0，x?0.y?by?b(1)F(y)?P{Y?y}?P{aX?b
2、?y}?P{X?}(a?0)?F()Yaay?b?y?b??當(dāng)?0，即y?b時(shí)，F(xiàn)(y)?1?ea.?Y?a???y?b當(dāng)?0，即y?b時(shí)，F(xiàn)(y)?0.??aY綜上所述：y?b???y?b，F(xiàn)(y)??1?ea，Y???0，y?b.2F(y)?P{Y?y}?P{X?y}Y?當(dāng)y?0時(shí)，F(xiàn)(y)?0.?Y????y??當(dāng)y?0時(shí)，F(xiàn)(y)?P{?y?x?y}?P{x?y}?P{x??y}?P{x?y}?F(y)?1?eY綜上所述：??1?e??y，y?0，F(xiàn)Y(y)????0，y?0.1/31應(yīng)用數(shù)理統(tǒng)計(jì)課后答案14.證明：?Cov(
3、?,?)?Cov(aX?b,cY?d)?acCov(X,Y)22D(?)?D(aX?b)?aD(X)同理：D(?)?cD(Y)Cov(?,?)acCov(X,Y)????????XYD(?)?D(?)acD(X)?D(Y)15.證明：?X~P(?)i?1，2.ii?X的特征函數(shù)為iit?(t)?e?i(e?1)Xi根據(jù)特征函數(shù)的性質(zhì)（2）得：2it??i?(e?1)?(t)?ei?12?Xii?1根據(jù)特征函數(shù)的性質(zhì)（5）得：X?X~P(???)1212第二章數(shù)理統(tǒng)計(jì)的基本概念8.解：設(shè)X為樣本，x為樣本的觀測(cè)值。由于數(shù)據(jù)已經(jīng)按照從小到
4、大的順序排列，于是經(jīng)驗(yàn)分布函數(shù)為：?0，x?(??，?3)，?1?，x?[?3，-2.6)，?8?1?，x?[?2.6，-1)，?4?3?，x?[-1，0.5)，8??1Fn(x)??，x?[0.5，1)，?2?5，x?[1，2.2)，?8??3x?[2.2，2.5),，?4??7x?[2.5，3)，，?8??1，x?[3，??)2/31應(yīng)用數(shù)理統(tǒng)計(jì)課后答案211.解：由題意知：X~N(0,0.5)i?1,2??10i標(biāo)準(zhǔn)化可知：2X~N(0,1)i?1,2??10i1022則：T?4?Xi~?(10)i?11022?P{?Xi?4}
5、?P{T?16}???(10)?16i?1經(jīng)查附表3，可得??0.10.12.解：由P定理2.4.1推廣可知：54?222E(X)??,D(X)?,E(S)??.n?X~E(?)i1121?E(X)?,D(X)?,E(S)?.22?n??16.解：（1）：z0.9??1.28，z0.975??1.96，z0.995??2.58.z0.999??3.10z0.1?1.28z0.005?2.58.22（2）：?(8)?17.535,?(15)?22.307.0.0250.12212?0.95(18)?9.391,?0.995(100)?(
6、z0.995?199)?66.479.2（3）：t0.9(10)??1.3722,t0,99(6)??3.1427,t0.95(60)?z0.95??0.5199.11（4）：F0.9(18,7)???0.48.F(7，18)2.080.111F0.99(7,10)???0.15.F(10,7)6.620.0111F0.95(10,10)???0.34.F(10,10)2.980.053/31應(yīng)用數(shù)理統(tǒng)計(jì)課后答案第三章參數(shù)估計(jì)1.解：??(??1)x,0?x?1,(1)f(x;?)??其中??1為未知參數(shù).?0,其他.矩估計(jì)：1???
7、1E(X)??x(??1)xdx?0??2???2X?1令E(X)?X解得?的矩估計(jì)為：1?X極大似然估計(jì)：設(shè)X,X,?,X.為抽自總體X的樣本，x,x,?,x.是樣本的一個(gè)觀測(cè)值，則似12n12n然函數(shù)：nn?n?L(?)??(??1)xi?(??1)?xi.(0?xi?1,i?1,2,?n)i?1i?1對(duì)數(shù)似然函數(shù)：nlnL(?)?nln(??1)???lnxi.i?1對(duì)?求導(dǎo)并令其為零，即得對(duì)數(shù)似然方程：ndlnL(?)n???lnxi?0.d???1i?1??n解得?的極大似然估計(jì)值：???1.n?lnxii?1??n則?的極
8、大似然估計(jì)量：???1.n?lnXii?1??x??e,x?0,(3)f(x;?)???0,x?0.矩估計(jì)：??1??xE(X)??x??edx?.0?4/31應(yīng)用數(shù)理統(tǒng)計(jì)課后答案?1令E(X)?X解得?的矩估計(jì)為：??
