1( = @ 1( > 2( > 3( 4( > 2( 45 % @ @ > > > @ > > = > 1( 180 2( 162 3 ...

112n2nE (2) nδ2∑ Xi ) = ∑ E(Xi =1 1. 2 (10. 2 (11. 1 (12P H (H0)= P(x = 01 , ) = 11p = 0/3⎛4⎞0 0 ⎛4⎞⎜ ⎟(0/3)( 0/7)+ ⎜ ⎟(0/3)1( 0/7)3= 6517⎝0⎠⎝1⎠n = 1001p =2x 1−z = 100 2 > 2 ⇒ x = 601×12 2100xZ =Z >− µ 0 x − 8=s 2n 8z = 2. 2 (13. 3 (14: . 1 (15Z 0/99 x − 8> 2/33 ⇒ x > 8 / 5814Z > Zα = φ−1( 1−Z : α) . 2 (16x − µ 0 1sZ = > φ−( 1−α) ⇒ x > φ−1( 1−α) + µs0nn. . 2 (17. 3 (18