《数理统计》课后题答案（西交大版）

数理统计习题答案

第一章

X?1n92?94?103?105?106n?xi??100i?152S2?1n??215n?xi?x?5??xi?100?i?1i?11.解： ?15???92?100?2??94?100?2??103?100?2??105?100?2??106?100?2?

? ?34解：子样平均数 X?1?l2. m*nixi

i?11 ?60?1?8?3?40?6?10?26?2?

?4 S2?1l子样方差2n?mix*i?x

i?1???1?8??1?4?2?4?0??3?24??1?0?6?2?4??2??2?

60??624?18.67

子样标准差

S?S2?4.32 3. 解：因为

yi?xi?ac 所以

x2i?a?cyi s21n x?n?xi?x

i?1??

x?1n?nxi

i?1?1n?na?cyi?a?cyi?i??2?1?n?a?cyi? ?1n ni?1

n??cyi?cy?1?n?i?1?2n??na??cyii?1???c2n?nyi?yi?1??2

?a?cnn?yii?1 2?a?cy为 s1n因y?n??yi?y?2 i?1所以 x?a?cy 成立

s222x?csy 成立 1

所以

Me?X?n?1??X?7??0?2???

R?X?n??X?1??3.21???4??7.21Me?X?n??2?1????X?8??1.24. 解：变换

yi?xi?2000

2 1697 -303 3 3030 1030 4 2424 424 5 2020 20 6 2909 909 7 1815 -185 8 2020 20 9 2310 310 i 1 1939 -61 xi yi 11n???61?303?103?042?4?209?09?185?? 20 310 y??yi9ni?1?240.4441n2sy??yi?y

ni?1??22122???61?240.444??303?240.444?1030?240.444????????9222 ?424?240.444???20?240.444???909?240.444??

??185?240.444?2??20?240.444?2??310?240.444?2???197032.247利用3题的结果可知

x?2000?y?2240.44422sx?sy?197032.247

5. 解：变换

yi?100?xi?80?

3 4 5 6 7 8 9 10 11 12 13 80.02 2 i 1 2 xi yi 79.98 80.04 80.02 80.04 80.03 80.03 80.04 79.97 80.05 80.03 80.02 80.00 -2 4 2 4 3 3 4 -3 5 3 2 0 1n113 y??yi??yi

ni?113i?11??2?4?2?4?3?3?4?3?5?3?2?0?2?

13?2.00?1n2 sy??yi?y

ni?1??2 2

21222?2?2.00?3?2?2.00?5?2.00?3?4?2.00?????????13?22 ?3??3?2.00????3?2.00??

??5.3077?利用3题的结果可知

y?80?80.02 1002sy2sx??5.3077?10?410000x?6. 解：变换yi?10?xi?27?

23.5 -35 2 26.1 -9 3 28.2 12 4 30.4 34 1 xi* yi mi 1l y??miyi

ni?1

1??35?2?9?3?12?4?34? 10??1.5?x?2y

y?27=26.85 1021l s??miyi?y

ni?1??21222???2??35?1.5?3??9?1.5?4?12?1.5?34?1.5????????? 10??440.25122sx?s?4.4025

100y7解： 身高 组中值 学生数 154?158 158?162 162?166 166?170 170?174 174?178 178?182 156 10 160 14 164 26 168 28 172 12 176 8 180 2 1lx??mixi*

ni?11?156?10?160?14?164?26?172?12?168?28?176?8?180?2?

100?166?

2

21l* s??mixi?x

ni?12??2122210?156?166?14?160?166?26?164?166?28?168?166?????????100?222 ?12??172?166??8??176?166??2??180?166????33.448解：将子样值重新排列（由小到大）

-4，-2.1，-2.1，-0.1，-0.1，0，0，1.2，1.2，2.01，2.22，3.2，3.21

?Me?X?n?1??X?7??0?2???

R?X?n??X?1??3.21???4??7.21 Me?X?n??2?1????X?8??1.21n11n2n1?xi?n2?xjn1i?1n2j?1nx?nx9解： x??1122

n1?n2n1?n21n1?n2xi?x s??n1?n2i?12??21?n1?n2n1in1?n2i?1?xn2jj?12i?x2??x??xi?12n1s1?x1??n1?n22??n?s22?nx?n2x2???11??n1?n2?222?x22n1?n2???nx?nx???11222?n1?n2?222?nx?n2x2?n1s1?n2s2nx?n2x2??11??11?n1?n2n1?n2?n1?n2?2?ns?ns?n1?n2211222n1?n1?n2?x1?n2?n1?n2?x2?n1x1?n2x222??2?n1?n2?22222n1s1?n2s2nnx?n1n2x2?2n1n2x1x2??121n1?n2?n1?n2?2

ns?ns??n1?n29 3 211222n1n2x1?x2??2

?n1?n2?28 0 7 9 6 4 5 0 4 2 10.某射手进行20次独立、重复的射手，击中靶子的环数如下表所示： 环数 频数 10 2 试写出子样的频数分布，再写出经验分布函数并作出其图形。 解：

环数 频数 频率

10 2 0.1 9 3 0.15 8 0 0 7 9 0.45 3

6 4 0.2 5 0 0 4 2 0.1

?0 x?4?0.1 4?x?6??0.3 6?x?7 F20?x???

0.75 7?x?9??0.9 9?x?10??1 x?1011.解： 区间划分 154?158 158?162 162?166 166?170 170?174 174?178 178?182 频数 10 14 26 28 12 8 2 频率 0.1 0.14 0.26 0.28 0.12 0.08 0.02 密度估计值 0.025 0.035 0.065 0.07 0.03 0.02 0.005

12. 解：

200150身高100500123456学生数789xi?P??? Exi?? Dxi?? i?1,2???,n, 1n1nn?EX?E?xi??Exi???ni?1ni?1n 1n1nn??DX?D?xi?2?Dxi?2?ni?1ni?1nn?b?a? i?1,2,???,n a?b13.解：xi?U?a,b? Exi? Dxi?2122 在此题中

xi?U??1,1? Exi?0 Dxi?1 i?1,2,???,n 3 4

1n1nEX?E?xi??Exi?0ni?1ni?1

1n1n1DX?D?xi?2?Dxi?ni?1ni?13n14.解：因为

Xi?N??,?2? EXi????0 DXi????1

所以

Xi????N?0,1? i?1,2,???,n

由?2分布定义可知

Y?1n2n?X???2?2??Xi????i?1?i?1?i????服从?2分布 所以

Y??2?n?

15. 解：因为

Xi?N?0,1? i?1,2,???,n X1?X2?X3?

EX1?X2?X3?0 DX1?X2?X333?1

所以

X1?X2?X33?N?0,1?

??X1?X22?X3?2?3?????1?

?X?X2同理 ?45?X6?2?3?????1? 由于?2分布的可加性，故

1?X?X22Y??12?X3??X4?X5?X6?23?3?????3?????2? 可知

C?13

16. 解：（1）因为 X2i?N?0,??

i?1,2,???,n

Xi??N?0,1?

n 所以 ??X?2Yi?1?i?????1?2??2?n?

F?Yy?Y1?y??P?Y1?y??P?1??2??2??

5

N?0,3?

y?2

? ?f??x?dx20?y?1fY1?y??FY'1?y??f?2?2??2

?????1?nx2??x2?n?n?e x?0因为 f?2?x???22?

?2?????0 x?0?n?1?y2?2y?2?e y?0n?n所以 fY1?y???22????n ?2?????0 y?0?（2） 因为 Xi?N0,?2

??i?1,2,???,n

Xi?n?N?0,1?

2?Xi?nY22????n? ???2所以 i?1????ny?nYny?FY2?y??P?Y2?y??P?22?2??????

?20?f??x?dx

2?ny?nfY2?y??FY'2?y??f?2?2?2

????n?1?nny22?2ny?2?e y?0 n?nn故 fY2?y????22??? ?2?????0 y?0?2（3）因为 Xi?N0,?

??i?1,2,???,n

?i?1nXi?N?0,1? n? 6

?nXi?Y32???所以 ???1? ?n??i?1n??y2?Y3?n?FY3?y??P?Y3?y??P?2?y???f?2?1??x?dx

?n??0?y?1fY3?y??FY'3?y??f?2?1??2?2

?n??n?x?1?2e x?0?f?2?1??x???2?x

?0 x?0?y??12e2n? y?0?故 fY3?y????2?ny

?0 y?0?2（4）因为 Xi?N0,?2

??i?1,2,???,n

? 所以

i?1nnXi?N?0,1?n?2Xi?Y4?2?????2???1??i?1n??

yy???Y4FY4?y??P?Y4?y??P?2?2???f?2?1??x?dx????0 ?y?1f?2?y??FY'4?y??f?2?1??2?2????y?2?1e2? y?0?故 fY4?y???2?y?

?0 y?0?217.解：因为 存在相互独立的U，V

X?t?n?

U?N?0,1? V??2?n?

使

X?U Vn

U2??2?1?

7

U2则 X2?1

Vn由定义可知

?2?F?1,n?

18解：因为 Xi?N0,?2 ??i?1,2,???,n

n

?Xii?1n??N?0,1? n??m ?X?2?1?i??????2?m?

i?nmnnXi所以

Yn1??Xii?1?i?1??t?m?nn??m?X2n?m2ii?n?1i???n?1?Xi?????m（2）因为

Xi??N?0,1? i?1,2,???,n?m

?n?2?Xi?2

?????n?i?1???n?m?2

i?n?1?Xi???2?????m?n?X2m?nX2?ii?1?i?????所以 Y2?i?1?nnn??mX2n?m2?F?iXn,m? i1i???i?n??n?1?????m19.解：用公式计算 ?20.01?90??90?2?90U0.01

查表得

U0.01?2.33

代入上式计算可得 ?20.01?90??90?31.26?121.26

20.解：因为

X??2?n? E?2?n 由?2分布的性质3可知

8

D?2?2n

X?n?N?0,1? 2n

?X?nc?n?P?X?c??P???

2n??2nc?n2n???X?nc?n?P????lim2n?n???2n故

?1?t2?c?n?edt???? 2n?2n?2?c?n?P?X?c?????

?2n?

第 二 章

1.

??e??x,x?0f(x)???0,x?0E(x)????????0f(x)?xdx??1???0?xe??xdx??xe??x??令1????0e??xd(?x)?e??x??0?1?1??x

从而有 2.

???1x

1).E(x)??k(1?p)x?1?k?1p?p?k(1?p)k?1x?1??p1??1??1?p???2?1p

1令

p＝X

p??所以有

1X

9

２）．其似然函数为

L(P)??(1?P)p?p(1?p)xi?1`nnXi?n?i?1ni?1

lnL(P)?nlnp?(?Xi?n)ln(1?p)i?1n

ndlnLn1??(?Xi?n)?0dpp1?pi?1

p??n解之得

?Xi?1n?i1X

３． 解：因为总体Ｘ服从Ｕ（a，b）所以

2a?b（a-b）n! D（X）=212r!?n?r?!2令E（X）=X D（X）=S，1nS2=?(Xi?X)2 ni?1a+b??X? ?2?2（a?b）??S2??12??a??X?3S ????b?X?3S E（X）?4. 解：（1）设1nni?1x,x2,?xn为样本观察值则似然函数为：

ni?iL(?)??(?xi)??1,0?xi?1,i?1,2,?,nlnL(?)?nln??（?-1）?lnxindlnLn??lnxi?0d???i?1

???????n?lnxi?1nnin解之得：

?lnxi?1i

（2）母体X的期望

E(x)??????xf(x)dx???xdx??01???1

而样本均值为：

10

1nX??xini?1令E(x)?X得?X??1?X 5.。解：其似然函数为：

??xi1??1?i?1L(?)???e??en2?(2?)i?1令1nlnL(?)??nln(2?)??xi?0nxi1n?i?1得：???x??i?11ni

xx??x（2）由于

E???????xe2??x?dx?2???0xe2???dx??xe??0????0e??dx??1n1n1E(?)?E(?xi)?E(x)??n????ini?1ni?1n

1n???xi所以ni?1 为?的无偏估计量。

?6. 解：其似然函数为：

kn?k(k?1)n???xni?(L(?)??xie)?xi(k?1)e??xi(k?1)!i?1i?1(k?1)!

L??nkdlnL(?)nk??d??n??k?nn?Xi???Xii?1i?1

?Xi?1i?0解得

??

?nk?Xi?1nk?Xi1f(x)?,0?x??,?

11

７．解：由题意知：均匀分布的母体平均数????0?2?2，

方差?2?(??0)2?212?12 用极大似然估计法求?得极大似然估计量 似然函数：L(?)??n1 0?minxi?maxxi?1?ni)1?i?ni??

(选取?使L达到最大

取???maxxi

1?i?n由以上结论当抽得容量为6的子样数值1.3，0.6，1.7，2.2，0.3，1.1，时???2?2.2即????2.2?2.22?1.1, ?2???12?12?0.4033 8. 解：取子样值为(x1,x2,?xn),(xi??)

则似然函数为： L(?)??ne?(xi??) xi??

i?1lnL(?)???n(xi??)???nxi?n?

i?1i?1要使似然函数最大，则需?取min(x1,x2,?,xn)

即?=min(x1,x2,?xn)

9. 解：取子样值(x1,x2,?,xn)(xi?0) n则其似然函数L(?)??n?e??x??ne??i?xii?1

i?1lnL(?)?nln????nxlnL(?)nn?n1i ??i?1d???xi ??? i?1?nxxii?1由题中数据可知

x?11000(365?5?245?15?150?25?100?35?70?45?45?55?25?65)?20 则 ???120?0.05

10. 解：（1）由题中子样值及题意知：

12

?1极差R?6.2?1.5?4.7 查表2-1得?0.4299 故??0.4299?4.7?2.0205

d5?1（2）平均极差R?0.115，查表知?0.3249 ??0.3249?0.115?0.0455

d10解：设u为其母体平均数的无偏估计，则应有??x 又因x????1(8?1?40?3?10?6?2?26)?4 60即知??4

12. 解：?X~N(?,1)

12?E(xi)?? ，D(xi)?1， (i?1,2) 则E(?1)?EX1?EX2??

33?13E(?2)?EX1?EX2??

44?11E(?3)?EX1?EX2??

22?所以三个估计量?1,?2,?3均为?的无偏估计

???2141415D(?)?D(X1?X2)?DX1?DX2???

3399999??51同理可得D(?2)?，D(?2)?

82?可知?3的方差最小也亦?2最有效。 13解：?X~P(?)?E(X)??,D(X)??

21n1n22E(S)?E[(X?X)]?[E(X)?nE(X)] ??iin?1i?1n?1i?1*2??11n?(n???)?? ?[?(???2)?n(??2)]?n?1n?1i?1n即S*是?的无偏估计

n1n11n又因为E(X)?E(?Xi)?E(?Xi)??EXi??

ni?1ni?1ni?12即X也是?的无偏估计。

又???[0,1] E(aX?(1??)S*)??E(X)?(1??)E(S*)????(1??)???

13

22

因此?X?(1??)S*也是?的无偏估计 14.解：由题意：X~N(?,?2)

因为E(?)?C?E(Xi?1?Xi)2?C?[D(Xi?1?Xi)?(E(Xi?1?Xi)2]

2i?1?n?12?C?[D(Xi?1)?D(Xi)?0]?C?2?2?2C(n?1)?2

i?1i?1?211要使E(?)??只需C? 所以当C?时?为?2的无偏估计。

2(n?1)2(n?1)n?1n?1?2215.证明：?参数?的无偏估计量为?，D?依赖于子样容量n 则???0,由切比雪夫不等式

????limD??0故有limp????????1 n??n???????即证?为?的相合估计量。

k16证明：设X服从B(N,p)，则分布律为 P(X?k)?CNPk(1?P)k (k?1,2,?N)

? 这时E(X)?NP D(X)?NP(1?P) EX2?DX?(EX)2?NP(1?P)?N2P2 例4中p??EXNPX??P（无偏） 所以E(P)?NNN???? DP?DXNP(1?P)P(1?P)?? 22NnNNn? 罗—克拉美下界满足

n1?KK?n?[LnCNPK(1?P)N?P]2CNPK(1?P)N?P IRk?0?pN ?n?[K?0?KK(LnCN?KLnP?(N?P)Ln(1?P))]2CNPK(1?P)N?K ?PEX22NEX?2EX2N2?2NEX?EX2KN?P2KKN?K ?n?[? ?n[2??] ]CNP(1?P)2P(1?P)1?PP(1?P)K?0PN

14

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