题目
设随机变量X的分布律为-10 3-|||-P 0.3 0.5 a,其中a为未知参数,则方差-10 3-|||-P 0.3 0.5 a_______.
设随机变量X的分布律为
,其中a为未知参数,则方差
_______.
题目解答
答案
一维离散型随机变量分布律的归一性,即
,则
,则
,
一维离散型随机变量X的数学期望为
,则
,
,
,则
,
则X的方差为
.
解析
步骤 1:确定未知参数a
根据离散型随机变量分布律的归一性,即所有概率之和等于1,我们有:
$0.3 + 0.5 + a = 1$
解得:$a = 0.2$
步骤 2:计算数学期望E(X)
根据数学期望的定义,我们有:
$E(X) = (-1) \times 0.3 + 0 \times 0.5 + 3 \times 0.2$
$E(X) = -0.3 + 0 + 0.6$
$E(X) = 0.3$
步骤 3:计算$E(X^2)$
根据$X^2$的分布律,我们有:
$P(X^2 = 0) = P(X = 0) = 0.5$
$P(X^2 = 1) = P(X = -1) = 0.3$
$P(X^2 = 9) = P(X = 3) = 0.2$
因此,$E(X^2) = 0 \times 0.5 + 1 \times 0.3 + 9 \times 0.2$
$E(X^2) = 0 + 0.3 + 1.8$
$E(X^2) = 2.1$
步骤 4:计算方差D(X)
根据方差的定义,我们有:
$D(X) = E(X^2) - [E(X)]^2$
$D(X) = 2.1 - (0.3)^2$
$D(X) = 2.1 - 0.09$
$D(X) = 2.01$
根据离散型随机变量分布律的归一性,即所有概率之和等于1,我们有:
$0.3 + 0.5 + a = 1$
解得:$a = 0.2$
步骤 2:计算数学期望E(X)
根据数学期望的定义,我们有:
$E(X) = (-1) \times 0.3 + 0 \times 0.5 + 3 \times 0.2$
$E(X) = -0.3 + 0 + 0.6$
$E(X) = 0.3$
步骤 3:计算$E(X^2)$
根据$X^2$的分布律,我们有:
$P(X^2 = 0) = P(X = 0) = 0.5$
$P(X^2 = 1) = P(X = -1) = 0.3$
$P(X^2 = 9) = P(X = 3) = 0.2$
因此,$E(X^2) = 0 \times 0.5 + 1 \times 0.3 + 9 \times 0.2$
$E(X^2) = 0 + 0.3 + 1.8$
$E(X^2) = 2.1$
步骤 4:计算方差D(X)
根据方差的定义,我们有:
$D(X) = E(X^2) - [E(X)]^2$
$D(X) = 2.1 - (0.3)^2$
$D(X) = 2.1 - 0.09$
$D(X) = 2.01$