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

94 4. 0/5 3 . Z 1 .P ( 0 < Z < 1/5)= 0/4332 (1)0/5(2) (3)0/5(4)z 10z 1− 20z 10z 11/5. 4(8− 9/6 −1029/6 −10P( −9/6 < X < 29/6)= P( < Z < ) = P( −1/96 < Z < 1/96)1010= 2φ1/96 −1= 0/95P ( −2 < Z < 2)= 2φ2−1= 2×0/9772 −1= 0/9544. 3 (9. . 4 (10 nq. δ =n = 200 > 20= φ1/62 = 0/9463> 5 ,np > 5,n > 20npq µ = np np = 200×0/3 = 60 > 5 ⇒ µ = 60(X −0/5)− µP(X ≥ 50)= P(Z ≥) = P(Z ≥δ49/50−60) = P(Z ≥ −1/62)200×0/3 × 0/7. 3(11. 1(12124 −1016 −10P( 4 < x < 16)= P( < Z < ) = P( −1