题目内容
(2012•温州二模)已知ξ旳分布列如表所示,若η=3ξ+2,则Eη=
.
| 15 |
| 2 |
| 15 |
| 2 |
| ξ | 1 | 2 | 3 | ||||
| p |
|
t |
|
分析:由题设先求出t,再求出Eξ,再由η=3ξ+2,利用Eη=3Eξ+2能求出结果.
解答:解:由题设知t=1-
-
=
,
Eξ=1×
+2×
+3×
=
,
∵η=3ξ+2,
∴Eη=3Eξ+2=3×
+2=
.
故答案为:
.
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 6 |
Eξ=1×
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 3 |
| 11 |
| 6 |
∵η=3ξ+2,
∴Eη=3Eξ+2=3×
| 11 |
| 6 |
| 15 |
| 2 |
故答案为:
| 15 |
| 2 |
点评:本题考查离散型随机变量的性质和应用,是基础题.解题时要认真审题,仔细解答.
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