e x> 例1设Xf(x)=1n! 用切比雪夫不等式证明 x≤0. n P{0<X<2(n+1)}≥ n+1 证EX= x +00 e +(n+ to dx=n+1 0n1 1+2 n+1 2 x EX xf-e dx e+(n+2 e-dx=(n+1)(n+2) n n! 所以DX=EX2-(EX)2=(n+2(n+1)-(n+1)2=n+1 从而P{0<X<2(+1)}=P{X-EXkn+1} n+1 n+1 (这里E=n+1) 欐率统计(ZYH) ▲区u概率统计（ZYH） 例1 设X~ , 0, ( ) ! 0 , 0, n x x e x f x n x −    =     用切比雪夫不等式证明 1 {0 2( 1)} +   +  n n P X n 证 0 e d ! n x x EX x x n + − =  2 2 0 e d ! n x x EX x x n + − =  所以 P{0  X  2(n + 1)} = P{| X − E X | n + 1} 2 ( 1) 1 1 + +  − n n + 1 = n n 2 2 2 DX EX EX n n n = − = + + − + ( ) ( 2)( 1) ( 1) = n + 1 = (n + 1)(n + 2) (这里 = + n 1) 1 0 0 e ( 1) e d ! ! n n x x x x n x n n + + + − − = − + +  2 1 0 0 e ( 2) e d ! ! n n x x x x n x n n + + + + − − = − + +  = n + 1 从而