1
概率论与数理统计(第二版
.
刘建亚)习题解答——第四章
4
?/p>
1
解:
E
(
X
)
=
1
´
0.25
+
2
´
0.4
+
3
´
0.2
+
4
´
0.1
+
5
´
0.05
=
2.3
4
?/p>
2
解:
?/p>
D
(
X
)
=
E
(
X
2
)
-
[
E
(
X
)]
2
?/p>
4
?/p>
3
解:
E(X)=0.3003
?/p>
E(X
2
)=0.4086
?/p>
D(X)=0.3184
?/p>
[
D(X)]
1/2
=0.5643
?/p>
4
?/p>
4
解:
E
(
X
*
)
=
E
X
-
E
(
X
)
=
1
E
[
X
-
E
(
X
)]
=
1
[
E
(
X
)
-
E
(
X
)]
=
0
D
(
X
)
D
(
X
)
D
(
X
)
2
D
(
X
*
)
=
E
(
X
*
)
2
-
[
E
(
X
*
)]
2
=
E
(
X
*
)
2
=
E
X
-
E
(
X
)
=
1
E
[
X
-
E
(
X
)]
2
=
1
D
(
X
)
=
1
D
(
X
)
D
(
X
)
D
(
X
)
4
?/p>
5
解:
+¥
1
x
E
(
X
)
=
-¥
xf
(
x
)
dx
=
-
1
p
1
-
x
2
dx
=
0
E
(
X
2
)
=
-+¥¥
x
2
f
(
x
)
dx
=
-
11
p
1
x
-
2
x
2
dx
=
01
p
21
x
-
2
x
2
dx
E(X)
E(X
2
)
D(X)
X
1
50
2501
1
X
2
50
2502
2
?/p>
D(X
1
)
<
D
(X
2
)
?/p>
用甲法测定的精度高?/p>
X
0
1
2
3
P
0.75
0.2045
0.0409
0.0045