武汉科技大学2020年《831概率论与数理统计》考研专业课真题试卷【答案】

5.
X ~ N ( 2,0.42 ) E( X 3)2
1.16
6. X1, X 2 , X3, X 4 Y (X1 X2)2 (X3 X4)2,
N (0, 22 )
C
1/8
CY 2 (2)
1. 1000
(9
5% 1
10
0.25%
90 )
2
1A
B
P(B) P(A)P(B | A) P( A)P(B | A)
3
1
12
0, x 1
F x) 1 x3, 1 x 1
1
2
1, x 1
1 (3) P 0 X 1) F (1) F(0)
2
3.
XY
,1
a,b, c, d, e
(X,Y)
2
2
XY
3
7
Y
X
y1 y2 y3 pi
x1
a1b 8
x2
1cd 8
1
pj
6
e
1 a 1 ,b 1 ,c 3,d 1 e 1
(5
24 12 8 4
X x1 x2 p1 3
44
Y y1 y2 y3 p111 623
4.
X ,Y )
6x, 0 x y 1
f (x, y)
;
0,
1 X,Y)
2
X,Y
1 x<0 x>1
0 x 1 fX (x)
XY
X
XY
fX (x), fY ( y)
fX (x) 0
1
f (x, y)dy 6xdy 6x(1 x). x
6x 6x2 , 0 x 1,
2
( + )= + +2Cov( , )=1+4+2*(-1)=3
2
Cov( - , + )= - =1-4= -3
2
Cov( X Y , X Y )
3
3
X Y,X Y
D(X Y ) D(X Y )
7* 3
21
2
X -Y, X +Y)
7 -3 -3 3
1
-3
21
-3
1
21
2
7.
X
(a 1)xa f (x)
fX (x)
0,
.
2 3
(1 ) (2 ) (1 )
y<0 y>1 fY (y) 0
1
0 y 1 fY (y)
f (x, y)dx
y
6xdx
0
3x 2
|
y 0
3y2.
2
3y2 , 0 y 1,
XY
Y
fY (y)
1
0,
.
2
f (1/2, 1/2) 3/2
XY
fX (1/2) fY (1/2) (3/2)*(3/4) 9/8 f (1/2, 1/2) 2
0,
1k
2X
F x)
3 p(0 X 1) .
(1)
+
f (x)dx
1 kx2dx 2 1 kx2dx
kx3 1 2
2k
1
-
-1
0
30 3
3
k3 2
1
(2)
x 1 F x) 0
1x1
F x)
x
f (r)dr
x 3 r 2dr 1 x3
02
2
x1
F x)
x
f (r)dr
1
f (r)dr
1 3 r 2dr 1
ce x , x
f (x)
( 0, c
)
0, x
p X< +a a 0
(C)
A. a
C.
a
B. a
D.
a
4.
X
C.
A. 2 f X ( 2y)
f X (x) Y 2X
y
B.
fX (
) 2
C.
1 2
fX
(
y) 2
Y
fY (y)
D.
-1 2
fX
(
y) 2
5.
XY
D
D={x, y | x2 y2 1}
B
1
(6
4
24 )
1. A,B
P(AB) P(AB) P(B) 1-p P( A) p
2.
P(A)=P(B)=P(C)=0.25 P(AC)=0 P(AB)=P(BC)=0.15 A B C
3.
XY
X ~ U(1, 4) ,Y
1 e
1y
5 ,y
0
f (y) 5
0,
D 2X 3Y 10)= 228
4. X N (2, 2 ) , P{2 x 4} 0.3 , P{x 0} ___0.2______
2020
831
A)
(6
4
24 )
1. P(A B) 1
(D)
A. A B
B. B A C. B A
D. P(B A) 0
2 f (x) F(x)
(B) A. f ( ) 1 f (x)dx
0
C. F ( ) F ( )
3
X
X
f (Fra Baidu bibliotek) f ( x)
1
B. F ( )
f (x)dx
20
D. F( ) 2F( ) 1
0
5
7
0 x1
X1, , Xn X
a
n
n
f (x1, x2 , , xn )
[(a 1)xai ] (a 1)n xai
i1
i1
3
n
ln f (x1, , xn ) n ln( a 1) a ln( xi )
2
i1
a
ln f a
n
n
a1
ln( xi ) =0
i1
3
n
a 1n
2
ln( xi )
i1
8.
7
A. X
[-1,1]
C. X Y
B. X Y D. COV (X ,Y 0
6. ( X1, X 2 , , X n )
N(2, 22)
X
A
A.
1n 4 i 1 (Xi
2)2 ~
2 (n)
B.
1n 4 i 1 (Xi
2)2 ~ F(n, 1)
C. X 2 ~ N (0, 1) 2/ n
D. X 2 ~ t(n) 2/ n
3
2
7
0.5 0.05 0.5 0.0025 0.02625
2
2
P(A)P(B | A) P(A)P(B | A)
P(A)P(B | A)
P(A| B)
P(B)
1 P(B) 1 [P(A)P(B | A) P(A)P(B | A)]
3
0.5 0.95
0.4878
1 0.02625
2
2.
X
kx2, 1 x 1 f (x)
x yz
z1 1
dx dy
0
0
1
zx
dx dy
2z
z2
1
z1 0
2
z2
11
F(z) f (x, y)dxdy dx dy 1 00 x yz
Z
fZ (z) F (z)
z, 0 z 1 2 z, 1 z 2 0,
11
6.
X ,Y )
V
1 4,
X -Y, X +Y)
( - )= + -2Cov( , )=1+4-2*(-1)= 7
S
L1, L2
L1, L2
,( )
S
Z
XY
L1 L2
L
Z min (X, Y)
2
z 0 F Z (z) P (Z z) P (min (X, Y) z) 0
2
z>0 F Z (z) P (Z z) P (min (X, Y) z) 1 P (min (X, Y)>z)
2
1 P (X>z, Y>z) 1 P (X>z)P (Y>z) 1
e xdx
e y dy 1 e ( )z
z
z
2
L
Z
d
(
)e ( )z , z 0
f Z (z) dz FZ (z) 0,
z0
2
9.
9
710(
cm).
4
7
5. X Y .
( X ,Y )
[0 1] f (x, y) 1, 0 x 1,0 y 1
Z z0
F(z) P( X Y z) f (x, y)dxdy
xyz
F(z) 0
Z XY f (x, y) 0
0 z1
z
zx
z2
F (z) f (x, y)dxdy dx dy
xyz
0
0
2
1z 2
F (z) f (x, y)dxdy