习题三
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
1. 证明:
由概率的非负性,知上式大于等于零,故得正.
2. 解:①,
0
1
0
1
2
1/56
10/56
10/56
5/56
20/56
10/56
3/28
15/28
5/14
3/8
5/8
1
②.
3. 解:①由概率的性质
又
②
当y>0时
当x>0时
4. 解:①
②
EMBED Equation.3
;
,
;
③
EMBED Equation.3
5.
6.①
②
讨论如下:
③
=
7. 证明:
EMBED Equation.3
故独立得证.
8.
0
1
P
9. 解:①
②
③
10. 解:①
②
故不独立.
11.
x y
1/24
1/8
1/6
1/8
3/8
1/2
1/12
1/4
1/3
1/4
3/4
1
12. 证明:必要性: 由:
充分性,若
从上式可得x与y独立.
13. 解:①
有实根的概率
EMBED Equation.3
=
②
14. 解:
当
时,
当
时,
当
时,
当
时
② 当
时,
Y<0时,
当X≥0时,
当X<0时,
EMBED Equation.3
15.解 1.由S(D)=
得X与Y的联合密度函数为
2. 由于0≤y≤1时,
从而 0≤y≤1时,
又 1≤y≤3时,
从而 1≤y≤3时,
;
又当Y<0时,Y>3时,f(x,y)=0,从而fY(y)
综上得:
此外,0≤X≤1时,
从而 0≤X≤1时,
当 1≤X≤2时,
从而 1≤X≤2时,
当X<0或X>2时,f(x,y)=0
从而
综上得:
16. 证明:
(1)
故
为均匀分布
又
故
为均匀分布.
17. 解:
故
18.
故独立.
不独立.
19.
不独立.
20.
21. 若
服从二元正态分布,则
EMBED Equation.3
因为
均无参数ρ,故可见,
不能由
决定.
22. 解:
EMBED Equation.3
均服从正态分布,但是
不服从正态分布.
23.
0 1 2 3 4
P 1/9 4/9 4/9 0 0
0 1 2 -1 -2
P 3/9 2/9 1/9 2/9 1/9
0 1 2 3 4
-2
-1
0
1
2
0 0 1/9 0 0
0 1/9 0 1/9 0
1/9 0 1/9 0 1/9
0 1/9 0 1/9 0
0 0 1/9 0 0
1/9
2/9
3/9
2/9
1/9
1/9 2/9 3/9 2/9 1/9
1
24.
2 3 4 5 6 7 8 9 10 11 12
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
25.证明:
当n=2时,
,k=0,1,2,……
由数学归纳法,设对n-1成立,即
服从参数为
的泊松分布,因为Xn服从参数为
的泊松分布,故
服从参数为
的泊松分布,即对n成立.
26.
i=1,2. ,Z=X1+ X2
Z>0时,
所以
27.由题意知X1~N(4,3), X2~N(2,1).且两者相互独立.
由X1=X+Y, X2=X-Y得X=
,Y=
,且X和Y服从正态分布
故
,
故X~N(3,1),Y~N(1,1)
即
,
28. 用数学归纳法进行推广,与25题类似.
Xi~N(
i,
),i=1,2……n.
29.
当0<z
2时,
当z>2时,
故
30.
XY
0
-1
1
P
0.8
0.1
0.1
X
0
1
P
0.6
0.4
Y
-1
0
1
P
0.4
0.2
0.4
=0×0.6+0.4×1+(-1) ×0.4+0×0.2+1×0.4
=0.4
31 由题意知
所以:
x>y y>x
32.证明: P(X=a
EMBED Equation.3 ,Y=b
)=P
, i,j=1,2…….
P(X= a
)=P
P(Y=bj)=P
EMBED Equation.3 j=1,2……
Y
0
1
2
…………..
19
20
P
…………..
33.
=8
34.因为可以看成是9重见利试验,EX=np=9
35.参考课本P84的证明过程.
36.
X2
X1
0
1
2
p
0
1
p
Cov(x1, x2)=E X1 X2-Ex1 ×Ex2
=
=
37.
所以:
因为
,所以X,Y独立.
故
38.
所以
因为 X1,Y1 独立。
所以 Cov(X1,Y1)=0
(也可以计算:Cov(X1,Y1)= Cov(X+Y,X-Y)
= Cov(X,X-Y)+Cov(Y,X-Y)
= Cov(X,X)-Cov(X,Y)+Cov(y,x)-Cov(Y,y)
=
=0)
39.
所以:
所以:X,Y不独立。
=………
40.
41.
①
②
当
时,
最小为
当
时,
42.解
①设投资组合的收益率为r
,则
所以:
当x=1 时,
故
所以,对任意x,有
,所以,任意组合P都有风险。
2 若
=1时
设投资组合中数为X,则
即
此时
=0
当
选投资组合中权数x,使得
,
,此时
③不卖座即0<x<1,能在0<x<1上得到比证券A和B的风险都小的投资组合,意味着
的最小值在0<x<1达到。
由
所以
所以
故为使0<x<1则
解得:
且
如果
,则上述等价于
如果
,则上述等价于
综上:
当
时,可在不卖座的情况下获得比DA和DB都小的风险投资组合。
43. ①Er=0.1×(-3%)+1%×0.105+2%×0.175+3%×0.26+4%×0.125+5%×0.13+6%×0.065+7%×0.04=2.755%
②P(r=-3%/rf=1.5%)=
P(r=1%/rf=1.5%)=
P(r=2%/rf=1.5%)=
P(r=3%/rf=1.5%)=
P(r=4%/rf=1.5%)=
P(r=5%/rf=1.5%)=
P(r=6%/rf=1.5%)=
P(r=7%/rf=1.5%)=
所以:E(r/ rf=1.5%)=0.05×(-3%)+0.1×1%
+0.2×2%+0.3×3%+0.15×4%+0.1×5%+0.05×6%+0.05×7%=3%
44.
,0<x≤y≤1
(0<x<1)
所以:
46.看成伯努利试验,X~b(120,
)→X~P(6)泊松分布or X~ N(6,5.7),A=“X≥10”
=采用泊松分布或正态分布近似计算
=0.0465(二项分布计算结果)
P(A)=0.022529+0.011264+0.005199+0.002228+0.000891+0.000334+0.000118+0.000039
+0.000012+0.000004+0.000001=0.042619---泊松分布
P(A)=1-F(10)=1-Ф0[(10-6)/σ] σ2=5.7
=1-Ф0[1.675415827737….]=1-0.95352 or 0.95244=0.04648 or 0.04756
显然,本题正态分布比泊松分布更准确。
47.X=开动生产的机床数 X~b(200,0.7) 所以X~N(140,42)
设以95%以上概率保证正常生产机器为K台,则
P(X≤K)≥0.95
所以
所以K=151台
所以 各电能≥K×15=2265个电能单位,以95%保证机器都正常
48.
X1, X2相互独立,
,X的均值为0,所以密度函数关于原点对称
(1)
=
(2)X=X1+…+XN~N(0,
)
所以 n=440.
� EMBED Equation.3 ���
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