上海海事大学概率论期末试题

z0.001 = 3.090 t0.025(28) = 2.0484 t0.05(15) = 1.7531
χ20.005(15) = 32.801 χ20.95(15) = 7.261 F0.025(16, 11) = 3.28
Φ(2) = 0.9772 z0.10 = 1.282 t0.05(28) = 1.7011 t0.05(16) = 1.7459
−
x¯)2
=
98.72592(
2).
µ, σ2
.
α = 0.05 , :
(1)
225( );
(2)
1002( 2).
8.
,
44 , 35 , 21 ,
?
100 , α = 0.05 ,
7
( B): 20080603
8 ( 1, 2, 5, 8
10 , 3, 4, 6, 7
15 )
1. P(AB) = P(A) + P(B) − P(A ∪ B) = 0.1,
44
1 3
100 3
58.08
35
1 3
100 3
36.75
108.06-100=8.06
21
1 3
100 3
13.23
: χ2 ≥ χ20.05(3 − 1) = 5.991. χ2 = 8.06 > 5.991,
α=
0.05
H0.
(1) P(A − B) = P(A) − P(AB) = 0.1;
(2)
P(A¯|B¯) =
1−P(A∪B) 1−P(B)
=
2 3
.
2. A1
1
, A2
1
, A3
1
,B
2
.ቤተ መጻሕፍቲ ባይዱ
P(A1)
=
(42) (62)
=
2 5
,
P(A2)
=
(21)(41) (62)
=
8 15
,
P(A3)
=
(22) (62)
=
1 15
5.
100
.
100 .
.
(1) 100 (2) 100
;
9000
.
6. (X1, · · · , Xn) θ>1
X
,X
f (x)
=
2θ(1 + 2x)−θ−1,
x > 0,
0,
,
.:
(1) θ (2) θ
7. ),
; .
16 N(µ, σ2),
,
x¯
=
1 16
xi
=
241.5(
s2
=
1 15
(xi
=
1 6
.
3. (1) B = 100;
(2)
P(X ≤ 150) =
1 3
;
(3) 3
19 27
.
Y
∼
b(3,
p),
p
=
1 3
.
P(Y
≥
1)
=
1
−
P(Y
=
0)
=
1−
8 27
=
4.
(1)
fX ( x)
=
1, 0,
0 < x < 1, ,
fY (y)
=
− ln(1 − y), 0,
0 < y < 1, ;
H0.
8. H0 : P(X =
) = P(X =
) = P(X =
)
=
1 3
.
Ai fi pi npi fi2/npi χ2 = fi2/npi − n
29
1 3
100 3
25.23
35
1 3
100 3
36.75
100.86-100=0.86
36
1 3
100 3
38.88
: χ2 ≥ χ20.05(3 − 1) = 5.991. χ2 = 0.86 < 5.991,
(2)
0<x<1
,
fY |X (y| x)
=
1 1−x
,
0,
x < y < 1, ;
(3)
E(XY) =
5 12
.
5. (1) Xi
i
, i = 1, · · · , 64.
E[
64 i=1
Xi]
=
6400,
D[
64 i=1
Xi]
=
640000;
(2)
, P(
64 i=1
Xi
≥
7000)
=
P(
X−6400 800
.
,
(1)
;
(2)
,
.
3.
(: )
X,
f (x)
=
Bx−2,
x > 50,
0,
.
(1)
B;
(2)
150
;
(3) .
3 3
,
150
1
.
6
4.
(X, Y)
f (x, y) =
1 2(2−x)
,
0 < x < y < 2,
0,
,
:
(1) X
Y
;
(2)
fY |X (y| x);
(3) E(XY).
χ20.005(16) = 34.267 χ20.95(16) = 7.962 F0.025(17, 12) = 3.14
1. A, B
, P(A) = 0.4, P(B) = 0.6, P(A ∪ B) = 0.9 :
(1) P(A − B); (2) P(A¯|B¯).
2.
5 ,4
;
2,
.
4
,5
2).
µ, σ2
.
α = 1% ,
(1)
2.12( );
(2)
0.022( 2).
8.
,
29 , 35 , 36 ,
?
100 , α = 0.05 ,
3
( A): 20080603
8 ( 1, 2, 5, 8
10 , 3, 4, 6, 7
15 )
1. P(AB) = P(A) + P(B) − P(A ∪ B) = 0.4,
.
P(B|A1)
=
(22) (62)
=
1 15
,
P(B|A2)
=
(32) (62)
=
1 5
,
P(B|A3)
=
(42) (62)
=
2 5
.
(1) (2) Bayes
,
P(A1 B)
=
P(B|A1)P(A1)
=
2 75
;
,
P(A1|B)
=
P(B|A1)P(A1) P(B|A1)P(A1)+P(B|A2)P(A2)+P(B|A3)P(A3)
ln(1+2Xi
)
8
7. (1) H0 : µ ≤ 225, H1 : µ > 225.
:
t
:
t
=
x¯−√µ0 s/ n
≥
t0.05(n − 1)
=
t0.05(15)
=
1.7531
.
t = 241.5−2√25 = 0.6685 < 1.7531.
98.7259/ 16
α = 0.05
H0.
(2) H0 : σ2 ≥ 1002, H1 : σ2 < 1002.
α=
0.05
H0.
5
( B): 20080603
8 ( 1, 2, 5, 8
10 , 3, 4, 6, 7
15 )
Φ(0.25) = 0.5987
: Φ(0.75) = 0.7734
Φ(1) = 0.8413
z0.05 = 1.645 t0.025(27) = 2.0518 t0.025(29) = 2.0452 t0.005(15) = 2.9467 χ20.995(15) = 4.601
:
t : |t| =
x¯−√µ0 s/ n
≥ t0.005(n − 1) = t0.005(15) = 2.9467 .
|t| =
2.125−2√.12 0.017/ 16
= 1.18 < 2.9467.
α = 0.01
H0.
(2) H0 : σ2 = 0.022, H1 : σ2 0.022.
:
χ2
1
( A): 20080603
8 ( 1, 2, 5, 8
10 , 3, 4, 6, 7
15 )
Φ(0.25) = 0.5987
: Φ(0.75) = 0.7734
Φ(1) = 0.8413
z0.05 = 1.645 t0.025(27) = 2.0518 t0.025(29) = 2.0452 t0.005(15) = 2.9467 χ20.995(15) = 4.601
.
; ,
3.
(: )
X,
f (x) =
Bx−2,
x > 100,
0,
.
(1)
B;
(2)
150
;
(3) .
3 3
,
150
1
.
2
4.
(X, Y)
f (x, y) =
1 1−x
,
0 < x < y < 1,
0,
,
:
(1) X
Y
;
(2)
fY |X (y| x);
(3) E(XY).
5.
100
.
64 .
z0.001 = 3.090 t0.025(28) = 2.0484 t0.05(15) = 1.7531
χ20.005(15) = 32.801 χ20.95(15) = 7.261 F0.025(16, 11) = 3.28
Φ(2) = 0.9772 z0.10 = 1.282 t0.05(28) = 1.7011 t0.05(16) = 1.7459
:
χ2
=
(n−1)s2 σ20
≤ χ20.995(n − 1) = χ20.995(15) = 4.601
χ2
=
(n−1)s2 σ20
≥
χ20.005(n − 1) = χ20.005(15) = 32.801 .
χ2
=
15×0.0172 0.022
=
11
∈
(4.601, 32.801).
α = 0.01
χ20.005(16) = 34.267 χ20.95(16) = 7.962 F0.025(17, 12) = 3.14
1. A, B
, P(A) = 0.5, P(B) = 0.7, P(A ∪ B) = 0.8 :
(1) P(A − B); (2) P(A¯|B¯).
2.
6
,4
.
2
,
.
2
.
(1) (2)
:
χ2
:
χ2
=
(n−1)s2 σ20
≤
χ20.95(n − 1)
=
χ20.95(15)
=
7.261
.
χ2
=
15×98.72592 1002
=
14.6202
>
7.261.
α = 0.01
H0.
8. H0 : P(X =
) = P(X =
) = P(X =
)
=
1 3
.
Ai fi pi npi fi2/npi χ2 = fi2/npi − n
(1) P(A − B) = P(A) − P(AB) = 0.3;
(2)
P(A¯|B¯) =
1−P(A∪B) 1−P(B)
=
1 4
.
2. A1
1
, A2
1
, A3
1
,B
2
.
P(A1) =
(52) (92)
=
5 18
,
P(A2)
=
(51)(41) (92)
=
5 9
,
P(A3)
=
(42) (92)
=
1 6
≥
9000−10000 1000
)
≈
Φ(−1)
=
1 − Φ(1) = 1 − 0.8413 = 0.1587.
6.
(1)
E(X) =
1 2(θ−1)
,
θˆ M M
=
1/2+X¯ X¯
;
(2) ln L(θ) = n ln(2θ) − (θ + 1)
n i=1
ln(1
+
2Xi),
θMLE
=
. n
n i=1
χ20.05(2) = 5.991 F0.025(11, 16) = 2.94 F0.025(7, 6) = 5.70
z0.0025 = 1.960 t0.05(27) = 1.7033 t0.05(29) = 1.6991 t0.005(16) = 2.9208 χ20.995(16) = 5.142 χ20.05(3) = 7.815 F0.025(12, 17) = 2.82 F0.025(6, 7) = 5.12
χ20.05(2) = 5.991 F0.025(11, 16) = 2.94 F0.025(7, 6) = 5.70
z0.0025 = 1.960 t0.05(27) = 1.7033 t0.05(29) = 1.6991 t0.005(16) = 2.9208 χ20.995(16) = 5.142 χ20.05(3) = 7.815 F0.025(12, 17) = 2.82 F0.025(6, 7) = 5.12
.
(1) 64 (2) 64
;
7000
.
6. (X1, · · · , Xn) θ>1
X
,X
f (x)
=
θ(1 +
x)−θ−1,
x > 0,
0,
,
.:
(1) θ (2) θ 7. 2.125( ),
:
; . 16 ,
N(µ, σ2),
x¯
=
1 16
xi
=
s2
=
1 15
(
xi
−
x¯)2
=
0.0172(
=
1 2−x
,
0,
x < y < 2, ;
(3)
E(XY) =
5 3
.
5. (1) Xi
i
, i = 1, · · · , 100.
E[
100 i=1
Xi]
=
10000
=
104,
D[
100 i=1
Xi]
=
1000000
=
106;
(2)
, P(
100 i=1
Xi
≥
9000)
=
P(
X−10000 1000
.
P(B|A1)
=
6 11
,
P(B|A2)
=
5 11
,
P(B|A3)
=
4 11
.
(1) (2) Bayes
,
P(A1 B)
=
P(B|A1)P(A1)
=
5 33
;
,
P(A1|B)
=
P(B|A1)P(A1) P(B|A1)P(A1)+P(B|A2)P(A2)+P(B|A3)P(A3)
=
15 46
.
3. (1) B = 50;
(2)
P(X ≤ 150) =
2 3
;
(3) 3
26 27
.
Y
∼
b(3,
p),
p
=
2 3
.
P(Y
≥
1)
=
1
−
P(Y
=
0)
=
1−
1 27
=
4.
(1)
fX ( x)
=
1 2
,
0,
0 < x < 2, ,
fY (y)
=
1 2
(ln
2
−
0,
ln(2
−
y)),
0 < y < 2, ;
(2)
0<x<2
,
fY |X (y| x)
≥
7000−6400 800
)
≈
1 − Φ(0.75)
=
1 − 0.7734 = 0.2266.
6.
(1)
E(X) =
1 θ−1
,
θˆ M M
=
1+X¯ X¯
;
4
(2) ln L(θ) = n ln θ − (θ + 1)
n i=1
ln(1
+
Xi),
θMLE
=
. n
n i=1
ln(1+Xi
)
7. (1) H0 : µ = 2.12, H1 : µ 2.12.