设随机变量X与Y相互独立， X在区间[ 3 , 9 ] 上服从均匀分布 -1 1-|||-Y~( dfrac (1)(3) 2,那么 D ( 2X-3Y ) =()A. 4B.10C.24

步骤 1：计算X的期望和方差
X在区间[3, 9]上服从均匀分布，其概率密度函数为
\[ f(x) = \left\{ \begin{array}{ll} \frac{1}{6}, & 3 < x < 9 \\ 0, & \text{else} \end{array} \right. \]
X的期望为
\[ E(X) = \int_{-\infty}^{+\infty} x f(x) dx = \int_{3}^{9} \frac{x}{6} dx = 6 \]
X的二阶矩为
\[ E(X^2) = \int_{-\infty}^{+\infty} x^2 f(x) dx = \int_{3}^{9} \frac{x^2}{6} dx = 40 \]
根据期望与方差的关系，X的方差为
\[ D(X) = E(X^2) - [E(X)]^2 = 40 - 6^2 = 4 \]

步骤 2：计算Y的期望和方差
Y的分布为
\[ Y = \left\{ \begin{array}{ll} -1, & \text{概率} \frac{1}{3} \\ 1, & \text{概率} \frac{2}{3} \end{array} \right. \]
Y的期望为
\[ E(Y) = -1 \times \frac{1}{3} + 1 \times \frac{2}{3} = \frac{1}{3} \]
Y的二阶矩为
\[ E(Y^2) = (-1)^2 \times \frac{1}{3} + 1^2 \times \frac{2}{3} = 1 \]
根据期望与方差的关系，Y的方差为
\[ D(Y) = E(Y^2) - [E(Y)]^2 = 1 - \left(\frac{1}{3}\right)^2 = \frac{8}{9} \]

步骤 3：计算D(2X-3Y)
由于X与Y相互独立，所以
\[ D(2X-3Y) = 4D(X) + 9D(Y) = 4 \times 4 + 9 \times \frac{8}{9} = 16 + 8 = 24 \]