概華论与款程统外 -2 0 1 3 例4设X~ 1 1 1 ,求D(2X3+5), 2 12 12 解D(2X3+5)=D(2X3)+D(5) =4D(X3) =4[E(X6)-(E(X3)2] (V-(x2 1261 解 (2 5) (2 ) (5) 3 3 D X + = D X + D 4 ( ) 3 = D X 4[ ( ) ( ( )) ] 6 3 2 = E X − E X 121 3 121 1 21 0 31 ( ) ( 2) 6 6 6 6 6 E X = −  +  +  +  , 6 493 = , (2 5). 121 121 21 312 0 1 3 ~ 3 +   − 例 4 设 X 求 D X