当0<x<2时,A(m)=(x,y)1(2x)-xy<x fs(x) 其他 (3)E(X)=(x212)=4/3,E()=4)h=0 E(XY)=[(/4)dy=0, cos(X, Y)=E(XY)-E(X)E(Y)=0 所以ξ与n不相关 3、由题设知 0 q p q 2pq p P(5+n=0,g=0) P(5+n=0)P(c=0) P(2+n=0,s=1)=p2=P(+n=0)P(=1) P(5+n=1,=0)=2pq2=P(+n=1)P(=0); P(5+n=1,=1)=2pq2=P(5+n=1)P(c=1); P(+1=2,s=0)=pq2=P(5+n=2)P(=0) P(+n=2,s=1)=p3=P(5+=2)P(s=1 所以5+n与5相互独立 4、1)F(1,1) +( 442丌44 2)P(5≤0,n≤1)=F(0,1) 3)F()=acnx+1,F,(y)=1 arctan y+1,s与n独立 P:(2)=1 0<z<1 0 ≤0 dx=e(1-e2)/2x≤0 f() ax=(1-e2)/2 7、设5 第i台彩电为次品且未被查出 i=1~2×10 0 其 他 E(5)=5×10°,D()=5×10°(1-5×10-) 经检验后的次品数n=∑5,E(n)=1,D(m)=1-5×10°, 由中心极限定理,近似地有n~N(,1-5×10-) P(>3)=1-P(≤3)≈1-Φ √1-5×10 1-Φ(2)=0.02284 当 0 x 2 时, − = = 0 其 他 1/(2 ) ( ) ( , ) ( ) x x y x f x f x y f y x (3) = = 2 0 2 E(X ) (x / 2)dx 4 / 3, = = − 2 0 ( ) ( / 4) 0, x x E Y dx y dy = = − 2 0 ( ) ( / 4) 0, x x E XY xdx y dy cos(X ,Y) = E(XY) − E(X )E(Y) = 0 所以 与 不相关. 3、由题设知 0 1 + 0 1 2 P q p P 2 q 2 pq 2 p ( 0, 0) ( 0) ( 0) 3 P + = = = q = P + = P = ; ( 0, 1) ( 0) ( 1) 2 P + = = = pq = P + = P = ; ( 1, 0) 2 ( 1) ( 0) 2 P + = = = pq = P + = P = ; ( 1, 1) 2 ( 1) ( 1) 2 P + = = = pq = P + = P = ; ( 2, 0) ( 2) ( 0) 2 P + = = = pq = P + = P = ; ( 2, 1) ( 2) ( 1) 3 P + = = = p = P + = P = . 所以 + 与 相互独立. 4、1)F(1,1)= 2 1 4 4 + ) 4 4 ( 2 1 + + 4 1 =16 1 ; 2)P( 0, 1)=F(0,1)= 8 3 ; 3)F (x) = arctan x 1 + 2 1 ,F ( y) = arctan y 1 + 2 1 , 与 独立. 5、 p (z) = − − − − 0 0 1 0 1 ( 1) 1 z e z e e z z z . 6、 = − = − = − − − − − − 0 2 (1 )/ 2 0 2 (1 )/ 2 0 ( ) 1 / 2 ( 2 ) 2 1 0 ( 2 ) 2 z e dx e z e dx e e z f z z z x z z x z 7、设 = 其 他 第 台彩电为次品且未被查出 0 1 i i 5 i = 1 ~ 210 6 ( ) 5 10− E i = , ( ) 5 10 (1 5 10 ) −6 −6 D i = − 经检验后的次品数 = = 5 2 10 i 1 i ,E() = 1, 6 ( ) 1 5 10− D = − , 由中心极限定理,近似地有 ~ (1, 1 5 10 ) −6 N − 1 (2) 0.0228. 1 5 10 3 1 ( 3) 1 ( 3) 1 6 − = − − = − − − P P