Логика. Множества. Вероятность. Лексаченко В.А. - 107 стр.

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=
1
, 2
1
,
5
− x
− 0, 5
x
2
, x ∈
[0; 1]
0, x /
∈ [0; 1]
.
3)
Mξ
=
∞
R
−∞
xf
ξ
(x)dx
=
1
R
0
1,
2
1
,
5
x
−
x
2
− 0,
5
x
3
dx
= 0
,
35.
4) Mξ
2
=
∞
R
−∞
x
2
f
ξ
(x
)
dx
=
1
R
0
1
,
2
1
, 5x
2
−
x
3
− 0
,
5
x
4
dx
= 0, 18
.
Dξ
=
Mξ
2
−
(
Mξ
)
2
= 0, 0575
.
5)
f
η
(y) =
∞
R
−∞
f
ξη
(
x, y
)dx =
R
∞
−∞
f
ξη
(
x, y
)
dx, y ∈
[0; 1]
R
∞
−∞
f
ξη
(
x, y
)dx, y /
∈ [0; 1]
=
=
1,
2
1−
y
R
0
(2
x
+ 3
y)
dx, y
∈ [0; 1]
0, y /
∈
[0; 1]
=
(
1
, 2
x
2
+ 3xy
1
−y
0
, y ∈
[0; 1]
0, y /∈ [0; 1]
=
=
1,
2(1 +
y −
2y
2
), y ∈ [0; 1]
0, y /
∈ [0; 1]
.
6) Mη
=
∞
R
−∞
yf
η
(
y)
dy =
1
R
0
1
,
2(
y
+ y
2
−
2
y
3
)dy
= 0
,
4
.
7)
Mη
2
=
∞
R
−∞
y
2
f
η
(
y
)
dy =
1
R
0
1, 2(y
2
+ y
3
−
2
y
4
)dy
= 0
, 22
Dη =
Mη
2
− (Mη
)
2
= 0
,
06
.
8)
M(ξη) =
∞
R
−∞
∞
R
−∞
xyf
ξη
(x, y)
dxdy
= 1
,
2
1
R
0
1
−x
R
0
(2x
2
y
+ 3xy
2
)dy
dx
=
= 1
,
2
1
R
0
(x
2
y
2
+
xy
3
)
1
−x
0
dx
= 1,
2
1
R
0
(
x
2
(1
− x
)
2
+ x
(1
− x
)
3
)
dx =
= 1
, 2
1
R
0
(x − 2
x
2
+
x
3
)dx
= 0,
1, K
ξη
= M
(ξη) −
MξMη
=
−0,
04
.
J
14 . Opredelit~ parametry normal~no$i sluqa$ino$i veliqiny
ξ :
m, σ , postonnu
k i
P
ξ
([a
; b
])
, esli
1)
f
ξ
(x
) = k
exp
−2x
2
+ 4
x
,
[a;
b
] = [0,
5; 1
,
5]
.
2) f
ξ
(
x) =
k
exp
−
x
2
/2
−
2
x
, [
a
;
b
] = [
−
3,
5; −
1
,
0].
3) f
ξ
(x
) = k exp
−
2
x
2
+ 2
x
,
[a
; b] = [−
0,
5; 0
,
5].
4) f
ξ
(x) =
k exp
−
2x
2
+ 8x
,
[
a
; b
] = [2, 0; 3,
0]
.
5)
f
ξ
(x) = k
exp
−
x
2
/
8
−
x/4
,
[a
;
b] = [
−3, 0; 1
,
0]
.
6) f
ξ
(
x
) =
k exp
−x
2
/2 + 3
x
, [a; b
] = [4, 0; 5
,
0].
7)
f
ξ
(x) =
k
exp
−8
x
2
+ 16
x
, [a;
b
] = [0
, 5; 1
,
5]
.
8) f
ξ
(
x
) = k exp
−x
2
/
2 − x
,
[a;
b
] = [−
1
,
5; 0
,
5]
.
9)
f
ξ
(
x) = k
exp
−2x
2
−
6x
,
[a
; b
] = [
−
1,
0; −0
,
5]
.
10)
f
ξ
(
x
) = k
exp
−
2
x
2
−
2x
, [
a; b] = [−
1
,
5; 0
,
5].
R e x e n i e dl f
ξ
(
x
) = k
exp
−9x
2
/2 + 9
x/
2
,
[a;
b] = [0; 1
,
0]
.
107

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