习题答案

习题答案

第 一 章

1. (1) $ S = \{ \frac{i}{n} \mid i = 0, 1, \cdots, 100n \} $，其中 n 为小班人数. (2) $ S = \{10, 11, \cdots\} $.

(3) S = {00, 100, 0100, 0101, 0110, 1100, 1010, 1011, 0111, 1101, 1110, 1111}, 其中 0 表示次品，1 表示正品. (4) S = \{(x, y) | x^2 + y^2 < 1\}.

1. (1) $ A\overline{B}\overline{C} $. (2) $ A\overline{B}\overline{C} $. (3) $ A\cup B\cup C $. (4) $ ABC $. (5) $ \overline{A}\overline{B}\overline{C} $.

(6) $ \overline{AB} \cup \overline{AC} \cup \overline{BC} $. (7) $ \overline{A} \cup \overline{B} \cup \overline{C} $. (8) $ AB \cup AC \cup BC $.

1. (1) $ P(A \cup B \cup C) = 5/8 $.

(2) $ P(A \cup B) = 11/15 $, $ P(\overline{A} \overline{B}) = 4/15 $, $ P(A \cup B \cup C) = 17/20 $, $ P(\overline{A} \overline{B} \overline{C}) = 3/20 $, $ P(\overline{A} \overline{B} C) = 7/60 $, $ P(\overline{A} \overline{B} \cup C) = 7/20 $.

(3)(i) $ P(A\bar{B})=1/2 $，(ii) $ P(A\bar{B})=3/8 $.

1. (1) $ \frac{113}{126} $. (2) $ \frac{1}{12} $. 6. (1) $ \frac{1}{12} $ (2) $ \frac{1}{20} $. 7. $ \frac{252}{2431} $.

1. (1) $ \frac{\binom{400}{90} \binom{1100}{110}}{\binom{1500}{200}} $ (2) $ 1 - \frac{\binom{1100}{200} + \binom{400}{1} \binom{1100}{199}}{\binom{1500}{200}} $. 9. $ \frac{13}{21} $. 10. 0.000 002 4.

1. 记 X 为最大个数， $ P\{X=1\}=\frac{6}{16} $， $ P\{X=2\}=\frac{9}{16} $， $ P\{X=3\}=\frac{1}{16} $.

1. $ \frac{1}{1960} $. 13. (1) $ \frac{4}{33} $. (2) $ \frac{10}{33} $. 14. (1) 0.25. (2) $ \frac{1}{3} $. 15. $ \frac{1}{3} $. 16. 0.18.

1. (1) $ \frac{28}{45} $. (2) $ \frac{1}{45} $. (3) $ \frac{16}{45} $. (4) $ \frac{1}{5} $ 18. (1) 0.3. (2) 0.6.

$$ (1)\frac{n}{n+m}\cdot\frac{N+1}{M+N+1}+\frac{m}{n+m}\cdot\frac{N}{M+N+1},(2)\quad53/99. $$

1. 3/5. 21. $ \frac{20}{21} $. 22. (1) $ \frac{3}{2}p - \frac{1}{2}p^{2} $. (2) $ \frac{2p}{p+1} $. 23. $ \frac{196}{197} $. 24. (1) 0.4. (2) 0.4856.

1. $ \frac{9}{13} $. 26. (1) 0.785. (2) 0.372.

1. (1) $ P(AB) = 0.72 $. (2) $ P(A \cup B) = 0.98 $. (3) 0.26.

29.（1）0.57.（2）0.0481.（3）0.0962.（4）0.6864.

31.（1）必然错.（2）必然错.（3）必然错.（4）可能对．32.p=0.5043.

1. (1) $ p_{1}p_{2}p_{3} + p_{1}p_{4} - p_{1}p_{2}p_{3}p_{4} $. (2) $ 2p^{2} + 2p^{3} - 5p^{4} + 2p^{5} $.

1. 0.9984, 3 只开关. 36. 0.6. 37. $ \frac{5}{9} $, $ \frac{16}{63} $, $ \frac{16}{35} $. 38. $ \frac{m}{m+n2^{r}} $

1. 0.8731, 0.1268, 0.0001.

$$ \frac{2\alpha p_{1}}{(3\alpha-1)p_{1}+1-\alpha} $$

第 二 章

1.

1. (1)

(2)

3.

1. (1) $ P\{X=k\}=pq^{k-1}, k=1,2,\cdots $

(2)

$$ P\{Y=k\}=\binom{k-1}{r-1}p^{r}q^{k-r},k=r,r+1,\cdots. $$

(3)

$$ P\{X=k\}=0.45(0.55)^{k-1},k=1,2,\cdots,p=\sum_{k=1}^{\infty}P\{X=2k\}=\frac{11}{31}. $$

1. (1)

$$ \begin{array}{r|ccc}X&1&2&3&\cdots\\\hline p_{k}&\frac{1}{3}&\frac{1}{3}\left(\frac{2}{3}\right)&\frac{1}{3}\left(\frac{2}{3}\right)^{2}&\cdots\end{array} $$

(2)

(3) 8/27, 38/81.

6.（1）0.0729.（2）0.00856.（3）0.99954.（4）0.40951.

7.（1）0.163.（2）0.353.8.（1）0.321.（2）0.243.

1. (1) $ 0.9^{10} \approx 0.349 $. (2) 0.581. (3) 0.590. (4) 0.343. (5) 0.692.

10.（1） $ \frac{1}{70} $（2）猜对的概率仅万分之三，此概率太小，按实际推断原理，认为他确有区分能力.

1. 0.0025. 12.（1）0.0298.（2）0.5665. 13.（1）0.2231.（2）0.9179.

14.（1）0.2388.（2）20.79分.

$$ P\{X\leqslant10\}=\sum_{k=0}^{10}\binom{5000}{k}(0.0015)^{k}(1-0.0015)^{5000-k},\quad P\{X\leqslant10\}\approx0.8622 $$

1. $ P\{X \geqslant 2\} \approx 0.0047 $

$$ F(x)=\{\begin{aligned}&0,&x<0,\\ &1-p,&0\leqslant x<1,\\ &1,&x\geqslant1.\end{aligned}.(2)F(x)=\{\begin{aligned}&0,&x<3,\\ &\frac{1}{10},&3\leqslant x<4,\\ &\frac{4}{10},&4\leqslant x<5,\\ &1,&x\geqslant5.\end{aligned}. $$

1. $ F(x) = \{ \begin{aligned} & 0, & x < 0, \\ &\frac{x}{a}, & 0 \leqslant x < a, \\ &1, & x \geqslant a. \end{aligned} . $

1. (1) $ 1 - e^{-1.2} $. (2) $ e^{-1.6} $. (3) $ e^{-1.2} - e^{-1.6} $. (4) $ 1 - e^{-1.2} + e^{-1.6} $. (5) 0.

1. (1) $ \ln 2 $, 1, $ \ln \frac{5}{4} $. (2) $ f(x)=\{\begin{aligned}&\frac{1}{x},&1

1. (1) $ F(x) = \{ \begin{aligned} & 0, & x < 1, \\ & 2(x + \frac{1}{x} - 2), & 1 \leqslant x < 2, \\ & 1, & x \geqslant 2. \end{aligned} . $

$$ F(x)=\{\begin{aligned}&0,&x<0,\\&\frac{x^{2}}{2},&0\leqslant x<1,\\&-1+2x-\frac{x^{2}}{2},&1\leqslant x<2,\\&1,&x\geqslant2.\end{aligned}. $$

1. (1) $ A = \frac{4}{b \sqrt{\pi b}}. $ (2) $ F_{T}(t) = \{ \begin{array}{ll} 0, & t < 0, \\ 1 - e^{-\frac{t}{241}}, & t \geqslant 0, \end{array} . $ $ P\{50 < T < 100\} = e^{-\frac{50}{241}} - e^{-\frac{100}{241}}. $

$$ \frac{232}{243}.\quad24.\quad P\{Y=k\}=\binom{5}{k}\mathrm{e}^{-2k}(1-\mathrm{e}^{-2})^{5-k},k=0,1,\cdots,5,0.5167.\quad25.\quad\frac{3}{5}. $$

1. (1) $ P\{2 < X \leqslant 5\} = 0.5328, P\{-4 < X \leqslant 10\} = 0.9996, P\{|X| > 2\} = 0.6977, P\{X > 3\} = 0.5. $ (2) c = 3. (3) $ d \leqslant 0.436 $.

1. (1) $ P\{X \leqslant 105\} = 0.3383 $, $ P\{100 < X \leqslant 120\} = 0.5952 $. (2) 129.74.

1. 0.0456. 29. $ \sigma=31.20 $ 30. 0.3204.

1. $ F(x)=\{\begin{aligned}&0,&x<0,\\ &0.2+0.8x/30,&0\leqslant x<30,\\ &1,&x\geqslant30.\end{aligned}. $

33.

1. (1) $ f_{Y}(y)=\{\begin{aligned}&\frac{1}{y},10,\\&0,&y\leqslant0.\end{aligned}. $

$$ f_{Y}(y)=\{\begin{array}{ll}\frac{1}{y}\sqrt{2\pi}\mathrm{e}^{-(\ln y)^{2}/2},&y>0,\\0,&y\leqslant0.\end{array}. $$

(2)

$$ f_{Y}(y)=\{\begin{array}{ll}\frac{1}{2\sqrt{\pi(y-1)}}e^{-(y-1)/4},&y>1,\\0,&y\leqslant1.\end{array}. $$

(3)

$$ f_{Y}(y)=\{\begin{aligned}&\sqrt{\frac{2}{\pi}}\mathrm{e}^{-y^{2}/2},y>0,\\ &0,&y\leqslant0.\end{aligned}. $$

1. (1)

$$ f_{Y}(y)=\frac{1}{3}\frac{1}{\sqrt[3]{y^{2}}}f(\sqrt[3]{y}),y\neq0. $$

$$ f_{Y}(y)=\{\begin{aligned}&\frac{1}{2\sqrt{y}}\mathrm{e}^{-\sqrt{y}},&\quad&y>0,\\ &0,&\quad&y\leqslant0.\end{aligned}. $$

(2)

1. $ f_{Y}(y)=\{\begin{aligned}&\frac{2}{\pi\sqrt{1-y^{2}}},0

1. $ f_{\Theta}(y)=\frac{9}{10\sqrt{\pi}}\mathrm{e}^{-\frac{81}{100}(y-37)^{2}} $

第 三 章

1. (1) 放回抽样的情况

(2) 不放回抽样的情况

2.（1）

(2)

$$ P\{X>Y\}=\frac{19}{35},P\{Y=2X\}=\frac{6}{35},P\{X+Y=3\}=\frac{4}{7},P\{X<3-Y\}=\frac{2}{7}. $$

1. (1) $ \frac{1}{8} $. (2) $ \frac{3}{8} $. (3) $ \frac{27}{32} $, (4) $ \frac{2}{3} $.

1. (2) $ \frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}.\quad5.\quad F_{X}(x)=\{\begin{aligned}&1-\mathrm{e}^{-x},x>0,\\ &0,\quad\text{其他}.\end{aligned}.\quad F_{Y}(y)=\{\begin{aligned}&1-\mathrm{e}^{-y},y>0,\\ &0,\quad\text{其他}.\end{aligned}. $

6.

7.

$$ f_{X}(x)=\{\begin{aligned}&2.4x^{2}(2-x),0\leqslant x\leqslant1,\\ &0,& 其他 ,\end{aligned}. $$

$$ f_{Y}(y)=\{\begin{aligned}&2.4y(3-4y+y^{2}),&0\leqslant y\leqslant1,\\ &0,& 其他 .\end{aligned}. $$

$$ f_{X}(x)=\{\begin{aligned}&\mathrm{e}^{-x},&x>0,\\ &0,& 其他 ,\end{aligned}.\quad f_{Y}(y)=\{\begin{aligned}&y\mathrm{e}^{-y},y>0,\\ &0,& 其他 .\end{aligned}. $$

1. (1) $ c = \frac{21}{4} $.

(2) $ f_{X}(x)=\{\begin{aligned}&\frac{21}{8}x^{2}(1-x^{4}),-1\leqslant x\leqslant1,\\&0,\quad 其他.\end{aligned}.\quad f_{Y}(y)=\{\begin{aligned}&\frac{7}{2}y^{5/2},0\leqslant y\leqslant1,\\&0,\quad 其他.\end{aligned}. $

1. (1)

(2)

1. (1) $ P\{X=n\}=\frac{14^n e^{-14}}{n!}, n=0,1,2,\cdots $

$$ P\{Y=m\}=\frac{\mathrm{e}^{-7.14}(7.14)^{m}}{m!},m=0,1,2,\cdots. $$

(2) 当 m=0,1,2, $ \cdots $ 时， $ P\{X=n|Y=m\}=\frac{e^{-6.86}(6.86)^{n-m}}{(n-m)!},n=m,m+1,\cdots; $

当 n=0,1,2, $ \cdots $ 时， $ P\{Y=m|X=n\}=\binom{n}{m}(0.51)^{m}(0.49)^{n-m},m=0,1,\cdots,n. $

(3) $ P\{Y=m \mid X=20\}=\binom{20}{m}(0.51)^{m}(0.49)^{20-m},m=0,1,2,\cdots,20. $

12.

1. (1) 当 0 < y ≤ 1 时，

$$ f_{X\mid Y}(x\mid y)=\{\begin{aligned}\frac{3}{2}x^{2}y^{-3/2},&\quad-\sqrt{y}

$$ f_{X\mid Y}(x\mid y=\frac{1}{2})=\{\begin{aligned}3\sqrt{2}x^{2},\quad&-\frac{1}{\sqrt{2}}

(2) 当 -1 < x < 1 时， $ f_{Y \mid X}(y \mid x) = \{ \begin{aligned} &\frac{2y}{1 - x^{4}}, x^{2} < y < 1, \\ &0, \quad y \text{ 取其他值}. \end{aligned} . $

$$ f_{Y\mid X}(y\mid x=\frac{1}{3})=\{\begin{aligned}&\frac{81}{40}y,&\frac{1}{9}

(3) $ P\{Y \geqslant \frac{1}{4} \mid X = \frac{1}{2}\} = 1 $, $ P\{Y \geqslant \frac{3}{4} \mid X = \frac{1}{2}\} = \frac{7}{15} $.

1. 当 $ |y|<1 $ 时， $ f_{X\mid Y}(x\mid y)=\{\begin{aligned}&\frac{1}{1-|y|},|y|

当 0<x<1 时， $ f_{Y\mid X}(y\mid x)=\{\begin{aligned}&\frac{1}{2x},|y|

1. (1) $ f(x, y) = \begin{cases} x, & 0 < y < 1/x, & 0 < x < 1, \\ 0, & \text{其他}. \end{cases} $

(2) $ f_{Y}(y)=\{\begin{aligned}&1/2,&&0Y\}=1/3 $

16.（1）放回抽样时相互独立，不放回抽样时，不独立.（2）不独立.

17.（2）X,Y 相互独立.

1. (1) $ f(x, y) = \{ \begin{aligned} & \frac{1}{2} e^{-y/2}, 0 < x < 1, y > 0, \\ & 0, \quad \text{其他}. \end{aligned} . $

(2) $ 1 - \sqrt{2\pi}[\Phi(1) - \Phi(0)] = 0 $. 1445.

1. $ f(x,y)=\frac{1}{2\pi}\mathrm{e}^{-(x^{2}+y^{2})/2} $

$$ \begin{array}{r|rrrrr}Z&0&1&2&\\ \hline p_{k}&\mathrm{e}^{-2}&\mathrm{e}^{-1/2}-\mathrm{e}^{-2}&1-\mathrm{e}^{-1/2}&\end{array} $$

1. (1) y > 0 时， $ f_{X \mid Y}(x \mid y) = \{ \begin{aligned} &\lambda e^{-\lambda x}, x > 0, \\ &0, & x \leqslant 0. \end{aligned} . $

(2)

$$ \begin{array}{c|c c}{Z}&{0}&{1}\\ \hline{p_{k}}&{\frac{\mu}{\lambda+\mu}}&{\frac{\lambda}{\lambda+\mu}}\\ \end{array}\quad F_{Z}(z)=\{\begin{aligned}{}&{{}0,}&{z<0,}\\ {}&{{}\frac{\mu}{\lambda+\mu},}&{0\leqslant z<1,}\\ {}&{{}1,}&{z\geqslant1.}\\ \end{aligned}. $$

1. (1) $ Z = X + Y $ 的密度为

$$ f_{z}(z)=\{\begin{aligned}&z^{2},&0

(2) $ Z=XY $的密度为

$$ f_{z}(z)=\{\begin{aligned}&2(1-z),&&0

1. $ f_{z}(z) = \{ \begin{aligned} & 1 - e^{-z}, & 0 < z < 1, \\ &(e-1) e^{-z}, & z \geqslant 1, \\ & 0, & \text{其他}. \end{aligned} . $

1. (1) $ f_{1}(x)=\{\begin{aligned}&\frac{x^{3}\mathrm{e}^{-x}}{3!},x>0,\\&0,\quad x\leqslant0.\end{aligned}. $ (2) $ f_{2}(x)=\{\begin{aligned}&\frac{x^{5}\mathrm{e}^{-x}}{5!},x>0,\\&0,\quad x\leqslant0.\end{aligned}. $

1. (1) 不独立, (2) $ f_{Z}(z)=\{\begin{aligned}&\frac{1}{2}z^{2}\mathrm{e}^{-z},&z>0,\\&0,& 其他.\end{aligned}. $

1. $ Z = X + Y $ 的密度为

$$ f_{z}(z)=\{\begin{aligned}&(z-2)e^{2-z},\quad&z>2,\\ &0,\quad& 其他 .\end{aligned}. $$

1. Z=X/Y 的密度为

$$ f_{Z}(z)=\{\begin{aligned}&\frac{1}{(z+1)^{2}},&z>0,\\ &0,&z\leqslant0,\end{aligned}. $$

1. $ f_{Z}(z)=\{\begin{aligned}&-\ln z,&0

1. (1) $ b = \frac{1}{1 - e^{-1}} $. (2) $ f_{X}(x) = \{ \begin{aligned} & \frac{e^{-x}}{1 - e^{-1}}, & 0 < x < 1, \\ & 0, & \text{其他}. \end{aligned} . $ $ f_{Y}(y) = \{ \begin{aligned} & e^{-y}, & y > 0, \\ & 0, & \text{其他}. \end{aligned} . $

(3)

$$ F_{U}(u)=\{\begin{aligned}&0,&u<0,\\ &\frac{(1-\mathrm{e}^{-u})^{2}}{1-\mathrm{e}^{-1}},&0\leqslant u<1,\\ &1-\mathrm{e}^{-u},&u\geqslant1.\end{aligned}. $$

1. (0.1587) $ ^{4} $=0.000 63.

31.

$$ F_{z}(z)=\{\begin{aligned}&(1-e^{-z^{2}/8})^{5},z\geqslant0,\\ &0,&z<0.\end{aligned}. $$

(2)

$$ 1-(1-\mathrm{e}^{-2})^{5}=0.5167. $$

1. (1) $ P\{X=2 \mid Y=2\}=0.2 $, $ P\{Y=3 \mid X=0\}=\frac{1}{3} $.

（2）

$$ \begin{array}{c|ccccc}{{{V}}}&{{{0}}}&{{{1}}}&{{{2}}}&{{{3}}}&{{{4}}}&{{{5}}} \\{{{\hline p_{k}}}}&{{{0}}}&{{{0.04}}}&{{{0.16}}}&{{{0.28}}}&{{{0.24}}}&{{{0.28}}} \\\end{array} $$

(3)

$$ \begin{array}{c|cccc}U&0&1&2&3\\\hline p_{k}&0.28&0.30&0.25&0.17\end{array} $$

(4)

$$ \begin{array}{|c|c|cc|l|l|c|c|}\hline{{\cal W}}&{0}&{1}&{2}&{3}&{4}&{5}&{6}&{7}&{8}\\ \hline{{p_{k}}}&{0}&{0.02}&{0.06}&{0.13}&{0.19}&{0.24}&{0.19}&{0.12}&{0.05}\\ \hline \end{array} $$

第 四 章

1. (1)

$$ \frac{X}{p_{k}}\left|\begin{array}{c}2\\ \frac{1}{8}\end{array}\right.\left|\begin{array}{c}3\\ \frac{5}{8}\end{array}\right.\left|\begin{array}{c}4\\ \frac{1}{8}\end{array}\right.\left|\begin{array}{c}9\\ \frac{1}{8}\end{array}\right|,E(X)=\frac{15}{4}. $$

(2)

$$ \frac{Y}{p_{k}}\left|\frac{2}{\frac{2}{30}}\frac{3}{\frac{15}{30}}\frac{4}{\frac{4}{30}}\frac{9}{\frac{9}{30}}\right|,E(Y)=\frac{73}{15}. $$

(3)

$$ \begin{array}{c|cccccccc}X&1&2&3&4&5&7&8&9&10&11&12\hline p_{k}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{36}&\frac{1}{36}&\frac{1}{36}&\frac{1}{36}&\frac{1}{36}\end{array},E(X)=\frac{49}{12}. $$

1. 1.0556. 3. $ \frac{25}{16} $. 5. 1500min.

1. (1) $ E(X) = -0.2 $, $ E(X^{2}) = 2.8 $, $ E(3X^{2} + 5) = 13.4 $. (2) $ E(1/(X + 1)) = \frac{1}{\lambda}(1 - e^{-\lambda}) $.

1. (1) $ E(Y)=2 $, $ E(Y)=1/3 $. (2) $ E(\max\{X_1, X_2, \cdots, X_n\})=\frac{n}{n+1} $,

$$ E(\min\{X_{1},X_{2},\cdots,X_{n}\})=\frac{1}{n+1}. $$

8.（1） $ E(X)=2,E(Y)=0. $ （2） $ -\frac{1}{15}. $ （3）5.

1. (1) $ E(X)=\frac{4}{5} $, $ E(Y)=\frac{3}{5} $, $ E(XY)=\frac{1}{2} $, $ E(X^{2}+Y^{2})=\frac{16}{15} $.

(2) $ E(X)=1 $, $ E(Y)=1 $, $ E(XY)=2 $.

1. (1) $ E[X^{2}/(X^{2}+Y^{2})]=\frac{1}{2} $. (2) $ \sqrt{\frac{\pi}{2}}\sigma $.

1. 33.64 元. $ 12.\frac{\pi}{12}(a^{2}+ab+b^{2}). $ 13. 45 V.

1. (1) $ E(X_{1}+X_{2})=\frac{3}{4} $. (2) $ E(2X_{1}-3X_{2}^{2})=\frac{5}{8} $. (3) $ E(X_{1}X_{2})=\frac{1}{8} $.

1. 1. $ 16.(n+1)/2 $. 18. $ E(X)=\sqrt{\frac{\pi}{2}}\sigma,D(X)=\frac{4-\pi}{2}\sigma^{2} $.

1. $ E(X) = \alpha\beta, D(X) = \alpha\beta^{2} $.

1. $ E(X)=\frac{1}{p}, D(X)=\frac{1-p}{p^{2}}. $ 21. $ E(A)=8.67, D(A)=21.42. $

22.（1） $ E(Y)=7,D(Y)=37.25 $

$$ Z_{1}\sim N(2080,65^{2}),Z_{2}\sim N(80,1525),P(X>Y)=0.9798,P(X+Y)>1400]=0.1539. $$

23.（1）1200，1225.（2）1282kg. 24.39袋.

1. (1) $ E(XY) = 1/4 $, $ E(X/Y) $ 不存在, $ E[\ln(XY)] = -2 $, $ E(|Y - X|) = 1/3 $.

(2) $ \rho_{AC}=\sqrt{6/7} $

1. (1) $ P\{X_{1}=2,X_{2}=2,X_{3}=5\}=0.00203,E(X_{1}X_{2}X_{3})=8,E(X_{1}-X_{2})=0, $ $ E(X_{1}-2X_{2})=-2. $

(2) 对于 $ E(Z) $，三种情况都有 $ E(Z)=29 $.

对于 $ D(Z) $: (i) X, Y 独立，则 $ D(Z)=109 $，(ii) X, Y 不相关，则 $ D(Z)=109 $， $ \rho_{XY}=0.25 $，则 $ \mathrm{Cov}(X,Y)=1.5, D(Z)=94 $.

27.（1）X，Y 不相互独立，也不是不相关的。（2）X，Y 不相互独立，但不相关.

（3）X，Y 不相互独立，但不相关.（4）X，Y 不是不相关的，因而一定也是不相互独立的.

(5) X, Y 相互独立，因此，X, Y 也是不相关的.

1. $ E(X)=\frac{2}{3}, E(Y)=0, \mathrm{Cov}(X,Y)=0. $

$$ E(X)=E(Y)=\frac{7}{6},\mathrm{Cov}(X,Y)=-\frac{1}{36},\rho_{XY}=\frac{-1}{11},D(X+Y)=\frac{5}{9}. $$

1. $ \frac{\alpha^{2}-\beta^{2}}{\alpha^{2}+\beta^{2}} $. 34. (1) $ \min E(W)=108 $. 35. $ f(x,y)=\frac{1}{3\sqrt{5}\pi}\exp\left[\frac{-8}{15}\left(\frac{x^{2}}{3}+\frac{xy}{4\sqrt{3}}+\frac{y^{2}}{4}\right)\right] $.

1. $ p \geqslant \frac{8}{9} $. 38. (1) $ \frac{1}{2} \ln 2 $. (2) a.

第五章

1. 0.2119. 2.（1）0.8944.（2）0.0019. 3.（1）0.1802.（2）n=443. 4.0.0787.

1. 0.0062. 6. 0.1075. 7. (1) 0.0003. (2) 0.5. 8. 0.9525.

1. (1) $ \overline{X} \sim N(2.2, 1.4^2 / 52) $, $ P\{\overline{X} < 2\} = 0.1515 $. (2) 0.0770.

1. 1.3427 g/km. 11. (1) 0.8968. (2) 0.7498. 12. 254. 13. 1537.

14.（1）0.8944.（2）0.1379.

第 六 章

1. 0.8293.

1. (1) 0.2628. (2) $ P\{\max\{X_1, X_2, X_3, X_4, X_5\} > 15\} = 0.2923 $, $ P\{\min\{X_1, X_2, X_3, X_4, X_5\} = 0.5785 $.

3: p=0.6744. 4. (1) C=1/3. (2) $ C=\sqrt{3/2} $

1. (1) $ f(x_1, x_2, \cdots, x_{10}) = \prod_{i=1}^{10} \frac{1}{\sqrt{2\pi}\sigma} e^{-(x_i - \mu)^2 / (2\sigma^2)} $, $ P\{\overline{X} < \mu\} = 1/2 $. (2) 0.431.

1. (1) $ P\{X_1 = x_1, X_2 = x_2, \cdots, X_n = x_n\} = p^{\sum_{i=1}^{n} x_i} (1 - p)^{n - \sum_{i=1}^{n} x_i} $.

(2) $ \binom{n}{k}p^{k}(1-p)^{n-k}, k=0,1,2,\cdots,n. $

$$ E(\bar{X})=p,D(\bar{X})=\frac{1}{n}p(1-p),E(S^{2})=p(1-p). $$

1. $ E(\bar{X}) = n, D(\bar{X}) = n / 5, E(S^{2}) = 2n. $

1. (1) $ X_1, X_2, \cdots, X_{10} $ 的联合密度为 $ \frac{1}{(2\pi\sigma^2)^5} e^{-\sum_{i=1}^{10} (x_i - \mu)^2 / (2\sigma^2)} $. (2) $ f_X(x) = \frac{\sqrt{5}}{\sqrt{\pi}\sigma} e^{-5(x - \mu)^2 / \sigma^2} $.

1. (1) p = 0.99. (2) $ D(S^{2}) = 2\sigma^{4} / 15 $. 11. 226.3333.

第 七 章

$$ \hat{\mu}=74.002,\widehat{\sigma^{2}}=6\times10^{-6},s^{2}=6.86\times10^{-6}. $$

1. 矩估计量为(1) $ \hat{\theta}=\frac{\overline{X}}{\overline{X}-c} $. (2) $ \hat{\theta}=\left(\frac{\overline{X}}{1-\overline{X}}\right)^2 $. (3) $ \hat{p}=\frac{\overline{X}}{m} $.

1. 最大似然估计量为(1) $ \hat{\theta} = \frac{n}{\sum_{i=1}^{n} \ln X_i - n \ln c} $. (2) $ \hat{\theta} = \frac{n^2}{\left( \sum_{i=1}^{n} \ln X_i \right)^2} $. (3) $ \hat{p} = \frac{\overline{X}}{m} $.

4.（1）矩估计值和最大似然估计值均为 $ \frac{5}{6} $。（2）矩估计量和最大似然估计量均为 $ \hat{\lambda}=X $。（3） $ \hat{p}=\frac{r}{x} $

5.（1）c与 $ \theta $的最大似然估计值分别为 $ \hat{c}=x_{1},\hat{\theta}=\bar{x}-x_{1} $

(2) c 与 $ \theta $ 的矩估计量分别为 $ \hat{c} = \overline{X} - \left[\frac{1}{n} \sum_{i=1}^{n} (X_i - \overline{X})^2\right]^{1/2}, \hat{\theta} = \left[\frac{1}{n} \sum_{i=1}^{n} (X_i - \overline{X})^2\right]^{1/2} $

1. 0.499. 7. (1) 由最大似然估计的性质 $ \hat{P}\{X=0\}=e^{-\bar{x}}. $ (2) 0.3253.

1. $ \hat{U} = e^{-1/\hat{\theta}} $, 其中 $ \hat{\theta} = -n / \sum_{i=1}^{n} \ln x_i $. (2) $ \hat{\theta} = 1 - \Phi(2 - \overline{x}) $. (3) $ \hat{\beta} = (3\overline{x}) / m - 1 $.

1. (1) $ c = \frac{1}{2(n-1)} $. (2) $ c = \frac{1}{n} $. 12. $ T_{1}, T_{3} $ 是无偏的、 $ T_{3} $ 较 $ T_{1} $ 为有效.

$$ 13.\quad(2)\hat{\theta}=X_{(n)}=\max\{X_{1},X_{2},\cdots,X_{n}\},E(\hat{\theta})=\frac{n}{n+1}\theta.\quad14.\quad a=\frac{n_{1}}{n_{1}+n_{2}},b=\frac{n_{2}}{n_{1}+n_{2}}. $$

1. 记 $ \frac{1}{\sigma_{0}^{2}}=\sum_{i=1}^{k}\frac{1}{\sigma_{i}^{2}}, a_{i}=\frac{\sigma_{0}^{2}}{\sigma_{i}^{2}}, i=1,2,\cdots,k. $

16.（1）（5.608，6.392）.（2）（5.558，6.442）

1. (1) (6.675, 6.681), $ (6.8 \times 10^{-6}, 6.5 \times 10^{-5}) $.

(2) $ (6.661, 6.667) $, $ (3.8 \times 10^{-6}, 5.06 \times 10^{-5}) $.

1. (7.4, 21.1).

1. $ \sigma^{2} $ 的置信区间为 $ \{\frac{\sum_{i=1}^{n}(X_{i}-\mu)^{2}}{\chi_{\alpha/2}^{2}(n)},\frac{\sum_{i=1}^{n}(X_{i}-\mu)^{2}}{\chi_{1-\alpha/2}^{2}(n)}\} $，(5.013,31.626)，(2.239,5.624).

1. (0.010,0.018). 21. (-0.002,0.006). 22. (-6.04,-5.96).

1. (0.222,3.601). 24. (0.101,0.244).

25.（1） $ \sigma $ 已知6.329； $ \sigma $ 未知6.356.（2）-0.0012.（3）2.84

1. 40527. 27. 39岁零4个月.

第八章

1. 接受 $ H_{0} $. 2. 接受 $ H_{0} $. 3. 认为不合格. 4. 认为显著大于 10.

1. 接受 $ H_{0} $，认为这批罐头是符合规定的. 6. 拒绝 $ H_{0} $. 7. 拒绝 $ H_{0} $.

1. 认为早上的身高比晚上高. 9. 拒绝 $ H_{0} $，认为 A 比 B 耐穿. 10. 接受 $ H_{0} $

1. 拒绝 $ H_{0} $，认为提纯后的群体比原群体整齐. 12. 拒绝 $ H_{0} $，认为偏大. 13. 接受 $ H_{0} $

1. 接受 $ H_{0} $. 15. 接受 $ H_{0} $. 16. 接受 $ H_{0} $.

1. 接受 $ H_{0} $，认为两者方差相等；接受 $ H_{0}^{\prime} $，认为所需天数相同.

1. 接受 $ H_{0} $，拒绝 $ H_{0}^{\prime} $，认为两者的可理解性有显著差异. 19. 接受 $ H_{0} $. 20. $ n \geqslant 7 $

1. (1) 接受 $ H_{0} $. (2) $ n \geqslant 7 $. 22. $ t = (\bar{x} - 2\bar{y}) / \sqrt{\sigma_{1}^{2} / n_{1} + 4\sigma_{2}^{2} / n_{2}} \geqslant z_{a} $

1. 认为服从泊松分布. 24. 接受 $ H_{0} $. 25. (2) 接受 $ H_{0} $. 26. 接受 $ H_{0} $.

1. 拒绝 $ H_{0} $，认为有显著改变. 28.（1） $ \hat{p}=0.6419 $.（2）接受 $ H_{0} $

1. 接受 $ H_{0} $，认为无显著差异. 30. 拒绝 $ H_{0} $，认为型号 A 比型号 B 使用时间长.

1. 接受 $ H_{0} $，认为差异不显著.

1. (1) (i) 取 $ \alpha=0.05 $ 时接受 $ H_{0} $; (ii) 取 $ \alpha=0.1 $ 时拒绝 $ H_{0} $; (iii) 拒绝 $ H_{0} $ 的最小显著性水平为 0.0808.

(2) p 值 = 0.4747. (3) p 值 = 0.0271，拒绝 $ H_{0} $. (4) p 值 = 0.0110.

第九章

1. 各总体均值间有显著差异；(6.75,18.45)，(-7.65,4.05)，(-20.25,-8.55).

1. 差异显著；(0.72,4.28)，(2.55,6.45)，(0.22,3.78). 3. 差异显著. 4. 差异显著.

1. 差异显著. 6. 只有浓度的影响是显著的. 7. 因素 A、因素 B 的影响均不显著.

1. $ \hat{y}=24.6287+0.05886x $

9.（2） $ \hat{y}=13.9584+12.5503x $ （3） $ \sigma^{2}=0.0432 $ （4）回归效果显著.

（5）（11.82，13.28）（6）（20.03，20.44）（7）（19.66，20.81）.

1. (2) $ \hat{y} = -0.104 + 0.988x $. (3) (13.29, 14.17).

1. $ \hat{y} = -3.85493 + 1.83396x $.

1. (1) 成绩关于奥运次数的回归方程 $ \hat{y}=9.168-0.1567x $，回归效果显著。预测值为 26.82 min.

（2）成绩关于年份的回归方程为 $ \hat{y}=105.4826-0.0392x $，回归效果显著.

13.（1） $ \hat{y}=1.896+0.53846x $ （2）b的置信水平为0.95的置信区间为（0.208，0.869）

1. $ \hat{y}=32.4556e^{-0.0867318x} $

$$ \hat{y}=19.0333+1.0086x-0.020381x^{2}. $$

1. (1) $ \hat{y}=9.9+0.575x_{1}+0.55x_{2}+1.15x_{3} $. (2) $ \hat{y}=9.9+0.575x_{1}+1.15x_{3} $.

第十二章

1. (1) $ F(x; \frac{1}{2}) = \{ \begin{aligned} &0, & x < 0, \\ &\frac{1}{2}, & 0 \leqslant x < 1, \\ &1, & x \geqslant 1. \end{aligned} . F(x; 1) = \{ \begin{aligned} &0, & x < -1, \\ &\frac{1}{2}, & -1 \leqslant x < 2, \\ &1, & x \geqslant 2. \end{aligned} . $

$$ \{\begin{aligned}0,\quad&x_{1}<0,\quad&-\infty

$$ F(x_{1},x_{2};\frac{1}{2},1)=\{\begin{aligned}&\frac{1}{2},&&0\leqslant x_{1}<1,\quad x_{2}\geqslant-1,\\ &\frac{1}{2},&&x_{1}\geqslant1,\quad-1\leqslant x_{2}<2,\\ &1,&&x_{1}\geqslant1,\quad x_{2}\geqslant2.\end{aligned}. $$

$$ \mu_{Y}(t)=F_{X}(x;t),R_{Y}(t_{1},t_{2})=F_{X}(x,x;t_{1},t_{2}). $$

$$ \mu_{X}(t)=\frac{1}{a t}\left(1-\mathrm{e}^{-\alpha t}\right),t>0,R_{X}\left(t_{1},t_{2}\right)=\frac{1}{a\left(t_{1}+t_{2}\right)}\left(1-\mathrm{e}^{-a\left(t_{1}+t_{2}\right)}\right),t_{1},t_{2}>0. $$

1. $ \mu_{X}(t)=a $, $ C_{X}(t_{1},t_{2})=\sigma^{2} $

$$ \mu_{Y}(t)=\mu_{X}(t)+\varphi(t),C_{Y}(t_{1},t_{2})=C_{X}(t_{1},t_{2}). $$

$$ R_{Y}(t_{1},t_{2})=R_{X}(t_{1}+a,t_{2}+a)-R_{X}(t_{1}+a,t_{2})-R_{X}(t_{1},t_{2}+a)+R_{X}(t_{1},t_{2}) $$

1. $ C_{Z}(t_{1},t_{2})=\sigma_{1}^{2}+(t_{1}+t_{2})\rho\sigma_{1}\sigma_{2}+t_{1}t_{2}\sigma_{2}^{2}. $

1. $ R_{X}(t_{1},t_{2})=C_{X}(t_{1},t_{2})=(1+t_{1}t_{2})\sigma^{2} $

$$ \mu_{Z}(t)=a(t)\mu_{X}(t)+b(t)\mu_{Y}(t)+c(t), $$

$$ C_{Z}(t_{1},t_{2})=a(t_{1})a(t_{2})C_{X}(t_{1},t_{2})+b(t_{1})b(t_{2})C_{Y}(t_{1},t_{2}),t_{1},t_{2}\in T. $$

1. (1) $ \sigma^{2}\min\{t_{1},t_{2}\},t_{1},t_{2}\geqslant0 $. (2) $ t_{1}t_{2}+\sigma^{2}\min\{t_{1},t_{2}\},t_{1},t_{2}\geqslant0 $.

(3) $ \sigma^{2}\min\{t_{1},t_{2}\},t_{1},t_{2}\geqslant0 $.

第十三章

1. 状态空间 $ I=\{1,2,\cdots,N\} $， $ p_{ij}=P\{X_{n}=j\mid X_{n-1}=i\}=\{\begin{aligned}&1/i,1\leqslant j\leqslant i,\\&0,\quad j>i,\end{aligned}. $ $ i=1,2,\cdots,N $.

$$ \mathbf{P}=\left[\begin{matrix}{1}&{}&{}&{}&{}&{}\\ {1/2}&{1/2}&{}&{}&{\mathbf{0}}&{}\\ {\vdots}&{\vdots}&{\ddots}&{}&{}&{}\\ {1/i}&{1/i}&{\cdots}&{1/i}&{}&{}\\ {\vdots}&{\vdots}&{}&{\vdots}&{\ddots}&{}\\ {1/N}&{1/N}&{\cdots}&{1/N}&{\cdots}&{1/N}\\ \end{matrix}\right]. $$

1. 状态空间 $ \{I=1,2,3,4,5,6\} $

(1)

$$ \mathbf{P}_{1}=\left[\begin{matrix}{1/6}&{1/6}&{\cdots}&{1/6}\\ {1/6}&{1/6}&{\cdots}&{1/6}\\ {\vdots}&{\vdots}&{}&{\vdots}\\ {1/6}&{1/6}&{\cdots}&{1/6}\\ \end{matrix}\right].\mathrm{~(2)~}\mathbf{P}_{2}=\left[\begin{matrix}{1/6}&{1/6}&{1/6}&{1/6}&{1/6}&{1/6}\\ {0}&{2/6}&{1/6}&{1/6}&{1/6}&{1/6}\\ {0}&{0}&{3/6}&{1/6}&{1/6}&{1/6}\\ {0}&{0}&{0}&{4/6}&{1/6}&{1/6}\\ {0}&{0}&{0}&{0}&{5/6}&{1/6}\\ {0}&{0}&{0}&{0}&{0}&{1}\\ \end{matrix}\right]. $$

1. 状态空间 $ I=\{1,2,\cdots\} $, $ p_{ij}=\{\begin{aligned}&p,&j=i+1,\\ &q,&j=i,\quad i,j=1,2,\cdots,\\ &0,& 其他.\end{aligned}. $

$$ \mathbf{P}=\left[\begin{matrix}{q}&{p}&{0}&{\cdots}&{}&{}\\ {0}&{q}&{p}&{0}&{\cdots}&{}\\ {0}&{0}&{q}&{p}&{0}&{\cdots}\\ {\vdots}&{\vdots}&{\vdots}&{\vdots}&{\vdots}&{\ddots}\\ \end{matrix}\right]. $$

1. 状态空间 $ I = \{0, 1, 2, \cdots, N\} $

$$ \mathbf{P}=\left[\begin{matrix}{1}&{0}&{0}&{0}&{\cdots}&{0}&{0}\\ {0}&{1-\alpha_{1}}&{\alpha_{1}}&{0}&{\cdots}&{0}&{0}\\ {0}&{0}&{1-\alpha_{2}}&{\alpha_{2}}&{\cdots}&{0}&{0}\\ {\vdots}&{\vdots}&{\vdots}&{\vdots}&{}&{}&{\vdots}\\ {0}&{0}&{0}&{0}&{\cdots}&{1-\alpha_{N-1}}&{\alpha_{N-1}}\\ {0}&{0}&{0}&{0}&{\cdots}&{0}&{1}\\ \end{matrix}\right], $$

其中 $ \alpha_{i}=\frac{2i(N-i)}{N(N-1)}\alpha,i=1,2,\cdots,N-1. $

5.（1）1/16.（3）7/16.（4）0.3993.

0 1 5月1日为晴天的条件下，5月3日为晴天的概率为 $ P_{00}(2)=0.4167 $; 5月5日为雨天的概率为 $ P_{01}(4)=0.5995 $.

1. $ P = \begin{bmatrix} 1/2 & 1/2 \\ 1/3 & 2/3 \end{bmatrix} $,

1. $ m=1,\pi=(2/3,1/3) $. 9. $ m=2,\pi=(4/25,9/25,12/25) $.

1. $ m=2, \pi_{j}=\frac{1-p/q}{1-(p/q)^{3}}\left(\frac{p}{q}\right)^{j-1}, j=1,2,3. $

第十四章

1. 是. 3. 是.

$$ \mu_{Y}(t)=\lambda L,R_{Y}(s,t)=\{\begin{matrix}{\lambda^{2}L^{2}+\lambda(L-|\tau|),}&{|\tau|\leqslant L,}\\ {\lambda^{2}L^{2},}&{|\tau|>L,}\\ \end{matrix}.\tau=t-s,s,t{\geqslant}0. $$

1. 是. 6. 不是. 8.(1) A/8, $ A^{2}/12 $; (2) A/8, $ A^{2}/12 $.

1. (2) $ R_{XY}(\tau) = a R_X (\tau - \tau_1) $.

1. (1) 5. (2) $ S_{X}(\omega)=4\left[\frac{1}{(\omega-\pi)^{2}+1}+\frac{1}{(\omega+\pi)^{2}+1}\right]+\pi\left[\delta(\omega-3\pi)+\delta(\omega+3\pi)\right] $

$$ \frac{1}{2}(\sqrt{2}-1),\quad14.\quad S_{X}(\omega)=\frac{4}{T\omega^{2}}\sin^{2}\frac{\omega T}{2},\quad15.\quad R_{X}(\tau)=\frac{4}{\pi}(1+\frac{\sin^{2}5\tau}{\tau^{2}}). $$

$$ 9.R_{XY}(\tau)=-R_{YX}(\tau)=\frac{1}{2}ab\sin\omega_{0}\tau,S_{XY}(\omega)=-S_{YX}(\omega)=\frac{\pi ab}{2}i\left[\delta(\omega+\omega_{0})-\delta(\omega-\omega_{0})\right]. $$

$$ S_{X Y}\left(\omega\right)=2\pi\mu_{X}\mu_{Y}\delta\left(\omega\right),S_{X Z}\left(\omega\right)=S_{X}\left(\omega\right)+2\pi\mu_{X}\mu_{Y}\delta\left(\omega\right). $$

选做习题

1. $ \frac{3(1-p_{1})^{5}}{3(1-p_{1})^{5}+2(1-p_{2})^{5}} $. 2. (1) 8/11. (2) 4/5. (3) 32/55.

1. 2p(1-p). 4. 甲:6/11,乙:5/11.

1. $ P(A)=1/3, P(B)=7/12, A, B $ 独立. 6. $ (1-p)/(2-p) $.

1. $ p_{1}p_{2}(1-p_{3})+p_{1}(1-p_{2})p_{3}+(1-p_{1})p_{2}p_{3}+p_{1}p_{2}p_{3} $

1. (1) $ P(F) = P(F|C_1)p_1 + P(F|\overline{C}_1)(1 - p_1) $,

$$ P\left(F\mid C_{1}\right)=p_{2}+p_{3}p_{5}+p_{4}p_{5}-p_{2}p_{3}p_{5}-p_{2}p_{4}p_{5}-p_{3}p_{4}p_{5}+p_{2}p_{3}p_{4}p_{5} $$

$$ P(F|\overline{C}_{1})=p_{4}p_{5}+p_{2}p_{3}p_{4}-p_{2}p_{3}p_{4}p_{5}, $$

(2) $ P(C_3|F) = \left[(1 - q_1 q_4 - q_2 q_5 + q_1 q_2 q_4 q_5) p_3\right]/P(F). $

9.

10\. (1)

（2）

1. (1) 0.632, (2) 3. 12. p=0.989 97. 13. (1) $ \frac{1}{11} $. (2) $ \frac{3}{55} $.

1. 0.5488, 0.5730.

1. (1) $ F(x) = \begin{cases} \frac{1}{2} e^x, & x < 0, \\ 1 - \frac{1}{2} e^{-x}, & x \geq 0. \end{cases} $

(2)

$$ \begin{array}{r|rrr|r}Y&-1&1&F_{Y}(y)=\{\begin{array}{ll}0,&y<-1,\\\frac{1}{2}&-1\leqslant y<1,\\1,&y\geqslant1.\end{array}.\\p_{k}&\frac{1}{2}&\frac{1}{2}&\end{array} $$

16.(1) $ k=\{\begin{aligned}&\lambda-1,\lambda,& 若 \lambda 是整数,\\&[\lambda],& 若 \lambda 不是整数.\end{aligned}. $

(2) $ k=\{\begin{aligned}&(n+1)p-1,&(n+1)p,\\ &(n+1)p,\quad& 若(n+1)p 是整数,\\ &(n+1)p,\quad& 若(n+1)p 不是整数.\end{aligned}. $

$$ 18.f_{Y}(y)=\{\begin{aligned}&2/3,&&0

1. (1)

(2) $ P\{X=1|Y=1\}=1, P\{Y=2|X=1\}=0 $

22.

1. $ \binom{n}{k}\left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right)^{k}\left(\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right)^{n-k} $

1. (1) $ P\{X=x,Y=y\}=\frac{\lambda^{x}\mu^{y}e^{-(\lambda+\mu)}}{x!y!},x,y=0,1,2,\cdots $

(2) $ P\{X+Y \leqslant 1\} = \mathrm{e}^{-(\lambda+\mu)}(1+\lambda+\mu) $.

$$ f(x,y)=\{\begin{aligned}&\frac{4}{\sqrt{3}},&(x,y)\in D,\\ &0,&(x,y)\notin D.\end{aligned}. $$

$$ F_{Y}(y)=\{\begin{aligned}&0,&y<0,\\&\frac{4}{\sqrt{3}}y-\frac{4}{3}y^{2},&0\leqslant y<\sqrt{3}/2,\\&1,&y\geqslant\sqrt{3}/2.\end{aligned}. $$

1. (1) $ f_{X}(x)=\{\begin{aligned}&\mathrm{e}^{-x},&x>0,\\ &0,& 其他,\end{aligned}.\quad f_{Y}(y)=\{\begin{aligned}&\frac{1}{(y+1)^{2}},&y>0,\\ &0,& 其他.\end{aligned}. $

(2) 当 y>0 时， $ f_{X\mid Y}(x\mid y)=\{\begin{aligned}&x(y+1)^{2}\mathrm{e}^{-x(y+1)},&x>0,\\&0,& 其他 ,\end{aligned}. $

当 x>0 时， $ f_{Y\mid X}(y\mid x)=\{\begin{aligned}&x\mathrm{e}^{-xy},&y>0,\\&0,& 其他 .\end{aligned}. $

1. (1)

(2) 1/2. (3) 5/6.

$$ P\{X=k,Y=i\}=\binom{k}{i}p^{i}(1-p)^{k-i}\frac{\lambda^{k}\mathrm{e}^{-\lambda}}{k!},\quad\begin{array}{l}k=0,1,2,\cdots,\\ i=0,1,2,\cdots,k.\end{array} $$

1. 1/2. 30. 0.19. 31. $ \Phi(\sqrt{2}) - \Phi(0) = 0.4207 $.

32.（1）0.8897.（2）0.2818.（3）0.9874.

1. (1) $ f_Z(z) = \begin{cases} \frac{1}{2\varepsilon} [1 - e^{-\frac{1}{2}(z+\varepsilon)^2}], & -\varepsilon < z < \varepsilon \\ \frac{1}{2\varepsilon} [e^{-\frac{1}{2}(z-\varepsilon)^2} - e^{-\frac{1}{2}(z+\varepsilon)^2}], & z \geq \varepsilon, \\ 0, & \text{其他}. \end{cases} $

1. $ f(z)=\{\begin{aligned}&-20\ln(20z),&0

1. (1) $ f_{X}(x)=\{\begin{aligned}&\mathrm{e}^{-x},&x>0,\\ &0,& 其他.\end{aligned}. $ $ f_{Y}(y)=\{\begin{aligned}&y\mathrm{e}^{-y},&y>0,\\ &0,& 其他.\end{aligned}. $

(2) X,Y 不是相互独立的. (3) $ f_{X+Y}(z)=\{\begin{aligned}&\mathrm{e}^{-z/2}-\mathrm{e}^{-z},&z>0,\\&0,& 其他.\end{aligned}. $

（4）对于 y>0, $ f_{X\mid Y}(x\mid y)=\{\begin{aligned}&1/y,&0

(5) $ P\{X>3 \mid Y<5\}=0.03082 $. (6) $ P\{X>3 \mid Y=5\}=2/5 $.

1. (1) np. (2) $ n(p + \alpha - p\alpha) $.

1. (1)

(2) 3/5. (3) $ Y \sim b(6,3/5) $, $ E(Y) = 3.6 $.

(4) $ P\{Y<4\}=0.456, P\{Y>4\}=0.233. $ (5) $ E(Z)=2 $

1. $ \frac{nr}{N} $. 39. 14.7. 40. $ \frac{\alpha}{\beta(\alpha+1)}+\frac{2}{\alpha^{2}} $. 41. (1) $ \frac{2}{5} $. (2) $ \frac{4}{3} $. 42. 0.7852.

43.（2）不是离散型也不是连续型随机变量，是混合型随机变量.

(3) $ P\{X=4\}=0, P\{X=3\}=0.316, P\{X<4\}=0.684, P\{X>6\}=0.135 $

1. 0.1469. 45. (1) 值域为 $ (0,0.75] $. (2) $ F_{Y}(y)=\{\begin{aligned}&0,&y<0,\\ &y,&0\leqslant y<0.75,\\ &1,&y\geqslant0.75.\end{aligned}. $

1. 最大似然估计量 = 矩估计量 = $ \frac{\overline{X}}{2} $，是无偏估计量.

1. $ \hat{\mu}_{1} = \overline{X}, \hat{\mu}_{2} = \overline{Y}, \widehat{\sigma}^{2} = \frac{1}{n_{1} + n_{2}} \left[ \sum_{i=1}^{n_{1}} (X_{i} - \overline{X})^{2} + \sum_{i=1}^{n_{2}} (Y_{i} - \overline{Y})^{2} \right]. $

1. $ \lambda $ 的对数似然方程为 $ \sum_{i=1}^{k}\frac{d_{i}(t_{i}-t_{i-1})}{\mathrm{e}^{\lambda(t_{i}-t_{i-1})}-1}-\sum_{i=2}^{k}d_{i}t_{i-1}-st_{k}=0. $

1. $ \hat{\lambda} = \frac{1}{T_0} \ln \frac{n}{n - k} $.

1. (1) $ L(p_1, p_2) = [(1 - p_1)p_2]^n_1 [(1 - p_2)p_1]^n_2 (1 - p_1 p_2)^{n_1 2} [p_1 p_2]^s $.

(2) $ \hat{p}_1 = 0.7150, \hat{p}_2 = 0.8290 $.

1. (1) $ \theta $ 的最大似然估计值为 13/32, $ \theta $ 的矩估计值为 5/12. (2) $ \hat{\beta} = \overline{x} / \alpha $

1. (2) $ \widehat{E(X)} = \exp\{\widehat{\mu} + \widehat{\sigma^2}/2\} $，其中 $ \widehat{\mu} = \frac{1}{n} \sum_{i=1}^{n} \ln x_i $， $ \widehat{\sigma^2} = \frac{1}{n} \sum_{i=1}^{n} (\ln x_i - \widehat{\mu})^2 $.

(3) $ \widehat{E(X)} = 28.3067 $.

1. $ \hat{\eta} = \left( \frac{T_m}{m} \right)^{1/\beta} $，其中 $ T_m = \sum_{i=1}^{m} x_i^\beta + (n-m) x_m^\beta $.

1. (1) $ \eta = 214.930 $. (2) p = 0.841.

1. $ a=\frac{n_1-1}{n_1+n_2-2} $, $ b=\frac{n_2-1}{n_1+n_2-2} $. 59. (2) $ \frac{2n\overline{X}}{\chi^2_a(2n)} $. (3) 3764.7.

1. (3) $ h_{\alpha/2} = (1 - \alpha/2)^{1/n} $, $ h_{1-\alpha/2} = (\alpha/2)^{1/n} $.

(4) $ (1-\alpha/2)^{-1/n} \max\{X_1, X_2, \cdots, X_n\}, (\frac{\alpha}{2})^{-1/n} \max\{X_1, X_2, \cdots, X_n\} $.

(5) (4.22,8.78).

1. 拒绝域为 $ \chi^{2}=\frac{2nx}{\theta_{0}}\geqslant\chi_{a/2}^{2}(2n) $ 或 $ \chi^{2}=\frac{2nx}{\theta_{0}}\leqslant\chi_{1-a/2}^{2}(2n) $; 拒绝域为 $ \chi^{2}>39.364 $ 或 $ \chi^{2}<12.401 $, 现在 $ \chi^{2}=24.83 $ 故接受 $ H_{0} $.

1. 认为无显著差异. 63. 拒绝 $ H_{0} $，认为可降低血压.

1. 认为是有偏爱的. 65. 认为 $ X \sim b(4, \theta) $.

1. 拒绝 $ H_{0} $，认为春季发生的案件数的均值比秋季的多. 67. 认为有显著差异.

1. (1) $ \hat{y}=13.487+1.065x $. (2) 认为回归效果显著. (3) $ \hat{y}\mid_{x=13}=27.332 $.

（4）在 x=13 处 $ \mu(x) $ 的置信水平为 0.95 的置信区间为 $ (27.332 \pm 1.244) $.

（5）在 x=13 处 Y 的新观测值 $ Y_{0} $ 的置信水平为 0.95 的预测区间为 $ (27.332 \pm 3.487) $.

1. (2) $ \hat{y} = 774.0125 - 0.35915x $. (3) 回归效果是非常显著的.