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

43 (4 (3 (1 (2 (1 %45 (290 (4 (426,23,25,29,31,2426 /5198 (3 162 (2180 (1 (3 (2 (4(4 (1 (3 (4 (325/5 (3 (2 (1 (527 (225 (1 20 3 , 3 , 4 (6 . . (4Y = 10 X + 4 . (4δ x = 2/1(4 84 (3. δ2Y = 25,µ Y =µ Xδ x= 41=2(3 48 (2 36 (124 Y (7µ X = 2/ 4 (2X µ X= 2(1 X 7 (8(3δx= 1(2δ x1=4(1

44 18,19,20,21,22 (922 /5(42/5 (3 2 (2 (1 (10 (4 (3 (2 (1 (11. . (3. . : , (12 100 (245 (4N (425∑ X ii= 1 (1 (3= 525X2i =i∑= 129 (3(1(2(420 n = 25 (13 21 (225 2∑(Xi − X)i=119 (1 1 25 (14 . 3 25 2 50( 30/5) (41,1,5,3,2,4,2,2 (4:(µ,δ2(30/25) (3) (20/5) (2(20/25) (1: (153 (31 (22 (1

45 20 100 (16( ) 150 -157157 -164164 -171171 -178178 -1851525302010. 167(2170(4166 / 3 (1168 / 2 δ 2 µ x a (17δ: xa(3δ 2(4(3(2(1aaa2a 64 (1816 (414 (3δ2δ 2: 12 (26 (1 0/75 17 (19 . 17 75 (1. 17 75 (2. 17 75 (3. 17 25 (4 90 15 (2090 (4 80 (370 (260 (1 X5 = −3,X4= −2,X3 = −1,X2 = 2,X1= −4 (21

4654b(440b(3. (2. (4 4∑3b(X2−1)2i=230b (27b(1: (22. (1. (3 100 (23x if ixi: . f i 0/2 0/1 0/4 0/34/7 (23/2 (1n6 (45 (3: (24 (4 0/818 (31 (21/2 (12 5 3 9 11 15 (2541865 (2 (411 (111 5 (3 (2611,13,12,12,15,11,10,14,13,13 (4230,235,235,250,245( 23567/5) (411 (417335184 (3 (2 (1: (x,S2) (27(23954) (3(23554) (2(23967/5) (1: 93711614 (287 (399 (28 (1 (29. (1

47 (2 (4 14 . . (2. (3. (4: (30 (1. (3∑fi= 40 (31 0/2 28 (4: 25 (3M3. (2. ∑ (X i − x)3=n(420 (222 (1 (32. (1. (3 120 – (33 80(496(3100 (280× 120 7 (3421 (43/5(3 . (4 12(314 (215 ,X4,X3,X2,X 1 3 (2( 1: 7(1 (353 , X3,X2,X 115(1K X ( 36. (2 . (1

48. (4 . ( 3 . 20 5 5 (37 17 (414(312/5 (29(1 n = 2 (38: .0 /25(40/5(31 (22(14X 8 16 (39: X 0 /25(40/5(31 (22(1: (H) (G) (X) (40. . (1. . (2(3() (412 4 3 (41. 16 . (1 4 (2. . 16(3 (4

49 . 2 (1α i= 360× ,, , = 360 × 0/45 = 162. 2 (2 4 (3 . 2 (4α i. . 3 (5. 2 25 + 2623 − 24 − 25 − 26 − 29 − 31⇒m = = 25 / 52= 360× : . 2 (6E(y)4α i = 360×= 48°30 y = ax + b x. δ y =aδxVar(y) = a2Var(x). . 1 (7E (y) = µ y = ax + b= aE(x) + b ⇒ 24 = 10E(x)+ 4 ⇒ µ x = E(x)= 2. 3 (8δ2y = 25 ⇒ δy= 5δy= aδx⇒ 5 = 10δx ⇒ δx=510=12. 3 (9

50X i123∑22 ∑(Xi−X)S =n−118 20219 20220 20221 20222 2022 ( − ) + ( − ) + ( − ) + ( − ) + ( − )S =5−125(2∑ X25 2 25 i )− =2−= 1∑(XX) ∑ Xiiii= 1i=1 nf in0/250/50/251fiXin0/2510/752f X2i in0/2522/254/552= 20−= 1925= 2/5. 3 (10. 4(11. 4 (12. 2 (13. 2 (14n∑ f i X iX =i=1= 2nn2n∑ f i X f i Xi∑ iδ2=i=1− (i=1)2= 4/5 − 22= 0/5n n150 -157157 -164164 -171171 -178178 -185. . 1 (15f i Fi15 1525 4030 7020 9010 100100. 1 (16

51 n 100= = 502 2n− Fi −1m =2ai + × lfi50−40m = 164 + × 7 = 166 / 330y = ax +b x . 3 (17a X δ y =αiaδ2a2δ2x δ y = xy = ax + b. 4-6-8-10-12 ( = 6 − 4 = 2) . 2 (18360×f ×= i 360 15= = 60°N 90. 3 (19. 1 (20. 4 (21[(22−1)2+ (( −1)2−1)2+ (( −2)2−1)2] b4b(X2)24b (X2∑ 3 −1= 3 ∑ −1)= 54i=2i=1 .X = m =( M < m

52X if inX2 ifif2 i X i60/2362072030/19109050/42540100040/316304801________2290n 2 2∑ fiXi− nxVar(x) =i=1n −12290−100×4/72Var(x) =100−1= 0/8182 222 ∑ f i X i − nx ( 20×36)+ ( 10×9)+ ( 40×25)+ ( 30×16)−100×4 / 7S === 0/818n −199. 115 . 3 (2513 . . 2 (26∑fX( X i i 124= = = 12/4) . n 10. X < M .. 1 (27∑ X 230+235 + 235 + 250+245X = i == 239n 5∑ −22 (X X)S = i ⇒n −1( 230−239)2+ ( 235−239)2+ ( 235−239)2+ ( 250−239)2+ ( 245−239)2= 67 / 55−1 . . 1 (28. 2 3-6-7-9 -11-147 + 9m = = 82. 4 (29: . 2 (30( = = ) . (1

53fi= ( < < ) . (2( < < ) . (3 i × ∑fi 1 (31F3 = 0/2×40 = 8 ⇒ F = F + F3⇒ F = 14 + 8 = 223 2 3. 1 (32 K =∑ f −3 i (x i x)Nδ3⇒K > 0⎯⎯→K = 0⎯⎯→K < 0⎯⎯→ n 1 1 2 1 1= + ⇒ = + ⇒ X = 96X H X1X2X H 80 120S′2SCV = = = 1⇒ X′2XX1 = X2= X3= X4= 1515 + 15 + 15 + 3X == 124n1X1+ n2X25 × 5 + 20×20X ==17n1+ n220+5CV xS = xµH ≤ G ≤ X8 = = 24. 3 (33. 1 (34: . 3 (35⇐ X ′ = X + K. 3 (36. 4 (37. 3 (38. 2 (39. 4 (40 4 3 . 1 (41 4 ( xi + 3)−12= 4xi: . 12 . . 4

54 2 5 (1 10 (472 (3120 (248 (1 (2 . 17 350/35 (2 0/52 (1 23 250/17 (40/6 (3 (30/17(4 . 12C 435(352 452(2120 12 0/ 52 (1 (4 ( ) 3 (412C 3(3n (211n(4 (3(2NnCNNn. 45(412(31412P 3(1N (5(21N(1, (6 34(1

55P(AU B)2 1P(B) = , P(A) =3 372(4(399 S( 816) (4P(A.B) 2(43 30/ 8 200 BA (7 1(21 (13= 2 , X =12 (88 9(-63) (3 (1014) (2 (618) (11 2P(B) = P(A) =3 32(39, 2 B (2A (9 (1 8 (10 . 31(4 (3(2 (1105 0/05 (11 10 (4 20 (3 2 (2 5 (1 0/91 S2 = 81 X = 75 (12 ( 20110/3) (4 P(A / B) 0 /833(4(45105) (3(25/980/9) (2P (A U B) = 0/9 P (B) = 0/ 80/752 (30/625 (2(20/989/1) (1P (A) = 0/6 (13: 0/551 (1

56 , 14 2 (14 (41 (3 51(2(13636. 5 (15111(4(3(2543 P (A / B) =P(A)P (A I B) = P(A)P(B)(2(4: P(AP (B/ A) =1(12A B (16I B) = 0P(B) (17 (4 (3 (2. .......(1(3 (1 5 (185 (420 (3108 (2120 (1 4 2 3 2 1 (19 . 5 : . 1341938(4(3(2(1794545 (20710(4(31919: B,A ,. (48(2195,P(A)=919B)(1P (B) = 0 / 0/3,P(AI = 0/. (3. (265 (21 (1 1 2 , 3 (22 .

576! (44× ! 3!(3 . 3 ! × 2!× 1!(23 ! × 3!6 4 B 3 3 A (23 . 1(41059,P(B)25,P(A(3P (A) = 0 / = 0/3 I B) = 0/0 /21(40/3 (3 (111(2(15221 B,A (24: P(A / B) 0/46 (20/69 (1 (2550 (465 (3 ) 256 (2336 (1 7 (26: . (:121(4221: P(AI B) 9(415 P(F)0 /25(4 615(31P(A / B) =3(3P (F / E) =0/75 (30/21621(24P(B) =5715P (F/ E) =(20/9 (20/ 7 16(14411P(A) = (2734(115P (E) = 0/9 (280/72 (175 25 (29 . 64 ( 75100) (4(6590) (3(6095) (2(5591) (1

58P (B/ A) =23514(464 (4(4.A? 0/95(4 P(A13I B)(324 (32 ( )631=2: (3 P(A)P(AI B)0/ 85(33 P(B) =412(2: 14 (26 ⎛6⎞ 1∑ ⎜ ⎟()x=0⎝x ⎠ 31 ( )2(231 P(A) =33 ⎛4⎞∑ ⎜ ⎟i=1⎝i⎠x( (3023(1 (318 (12)6−x3 B, AP (B/ A) = 0/ 90/045P (A) = 0/05(21181(1(1(32(33 (34 .. %0/93(485 و%‏ 98 900 / 20 0/30 0/و و 50‏%و .56(41 13 20/929(30/ 891(2 45(334:0/109 (1A, B P(AU B)(223(1(35

59 4 3 . 21 (1 . 3 4 . . (. (0/70/3 , (2 . %30 . . %20 ,

60 2! . 1 (1 . 4! 4 P ( )P ( / ) =P(H) = 3P(T)P(H) = 3 × P(T) = 3×= P(I )1434. 4× ! 2!35=10035=100=52100P(H) + P(T) = 1⇒3p(T)+ P(T) = 1⇒4p(T)= 1⇒P(T) =1P(A I B) = P(A) ⋅ P(B) = ×3=. 23=293552 . 2 (2. 3 (3. 3 (4. 3 (5. 1 (614B A . 4 (71 2 2 7U B) = P(A) + P(B) − P(A I B) = + − =3 3 9 9 . 1 (8P(AP (I )P()1: . ( 1− ) k2( µ − kδ,µ + kδ)

611 81−= ⇒ k = 3 ⇒ µ − kδ〈X〈µ + kδ ⇒ 12 − 6〈X〈12 + 6 ⇒ ( 6,18)k2 9P(AI B) = P(A) ⋅P(AI B)= P(A)P(B)P(B) . 3 (9=13×23=29: . . . 1 (10. . . E (x)= µ = np = 200 × 0/05 = 1011 11−= 0/91⇒0/09 = ⇒k2k2 k10X ± kδ ⇒ 75 ± × 9 = ( 45,105)3P(A I B) 0/5= = = 0/625P(B) 0/8. 4 (11. = 0/3 ⇒ k =103. 3 (12. 2 (13P(A U B) = P(A) + P(B) − P(A I B) ⇒ P(A I B) = 0/6 + 0/8 −0/9 = 0/5P(A / B) 6 . 4 (14 = n(S) = (n −1 )! = ( 5 −1)!= 24. 12 . 1 (15 2! . ( 4 −1)! 4

62n(A) = ( 4 −1)!× 2!= 12n(A)P(A) =n(S)=1224=125 ! =120. 1 (16. 2 (17. 1 (18P( 11) = ×2251+ ×249=210 2+2918 + 20= =903890=1945. 2 (19P ( 1 P(A U B) = P(A) + P(B) − P(A I B)P(A) ⋅ P(B) = 0/3 × 0/5 = 0/15: . 1 (201 2×9=2 5=19 1945. 2 (21⇒ P(A I B) = P(A) + P(B) − P(A U B) = 0/5 + 0/3 −0/65 = 0/15. BA P(AI B) = P(A) ⋅ P(B) . 3 (22 1 2 3! . 3 ! × 4! 6 B 4P (1) P() =P () 3 . 4! 1 A 3 ⋅ )1 . 1 (23

63 P ( ) = P( )3 4 3 6 1P ( ) = P( ) + P ( ) = × + + =6 6 6 10 2. 3 (24P(A I B) = P(B) − P(A I B) = 0/3 −0/2 = 0/09P(A / B)P(A I B) 0/09= = = 0/3P(B) 0/3. . 2 (258 ⎛8⎞8!C3 = ⎜ ⎟ = = 56 . ⎝3⎠3!(8 − 3)!. . 4 (26C k N−K= x CP(x) n − xCN n⎛2⎞⎛7⎞⎜ ⎟⎜⎟2 0 1×1 1=⎝ ⎠⎝⎠= = =⎛9⎞9!9×8×7!⎜ ⎟⎝2⎠2!× 7!2×7!136. 2 (274 1P(B) = 1 − P(B)= 1−=5 5P(A I B)1 1 1P(A / B) = ⇒ P(A I B) = P(A / B) ⋅ P(B)= × =P(B)3 5 151 1 1 7P(AU B) = P(A) + P(B) − P(A I B)= + − =3 5 15 15. 4 (28P (E) = 1 − P(E) = 1−0/9 = 0/1P(F)= P(F I E) + P(F I E) = P(F/ E) ⋅ P(E) + P(F/ E) ⋅ P(E)⇒ 0 / 2 × 0/9 + 0/7 × 0/1 = 0/25. 1 (29

641 3 1 3 11−= ⇒ = 1−= ⇒ k = 2k2 4 k2 4 4U = µ + kδ = 75 + 2×8 = 91L = µ − kδ = 75 − 2×8 = 59P(B) = 1−P(B) = 1−34P(A I B)= =P(A)14P(A I B) = P(A) + P(B) − P(A U B) =P(B / A). =∑ f (x) =1=11213=1413+14−12. 4 (304 + 3 − 6= =12112. 2 (31 . 1 (32P (A I B) = P(A) ⋅ P(B / A) = 0 / 05×0/9 = 0/0450 / 5×0/85 + 0/3 × 0/9 + 0/2×0/98 = 0/89112+131− ×213=56−16=46=23. 2 (33. 2 (34. 1 (35P(AU B) = P(A) + P(B) − P(A I B) = P(A) + P(B) − P(A)P(B)

65 4 1 33P(W 1 ) =7 3 2 4P(B1 ) =47 : W 1 ( (1 : W 2: B 1P(W2) = P(W 1)⋅ P(W2/ W 1)+ P(B1)⋅ P(W2/ B1)3 5 4 4 15 + 16 31P(W2) = × + × = =7 8 7 8 56 563 5×P(W 1)⋅ P(W1/ W2) 7 8 15P(W2 / W 1)== =P(W2) 31 3156: (: (2P(S i ) ⋅ P(A /S i ) 0/3 × 0/3 0/09P (S i / A) === = 0/36k0/3 × 0/3 + 0/7×0/2 0/09 + 0/14∑ P(S i ) ⋅ P(A / S i )i=1

66 . X (1 75(46416 Cov(x,XP(X)XP(x)-0/5 (4P(X= 20140/5 (3) . (338(21 (412516(1(3X (2 (2y) = −10, Var(y) = 25 , Var(x) -0/025 (263(164= 16 (3y x 0/025 (1X (41314(2(4 5000 . (515912 . 2000 0/00323 (4 -1142X01412/5 (3 (2X313138616218 (4: 0/001 11 (2E(1(310 (1[(X− µ)(X − µ) ] (6 (1 (3

67X = xP (X =x) (435 (4 E (X − E(X))2= E(X) (7 (3 (2: : 34 (3 (1X (8y = 2x−1 17 (233 (1: Y = −2 x +1 E(X2 ) = 8 E (X) =10 (4 3/50 (3 7 (21/75 ( 12/ 5 (9: X X . 3 (10XP (X =: y =100/2x)XP (X =1 /75(49(42x + 5 − 3 2 −8150/3013x)(4200/41a018250/1 121− 2a382381/5 (3349(3 (23 (116 X (11E(2 x + 1 a .1412(2(41613(1(3) (12288328(2(4 δ 2 x 129 − 2 2δ2316(33181/25 (210(1830(38 (130/75 (1δ 2 x 8δ2− 4(222δ(14x 2 (-11) x (15. (1

6813Cov(x, y)Cov(x, y)> 1= 0(2(4Cov(x, y)Cov(x, y) 3 y x (167 (46 (3: 5 (2= 1< 0x + y(1(38 (15 . 0/80 (175 . 0/00512 (40/0064 (3 0/0016 (20/128 (1 0/5 Y X δ 2 X Y (18δ2(42Fy (Y) Fx (X) f (x, y)f (x, y) >. ≥ fx (X)fx (X)2δx, y(3 22δ(2 X − Y 3 δ22F(x, y): ( ) (2(4f (x, y) ≤ fx (X) + f y (Y)f (x, y) =fx (X)fy (Y) . 36 8 (46 (3: 5 (2 Cov (x, y) =(x)δxδy(4 F(x)Cov(x, y)= 0(3Cov(x, y)< 0(2X,Y(1(1(3( X −1)(Y− 2)Cov(x, y)4 (1x, y > 0(1(19(20(21: (220 1 20/230/480/29

69 1000 δx= 2µx 770 (4= 5 . δy 710 (3n − 2 () . (4Y = ax += 1µ y 480 (2b = 0 ( a = −0 / 5,b = 2/5( a = 0 / 5,b= 2/5)) ( a = 0/5,b = −2/5n≥ 2 () 1 (3: 230 (1x, y b,a( a = 0 / 5,b= 2/5)( a = 0/5,b = −2/5)( a = −0/5,b= −2/5)) n = 1( a = 0 / 5,b= 2/5)2χ(23(1(2(3(4 x (24 . 0 (2-1 (1xy-1101818128023818. . Var(y). y, x y, x . Cov(x, y) V = 2x+ (1y(( (. . 5 5 (2 . 1/1 . (: . 1. (

70XP(x)XP(x)X 2 P(x)−-1141414014002 3Var(x) =2[ ]9 ∑ X P(x) − E(x) = − ( )2 =x8 8 . 2 (13µ = E(x) = ∑ XP(x) =x 898−964. 1 (263=64. 4 (3Cov(x, y) −10r = = = −0/5δxδy4×5. . 4 (41 1 1∑ P(Xi) = 1 ⇒ P(X = 2)= 1−( + + ) =4 3 6µ = E(x) = XP(x) 5000 0/001 2000 0/003 1114∑ = × + × =x. 2 (5. 4 (6. . 3 (7µ = E(x) = ∑ XP(x) = 17xy = ax + b = 2 × 17 −1=331383838[ E(x) ]2= 8 − 2/521/75Var (x) = E(X2) −=218284813898. 1 (8. 2 (9

71Var(y) = a2Var(x) = ( −2)2× 1/75 = 7. 1 1S2= npq = 3×× =2 2 . . 3 (1034. 4 (1111 1 1 1∑ p(x) = 1 ⇒ + a + 1−2a+ = 1⇒+ = a ⇒ a =x 36 3 6 2. 4 (12E(x)= ∑xP(x) = 0×x18+ 1×38+ 2×38E( 2x+ 1)= 2E(x)+ 1=2×1/5 + 1=4 =1E(x2) =2∑ x P(x) = 0×+ 1×x8δ2E(x2x = ) −38+ 3 ×328[ E(x) ]2= 3 −1/52= 0/75V ( 2x+ 5 − 3 2 − 9)= 2V(x)= 2δ2f (x)=Cov(x, y)120−1 < x < 1 1 1 1 x21E(x) =1∫− 1 xdx = × − 1 =2 2 2 431 1E(x ) =3∫ x dx = 0−1238= Cov(x,x2) = E(X3) − E(x)E(x2)−12= = 1/583 1 24+ 4×+ 9×= = 38 8 81= 04. 1 (13. 1 (14. 4 (15. 1 (16

72Cov(x, y)P (x,y) =δ x δ y=Cov(x, y)⇒ Cov(x, y) = 1V(x + y) = V(x) + V(y) + 2Cov(x,y) = 3 + 3 + 2 = 8P = 0/ 8n = 5P (x,y)=Cov(x, y)δ x δ y3⎛5⎞P (X = 1 ) = ⎜ ⎟(0/8)(0/2)4= 0/0064⎝1⎠⇒ 0/5 =Cov(x, y)Cov(x, y)V(x − y) = V(x) + V(y) − 2Cov(x,y) = δ2+ δ2− δ2= δ2 x, yδδ⇒. 3 (17. 3 (18= 0/5δ2 . 1 (19F (x, y) =Fx (X)Fy (Y): . 2 (20[ − 1 )(y − 2)] = E(x −1)E(y− 2)= [ E(x) −1][ . E(y) − 2] = ( 6 −1)(3 − 2)= 5E (x. 3 (21 . 3 (22 1000 n = 0 / 71×1000 = 710:E(Y) = 0⇒E(ax + b) = aE(x) + b = 5a+ b = 0F = 0 / 23 + 0/48 = 0/71 1V(y) = 1⇒V(ax + b) = 1⇒a2V(x) = 1⇒4a2= 1⇒a = ±2155a+ b = 0⇒5×( ± ) + b = 0⇒b = ±22. 2 (23. 2 (24

73xy012f (y) ( (11 2 3 6 x-18 8 8 82 1 2 1 4 211 2 f (x) = = =108 4 8 4 8 488 8 y -1 12 1 4f (x)8 4 86 3 2 1f (y) = =1 1 2 58 4 8 4E(x) = ∑ xp(x) = 0×+ 1×+ 2×=4 4 4 4(3 1 −2E(y) = ∑ yp(y) = −1×+ 1×=4 4 41 2 3 1E(xy) = ∑ ∑ xyf (x, y) = −1×0×+ ( −1)× 1×+ ( −1)× 2×+ 1×× 0+1×1×0+x y8 8 8 81 6 31×2×= − = −8 8 43 2 5 − 2 1Cov(x, y) = E(xy) − E(x)E(y) = − − ( − )( ) = = −4 4 4 16 81 1 2 9E(x2) = ∑ x2f (x)= 0×+ 1×+ 4×=(4 4 4 49 5 11Var(x) = E(x2) − [ E(x) ]2= − ( )2=4 4 63 1E(y2) = ∑ y2f (y) == ( −1)2× + 1×= 14 423[ ]2 − 2Var(y) = E(y ) − E(y) = 1−( )2=4 411 3 1Var( 2 x + y) = 22Var(x) + Var(y) − 2×2×1×Cov(x, y)= 4×+ + 2×2×= 46 4 85 4 5 4 40 4P ( ) = × + × = =(210 9 10 9 90 9 X 1/1 -14 5P ( ) = 1 − =4 59 9P(x)4 59 9E(x) = ∑ XP(x) = 1 / 1×+ ( −1×) = −0/067 x9 92 25[ ]2 ⎡ 2 41 1 12 ⎤δ X P(x)E(x) / ( ) [ 0/067] 2x = ∑ − =⎢× + − × − − = 1/088x9 9⎥⎣⎦012

74 (4P(X= 0).. (4 (1 (31n =100 P =100−100e (2 (3 (1X (2 −1e(2 (1: 4 . 6 (3115 ( )62(414 ( )42(31 ( )62(214 ( )62 15 (4 6 0/205 (40/3 (30/15 (2(10/65 (1 25 0/3 (55 (4P (X = 1 ) = 0/0/147 (45/2 (3: 12/5 (2368 0/92 (30/184 (2: 7/5 (1X (6P(X= 2)0/736 (1 . 20 (7 . 0/01 0 /021 2(4 0/025 (3 0/012 (20/016 (1

75 a . 3 (8: a . 78(414(318(2b 3 6 n (91n()n2(4: .n( 21 ). : X 3/75 (4(3x 11− n()n2X(212n1− ( 21 )(1(1 (10⎛20⎞P(X = x) = f (x) = ⎜ ⎟(0/25)x( 0/75)20−x⎝ x ⎠3/25 (32/75 (2. 4 20 P80 (440 (3100 (2X2/25 (1 (11: n 20 (1: n . 3/2 4 (12a(b40 (4 a− b)−nb 30 (3n, p (4n a − b( )b1p = , nn20 (210 (1. (13(3n a( )−nb(2a( )nb(1. (14: (4 X 1n(3−1e(2X1,X2,X3e −n(1 (15

76 T = X1X2X3P(TP(T . = 0)= 1−P3[ P( 1−)] 3= 0)= P(2(40 < P < 1 P(TP(T[ P( 1−)] 3= 0)= P= 0 ) = P( 1−P)3 3 2 (16 . 45(415(3 . (17 ، 3 18(414(3 . (1816(414(33516(2(22512(1(3(1(1 , 0/001 (190/180 (4: 3 20000/291 (313(20/369 (212(10/528 (1 . 3 8 (2041351715(4 (3(2(156565628 10 , 4 70 (21: .

77 ⎛4⎞⎛66⎞⎛4⎞⎛66⎞⎛4⎞⎛66⎞⎛4⎞⎛66⎞⎜ ⎟⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎜⎟⎜ ⎟⎜⎟⎝0⎠⎝6⎠ ⎝4⎠⎝6⎠ ⎝0⎠⎝10⎠−⎝0⎠⎝10(4(3(2 1⎠(1⎛70⎞⎛70⎞⎛70⎞⎛70⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝10⎠⎝10⎠⎝10⎠⎝10⎠ X (22: 21030 (270 (4. (17 (3 (23 (4 (4 (3 (2. ...... (1: (24 (3 (2 (1 2 3 10 (25 . ( ) 35(478P(X≥ 2)1− e−2(4 602830(3(2(1787878 µ = 2 X (261− 2e−2(32 1000. 226−e(41− 4e−2(21− 3e−2(1 (27 3000: 2 12e(3−2e(2−6e(1

78 8(4x = 2 , x = 1 − 2 2δ2(3 8δ2− 4 n . nx (28 (2nλ22δ(1 (29 e − n (40 (3e−2(2(1 .0/1 (30 . 0/0081(4: 0/00081(30/004863 2 (2x2n0/0486(1 (31: 3 1P(x= ) = (4 E(x) = 1 (3 P (x = 1)= e (2 Var(x) = 3 (12 2λx32 3.λ = 1/ 7 (32 x 1032 6 (4P(x= 0 ) = P(x = 1)4 (419 (35 (2 . 3 (3x 2 (21 (1 (33E(x2 )1 (1 (34 5 2 . 0/ 74(40/ 21(3: 0/19(20/02(1

797(49 ( E(X) = λ)100−1−e λ(4 2 1P(B) = , P(A) = B A (353 3 P(AU B) 21(393λ 1−e− 100λ(3(2X: 100−e λ(21 (1 (36P(X≥100)e −100λ(11 . λ = (378 8 1−1− e 8(4 1− e−1( µ ± 3(3−1e(21−e 8δ) x (38x1 −f (x) = e 5 x ≥05(⎛ 2 4 ⎞ ⎛ 2 4 ⎞10, 20)(4 ( −10,20 ) (3 ⎜ , ⎟ (2 ⎜ − , ⎟ (1⎝ 5 5 ⎠ ⎝ 5 5 ⎠: f (x) = 3e−3 x X ≥0 (3913e−1 − e−3(4(419:e−3 − e−1(3 (31− (23P(1 < x < 3)e−9 − e−3(2− 3(1(1 (40e−3 − e−9 (1 10 (41 5 . 5 (. )

801− e − 1/5(4 f (X)e −1/ 5= θe−θx, X− x(4(3e −0/ 5(21− e − 0/ 5 (1> 0 θ (42x: (31−xn (21x(1 5 . 7 10 (1: 5 (. (

81 . . 3 (1λ x = δ 2 x = np. 2 ( 2 . 1λ = np = 100×= 1100λxe−λ10e−1P(x) = ⇒ P(x = 0)= = e−1x!1. λ = np . 4 (3⎛n⎞P(X = x) ⎜ ⎟Pxqn−x⎝ x⎠⎛6⎞1 4 1 6 4 1= ⎜ ⎟() ( 1−)−= 15()6⎝4⎠2 2 2: . 2 (4⎛n⎞P(x) = ⎜ ⎟Pxqn−1⎝ x ⎠⎛15⎞1 6 1 15 1 15 1115 6 ⎛ ⎞ 15 !P(x) = ⎜ ⎟() ( − )−= ⎜ ⎟() = × ( )15= 0/15⎝6⎠ 2 2 ⎝6⎠ 2 96 ! ! 2. . 1 (5µ = np = 25 × 0/3 = 7 / 5λ = S2= 1λxe−λ1e−1P(x) = ⇒ P( 1)= = e−1= 0/368X!1!12e−10/368P ( 2)= = = 0/1842!2. 2 (6

82n = 3µ = np = 20×0/01=0/2P(x = 2)=xC n x P201n x ⎛ ⎞( − P)−= ⎜ ⎟0/012× 0/9918= 0/016⎝2⎠1P(a > µ) = P(b > µ) = P(d > µ) =2P(a < µ) = 1−P(a > µ) = 1−P(µ < a < d < b). 1 (7. 4 (8 ⎛3⎞1 3 1 0 1P(X = 3 ) = ⎜ ⎟() ( ) =⎝3⎠2 2 81 7P(a < µ ) = 1 − =8 8E(x) = 6 = nP1V(x) = 3 = nPq ⇒ 6q= 3 ⇒ q =2⎛n⎞P(x ) P(x ) P0qn 1≥ 1 = 1−= 0 = 1−⎜ ⎟ = 1−( )n⎝0⎠ 21 3 15Var (x) = npq = 20 × × = = 3 / 754 2 4Var(x)q = :E(x)µ = np = 20: 1P =2. 1 (9. 4 (10 . 2 (1116Var(x) = npq = 16 ⇒ 20q= 16 ⇒ q = = 0/8 ⇒ p = 0/22020np = 20⇒n = = 1000/2. 2 (12

83E(x) = np = 43 / 2Var(x) = npq = 3 / 2 ⇒ 4q= 3 / 2 ⇒ q = ⇒ p =4n ×210= 4 ⇒ n = 20⎛n⎞P(X = 0)= ⎜ ⎟p0qn⎝0⎠E(X) = np = aV(x) = npq = b ⇒ aq = b ⇒ q =P(X = 0)=nq a= ( )−nbpba n 210. 2 (13 . 2 (14. 1λ = np = n × = 1n P(T = 0)= P(A)= 1−P(A ′)⎛3⎞= 1−⎜ ⎟p3q0= 1−p3⎝3⎠e−110P(X = 0)= = e−10!⎛3⎞⎛2⎞⎜ ⎟⎜⎟1 1P(A) =⎝ ⎠⎝⎠=⎛5⎞⎜ ⎟⎝2⎠610=35. 2 (15. xi :P(A). xi : P(A ′). 2 (16 : P(A). . 1 (17 . 4 (18.

84⎛2⎞⎜ ⎟2P(A) =⎝ ⎠=⎛4⎞⎜ ⎟⎝2⎠16 : P(A) n > 20 np ≤ 5 . 4 (19λ = np = 0/001×2000 = 2 < 5λxe−λP(x) =X!⇒ P( 3)=32e−23!= 0/18. : . 1 (20⎛3⎞⎛5⎞⎜ ⎟⎜⎟CkCN−K1 1 3×5 15 15P(x) = x n − x =⎝ ⎠⎝⎠= = =CN ⎛8⎞8×7×6!56 28n ⎜ ⎟⎝2⎠2!6!2. . 1 ( 21⎛4⎞⎛66⎞⎜ ⎟⎜⎟⎝0⎠⎝10P ( ) =1− P ( ) = 1−⎠⎛70⎞⎜ ⎟⎝10⎠. . . 3 (22210 µ = 210⇒ µ = = 7 ⇒ µ x = δx= 730. 4 (23 . 4 (24Ck N −K= x CP(x) n − xCNn⎛10⎞⎛3⎞⎜ ⎟⎜⎟1 1=⎝ ⎠⎝⎠⎛13⎞⎜ ⎟⎝2⎠10×3= =13×12260156=3078. . 1 (25

85e−λλxP(x) =X!P(X ≥ 2)= 1−P(X < 2)= 1−P(X ≤ 1)= 1−e−2⇒ 1−(0!P =21000×02212e−2+1!=13000121n = 3000 , λ = 3000×= 13000e−112e−2P(X = 2)= =2!2 () λP(x = 1)= e−λλ−2) = 1−3ex = 1 . 1 (26[ P(X = 0)+ P(X = 1)]: . 3 (27λ = np . 1 (28 . ∂p x (x)= −e−λλ + e−λ⇒ −e−λλ + e−λ∂λ2P = n → ∞ ⇒ p →0np(x = 0)= e−22λ = n × = 2n. = 0⇒e−λ(λ −1)= 0⇒λ = 1. 2 (29 . . 2 (30⎛5−1⎞P (x) = ⎜ ⎟ × 0/13× 0/92= 0/00486⎝3−1⎠. 3 (31

861101203233322333222=⇒=⇒=−=−⇒−×=−−==−==E(x)λλ)(λλλ!λλe!λλe!λλe)P(x!λλe)P(x3232322(32 .:λE(x)(x)V == λ x:532310=++=x3333332(33.[ ] 22121221110=⇒−⇒=−=⇒====⇒−=−⇒−==−==)E(x)E(xE(x))E(xVar(x)λVar(x)E(x)λλλeλeλλe)P(x,λe)P(x343434(34 4.740244436343224314343143444343334343224343143434343210543712/!)/()/()/()/(/(/e!)/(/e!)/(/e!)/(/e!)/(/e/e)P(x)P(x)P(x)P(x)P(x)P(x//λ≅+++++−=−+−+−+−+−==+=+=+=+==

87f (x) =1µ = δ =λ . . 3 (37: 1−8×P(a ≤ x ≤ b) = e−λa− e−λb⇒ P(. < x < 8)= e0− e 8 = 1−e−11. µ = δ = λ1 1 1λ = ⇒ µ = δ = = = 55 λ 15µ ± 3δ= 5 ± ( 3×5)= ( −1020, )δxλe− λx1 = µ x = =λX ≥. 3 (38. λ = 3 . 4 (39130 , λ >0P(a ≤ x ≤ b) = e−λa− e−λb⇒ f (x)X1e − 88X ≥00 . 1 (40= e−3×1− e−3×3= e−3− e−9: . 2 (41=0λ = 10f (x) == 0110−xe 10P(x > a + b x > a) = P(x > b)P(x > 10x> 5)= P(x > 5)P(x ≥ a) = e−λa⇒ P(x > 5)= e−0/50< x < ∞ . 1 (42

88f (x,θ) =nθn− ∑ x i1θne−θ∑xi⇒ Lnf (x,θ) = nLnθ − θ∑xin= 0⇒n − θ∑x i1n 1= 0⇒θˆ = ⇒ θˆ =∑ xi xθLnf (x,θ)dθ=n = 5 ,X = 5,K= 7,N = 10 . (1⎛7⎞⎛3⎞7!3!⎜ ⎟⎜⎟ ×CkCN−K5 0 5 7 − 5 0 3 −01= − ⎝ ⎠⎝⎠ !( )! !( )!P(x) x n x = ==CN ⎛10⎞10!12n ⎜ ⎟⎝5⎠5!(10−5)!k 7µ = E(x) = n = 5×= 3 / 5N 102k N − k N − n 7 10−7 10−5δ = Var(x) = n ⋅ ⋅ = 5×× × = 0/58N N N −110 10 10−1

89 P( − a < z < a)0/921=2(4 6 100 a 6 100 (21(32 1/96(2 Z (10/ 68 (1X (2X −100 6 1 (11001 (4 1 (36 . 25 5 (3. δP ( −a< Z < 11 (4 5 (3(25= 1 µ =10 0 /68260 /6836(4(40/8413 (30/8413 (3a) P(Z < −a)34 3=8 0/9587 (2P(X> 9)0/9587 (225X(1X (4P(9 < X < 11)0/3413 (1,4 (50/3264 (1 Z (6 13(4(3(248Z 1 P (Z ≤ Z1 ) = 0/65(2P (Z ≥ Z1 ) = 0/555(18Z (1(7

- 9/6 90P ( 1/5 ≥ Z ≥ Z1 ) = 0/3(4P ( −2 ≤ Z ≤ Z1)= 0/75 , 100 10 (8(3 29/695 (465 (347/5 (25 (1: Z = 2 Z = −2 (90/6826 (40/9544 (30/98 (20/9973 (1 . , %30 (10 , 200 50 0 /9463(40/9382 (30/4474 (20/4382 (1 ( µ − 2 δ,µ + 2δ) (11% 68(4%95 (3%99/7 (2%75 (1 δ2= 36,µ =10 X (12 P(4 < x < 16)0 /26(40/84 (30/34 (20/68 (1 150 . %70 (13 , 0/6 0 /0035(40/5 (30/5993 (20/0062 (1 Y = 10Z− 5 Z (14-5 (415 (35 (210 (1: Y (14) (155 (410 (3100 (295 (1 9 10 X (16: X = 12

9122− (4(399 1818(40/6587 (49(30/1587 (3 5 22− (2(133() (17: 3(2: 0/8413 (218(13P(0 < Z < 1) (180/3413 (1 4 10 X (190/95 (40/4174 (3: P(9 < X < 11)n =160/6826 (20/8413 (1 . S 0/99 (4P ( x − µ < (40/89 (3δ) δ 0/95 (3: 0/95 (2x + 3δxx − 3δ 0/75 (1(20µ (21 0/75 (20/68 (1 Y,X (22X 16 . 48 N(0,1)(4: N(0 , 2 + 2)X − Y . Y 8(3N(0,12)(2N(0,4) Z n x 2 Y (23(1x2(n−1)(4 x2( 1)(3ZnY t(n) (2 t(n−1)(1

92 500 (24 . %30 . 0/05 %95 70 (481 (3197 (2( Z0 / 975 = 1/96)323 (1 . 16 12 1 40 18 10 50 (2 , . 8 . 60 52

93 1 3P ( −a< z < a) = 2 φ a −1= ⇒ φ a = ⇒ a = 0/682 4 . 1 (1X − µ Z = . 3 (2δ. δxδ = xn= 5 = 125X − µZ =δX . 4 (3 . 4(4. 9 −1011−10P(9 < x < 11)= P( < Z < ) = P( −1 9 ) = P(Z > ) = P(Z > −1)= 1−φ−1= φ = 0/84131P(Z < −a)= 1−φ a =P( −a< Z < a) = 2φ a38⇒ φ a−1= 2×58=58−1=28=14. 3(5. 3(61 1P( − a < Z < a) = 2 P( −a< Z < 0)= 2×=8 4 Z 1 0/5 1 . 1(7. 0/5 Z 1 2 . -1− a001a

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

95P − PP(P < 0/6)= P(Z < ) = P(Z

96E(x − y) = E(x) − E(y) = 02S21 SV(x − y) = V(x) + V(y) = +2=n1n28164+ = 18 ( ) . 2 (23. t Z n = Z ~ t(n)Y yn24Pˆ = 0/ 3z2pˆqˆn =d2d = 0/05( 1/96)2( 0/3)× ( 0/7)=≅ 323( 0/05)2. 1 (24 µ =12 δ2= 16 ⇒ δ = 418 −12P (X > 18 ) = P(Z > ) = P(Z > 1/5)= 1−φ1/5 = 1−0/9332 = 0/066410−12P (X < 10 ) = P(Z < ) = P(Z < 0/5)= φ0/5 = 0/30854= 40 × 0/ 3085 = 12/34 U = P + Zα δP= 0/4 + 1/96×0/08 = 0/57532(2X = 50 δ = 8 n = 60X − µ 52 − 50P(X > 52)= P(Z > ) = P(Z > ) = P(Z > 1/93)δ8n60⇒ 1−φ1/93 == 1−0/9732 = 0/0268(1

97 X . δ = 2 µ =10 (1 (1014) X n =16 0/5228 (45∑ X2 = i1S =0/ 4120 (40/0228 (35∑ X i1S = 2/ 58= 20,0/4772 (20/5 (1n =15 (2 . (3S = 2/15(2S =1/9(1 0/95 (3( 1/572/29) (4(1/894/07) (3(2/293/47) (2(1/573/47) (1 S 2 p S1 = S2= 2,n2= 17,n1= 10 (4148495(4(32 (2 4 (1275 (5F (4χ 2 (3t (2 . 3/253/355 98 δ2P1−P2: S = 3 , X =1 9(-0/1152/115) (2(-2/3554/355) (4 P1( 1−P1) P2( 1−P2)= +n1n2 (1t (6 (-0/072/07) (1(-2/254/25) (3P 2 P 1 (7P1 − P 2 (22 P1( 1−P1) + P2( 1−Pδ =2P 1−P2n1+ n2)(1

98δ2P1−P2P1( 1−P1) + P2( 1−P2)=n1n2 (4(42 P1P2+ ( 1−P1)( 1−P2)δ =P 1−P2n1+ n2 (8 (3 (2(3 (1: X (9µ , (1.δn , (2δnδn.δn µ µ , (3. µ , (4 – –.: (10 – (2 – (4 – – (1 – – (3 . 20 , n =100 (110 /038 %95 (40/068 (3: n 1 + n2n1n2(20/078 (2X1− Xt =21 1Sp +n1n20/039 (1 (12n 1 + n 2 (1

99. (4 n1 + n2−2 (3 X X (131 (230 (430 (1 (3 %65 n = 400 (14 . 0/40 (40/35 (3: 0/65 (20/024 (1n 1 X 2 X 1 (152δ2: δ2x 1−x 22δ21 δ−2(3n1n2 n 2 δ 1 +2 (41+2(2 δ2 21 + δ2(1n1n2n1n2 5 30 100 (16 10 25 200 3 /4472 (4(41/96 (3δ2δ2: 5/77 (2 (n) , δx48 (336 (2= 12 δx15/81 (1= 2 (1724 (1 (18. (4. (3. 2 (2: . 4 (11: f (x) = , 2 < x < 4 (19219 (4 1 (3(23 (13

100X . (1525) X (20 14(4 (335: () Y = ax +1( b − a)(4P ( x − µ < δ)0/68 (4 (3 34(2( b − a)2(21(12b (21( a + b)2 δ µ (220/75 (30/95 (2(1 b δ 2 µ a2δ2+ b20/34 (1X: a (23≠ 0 a µ + b (1aδ 2 aµ (2a2 δ2 ( 4 a µ + b (3 (24 (4 (3 (2 (1 1(418: (3f (x)141b − aP(XY> 0)(2a ≤ x ≤ x, y2. δ12(25 300 ( ) χ 2 X (26=0(1

101: P(X< 300): (4 E(Z4 )1(312(2 14(1Z (274 (43 (32 (21 (1 . 15 (1 X : (010) 7 (. X (800 . %5 (2. 2δ 5 , δ %99 (3216,221,220,218,215:

102 X − µ. Z = . 1 (1δn10−1014 −10P ( 10

103 . 2 (7. 3 (8. 2 (9. 4 (10. 2 (11x 20P( 1−P) 0/2×0/8P = = = 0/2 δ P = = = 0/04n 100n 100d = P − P = Zα δP= 1/96×0/04 = 0/07842. 3 (12( ) . 4 (13. . 1 (14P( 1−p) 0/65×( 1−0/65)S = == 0/024n400. 4 (15. Z 30 . 2 (16X1− X230−25Z = == 5 / 77S225 1001 S+2 +n100 2001 n212=Sx⇒ =Sx 2=2Sx n ( ) ( ) = 36n Sx 2. 2 (17. 1 (18: . 2 (19(b − a)2( 4 − 2)21Var(x) = = =12 12 3. d = b = 25 c = µ = 20 . 1 (20

104a + b 15 + 25µ = E(x) = = = 202 2d − c 25 − 20P(c < x < d) = = =b − a 25 −15510=12. 3 (21. 4 (22. . 3 (23E (ax + b) = aE(x) + b = aµ + bV (ax + b) = a2V(x) = a2δ2P(XY> 0 ) = P(X > 0,Y> 0)+ P(X < 0,Y< 0). 1 (24. 1 (25: P(XY> 0 ) = P(X > 0)P(Y> 0)+ P(X < 0)P(Y< 0): 1 1 1 1 1 1P(XY > 0 ) = × + × = + =2 2 2 2 4 412x, y X 2 . 2 (26X ~ X2(n) ⇒ E(X) =P(X < 300)=Z2~ χ2( 1xP(n, V(X)= 2n− µ 300−300< ) = P(Z < 0)=δ 60012: X 2 ) Z . 3 (27: V(Z) = E(Z2) − Z2(Z) ⇒ 1 = E(Z2) −0⇒E(Z2) = 1V(Z2) = E(Z4) − E2(Z2) ⇒ 2 = E(Z4) −1⇒E(Z4) = 3E(z2) = 1,Var(z2) = 2

105 f (x) =1b − a0a < x

106(30653216125751612520699005641212217518602145641222156426152218523764611222218510901/δ//δ////),nα(X)S(nU///),nα(X)S(nL/nninxiXSnniXiX

107 (4 (3 (2H 1 : µ < µ 0 H 1 H 0 (1 (1H0 : µ ≥ µ 0 (2 , t ≤ −t a (4 t < t a (3 Z < −Z a (2 Z < Z a (1. S = 3 / 2,X= 28/8,n = 9 H0: µ = 28 (3H1: µ ≠ 28 0 /75 (49 (33 (21 (1 100» H 1 (4«...... 100» H 0 « . (2. (4. (1. (3 (5 α (1β (2. (3. (4 %20 (6 60 400 .H 0 : P < 0/2 %99 H0 : P0/2

108Z = 2/ 56 (2Z = −256 (4( P(Z < 2 / 33)= 0/99. H 0 H 0 (2. H 1 H 1 (4H 0 H 1H 0 H 0 N( 0,δ2) . Z = 256 (1Z = −2/ 56 , (7(2(4(3. H 0 H 0 (1. ) 2H 1 H 1 (3: (8H 0 H 1H 0 H 0 (1(3X1,X2,...,X n y = n 2∑ Xii=0δ 2 / n (4nδ (3δ 2 (2 (1 n t T (10 T 2 X 2 (n) (4 X2(n −1)(3 F(1,n) (2 F(n,1)(111 θ1 ≠ θ 2 θ 1 = θ 2 (11 . θ1 ≠ θ 2 θ 1 = θ 2. θ1 ≠ θ 2 θ 1 = θ 2. θ 1 = θ 2 (1 (2 (3. θ 1 = θ 2 (4 H 0 : P = 0/5 X ~ bin( 4,P) (12 . { x x = 0,1 } H : P = 0 31 /(9

1090/31250 (40/343 (3 . 100 0/3483 (2H0 : P = 120/6517 (1 (13 20 (4. 15 (3t: 10 (2t: 5 (1 (14n −1n1 + n2− 2 (1 (2 (3.t (42 = 4 µ X δHx > 8 / 640 : µ = 8.(4 64 . x > 8 / 64( α = 0/01)(3 ( Z0 / 995 = 2/57)x > 8 / 58δ 2 µ H >1 : µ µ 0 (2H 1 : µ > 8 / 7( Z0 / 99 = 2/32)x > 8 / 58(1X n ,..., X 2 , X 1 0 : µ µ0 . H =x > µ 0 +δφ−1( 1−α)nx < µ 0 +δφ−1( 1−α)n(2(4x > µ 0 −δφ−1( 1−α)nx < µ 0 −δφ−1( 1−α)n(15(16 α = 0/05 (17. 0/05 H0 (1. 0/05 H0 (2(1(3

110. 0/05 H 1 H0 (3. 0/05 H 1 H0 (4 (18: H0 α = 0/05 . X012345P(X / θ0)0/050/050/850/030/020P(X / θ0)0/090/050/020/010/010/01X = 3(4X = 0 5(3X = 2(2X = 1(1

111 . H 0 H 0 . 3 (1 . 4 (2t 30. X − µ 0 28/8 − 28t = = = 0/75S 3 / 2nP0= 0/2xP = Pˆ = =nP − PZ =P 0(1−0n604009= 0/15t . 4 (30 0/15 −0/2 −0/05= = = −2/5P ) 0/2×0/8 0/02400. H 0 2/33 . . 4 (4. 1 (5. 3 (6− 2/ 5β = P( ) = P ( Var(x) = E(X2) −[ E(X) ]2⇒ δ = E(X )22H0/. 32 (7 H 1 ). 1 (8. 3 (9

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

113 3 2 (4 Y = 3 +0/4X0〈r≤14 (4 (1r =1 (3 40 (3Y X r = −1(2r =0 (1 (2Y = 4/ 6 (2X 4/86 (1 r (3(4r = −1(3 Y −1 ≤ r〈0(2Xr〉1(1 (4 (20 /3(41 (1-1 (3 n = 20 (5Sxy = 876: . . xyXY11 r32110 42S xx = 618 / 75ŷ = 1 / 42x+ 0/085ŷ = 0 / 085x+ 1/42120r61=0 y, xy, x(2(4(2(4: y = α + βx +x =1/ 75y = 2/ 4 ŷ = 1 / 42x−0/085ŷ = 0 / 085x−1/42(1(3y, x (6. . y, xy, x(1(3y, x (7e -1 (20/5 (41 (1 (3y, x (8

114βˆ = rδ2/ δ2 y xβˆ = rδ y / δ x y x 2 1−21−3(4(4: −1y, x1(2(4(3(3: y, xβˆ = rδ x / δ yβˆ = rδ2/ δ2 x y(1(3y = 1+x 2 12(2 −1(213 x + y = 813(1(9(10 (11y µ x %70 %3 (4(1. (1%51 (3= 40 ŷ = 80+ 2x(4. (2. (3. (4y, x %70 (2y, x : ŷ = 2x . x (3x ŷ = 80+ 4xy : x %49 (1y, x δ2 2y = 4δ x µ y= 80y (21ŷ = 40+2y, x Bˆ rx (1 . . y′= 100/000y x′= 1000x y, x(12(13(14

115 (4( r = 100r′)X(3 Y,Y( B = B ′) (2 ( r = r′)(1 X (15. (1. 1 (2. (3. -1 +1 (4 0/2 Y,X (16 Xr′1 (4Y 0/6 (3 Y = 10 y r = 10r′(4Y ′ = a′+ b′x′r = r′(3: 3 Y + 2, X0/2 (20/1 (1r X,Y 5X ′ = 103 xX′b = 10b′ (2Y = a +y′bxb = b′(1(170/2 X Y 0/8 Y X (180/16 (4 δ 22 , δ 1 0/8 (3X,Y 0/4 (22 X 1 kX 2 k .U = +δ−1rδ2(4δ1rδ2. (30/2 (1X 1 , X 2 r rδ−2δ1. (2V = X 1rδ2δ1(19. ...... r =0 (20(2. (4(1. (1. (3

116: (Y) (X) 9 (1X11/5121316/515/5151312/517Y121412/51516/517 16/5 14/5 16. (. (

117 . X y . r=1 . 4 (1X . 2 (2X y = 3 + 0/4x⇒ 4/6 = 3 + 0/4x⇒ x = 4. -1 +1 . 1 (3 y X . 3 (4. ∑ x 12 ∑ y 6x = i = = 4 y = i = = 2n 3n 3∑(X − X)(y − y)r = i i ⇒∑(X − X)2∑(y − y)2i i[(2 − 4)(3 − 2)+ ( 4 − 4)(2 − 2)+ ( 6 − 4)(1−2)]= −1⎡(2 − 4)2+ ( 4 − 4)2+ ( 6 − 4)2⎤⋅ ⎡(3 − 2)2+ ( 2 − 2)2+ ( 1−2)2⎤⎢⎣⎥⎦ ⎢⎣⎥⎦. 1 (5Sxyβ =Sxx876= = 1/4157 ≈1/42618/75αˆ = y − βˆx = 2/42 −1/42(1/75)= −0/085ŷ = βˆx + αˆ = 1/42x−0/085. 3 (6 y x . 3 (7. . 4 (8

118E(y /r2= 0/δyx) = µ y − r (x − µ xδx). 1 (9. 2 (10. 2 (1149. x y . 1 (12δy2δxµ y / x = y + ρ (x − x) = 80+ρ (x − 40)δxδx⇒ ŷ = 80+2(x− 40)⇒ ŷ = 2x. 3 (13. 1 (14. 4 (15. 2 (16ρ x,y = 0/ 23y + 2 = zCov(z, x) Cov( 3y+ 2,x) 3Cov(x,y)ρ z,x = === ρ x,y = 0/2S z S x 3S y S x 3S x S y. 3 (17. 2 (18r = r1 × r2= 0/2×0/8 = 0/4. Cov(u, v)= 0 . 4 (19u, v

1192121210212101212121212121212101210121rSSSrSSkS)Sk(rS)V(x)x,kCov(xδrδ)x,Cov(xδδ)x,Cov(xrδ xδ x)x,Cov(x,xρ x)V(x)x,Cov(x)x,kxCov(x−=−=⇒=+⇒=+=⇒=⇒==+⇒=+202020. 3 (20 4. (18914913414912611/nyynxxniinii======∑∑==xy2y2x iyx138144132/251211/51681961441412162/5156/2516912/513247/5225272/251516/5255/75272/25240/2516/515/52552892251715214/5272/2516916/513181/25210/25156/2514/512/527225628916171894/520211797134126

1203314917972212=×−=−= ∑=nxxSniixx59248914920212212//nyySniiyy =×−=−= ∑=36188914149518941///nxyyxSniiixy =××−=−= ∑=175560171455608914556033361864405924333618/x/aβxy///βxya//SxxSxyβ///SxxSyySxyr+=+==×−=−=====×==

121 82-8383 60: ( ) 20: µ + δµ (10/68 (40/34 (30/28 (20/14 (10/5 0/35 (2 40 .6 (48 (312 (215 (1 ( Q 1 ) 446881010121416 (36 (46/25 (3 1/5 12/5 X 324 (4 ⎛4⎞⎜ ⎟⎝2⎠(4⎛n⎞⎜⎝0⎟ −⎠4/5 (210,...,X2,X2X10−12, X2−12, X1−1625 (3⎛n⎞ ⎛n⎞⎜ ⎟ + ⎜⎝1⎠ ⎝2x1 + x2+ x3= 4⎛7⎞⎜ ⎟⎝4⎠(3⎟ −⎠624 (25/5 (11 (4 325 (1 (5⎛n⎞... − ⎜ ⎟ = 0 n⎝n⎠⎛n⎞⎛n⎞⎛n⎞⎜ ⎟ + ⎜ ⎟ + ... + ⎜ ⎟ = 2n⎝0⎠ ⎝1⎠ ⎝n⎠⎛n⎞ ⎛n−1⎞⎛n−1⎞⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟⎝ x⎠⎝ x ⎠ ⎝ x −1⎠⎛n⎞ ⎛n⎞⎜ ⎟ = ⎜ ⎟⎝ x⎠⎝n− x⎠ (1(2(3(4 (6⎛6⎞⎜ ⎟⎝4⎠(2⎛4⎞⎜ ⎟⎝3⎠ ( ) 5 9 8 5 32 (7 (1

1225× ! 3!(4⎛5⎞5!× ⎜ ⎟⎝ 2 ⎠(3 23× 35× ( 10!)2!3!5!10!23× 35×3!× 5!(2(4x2y3z5× ! 2!(25 ⎛5⎞4!× ⎜ ⎟⎝ 2 ⎠( x + 2 y + 3z)(110 (810!23× 35×5!23× 35× (10!) 2 . 4 5 (959 . (439(34 5 3 4 (10 . . 3477(44277(3 X . () (11f (x)29x⎛ 1 ⎞= K⎜⎟⎝ 2 ⎠1K =4(4(418x = 012 , , , 3,... ( F(x) )X 1K =2(3 692477(2(2P(X≥ 1016(2492877(1(3(1(1) K (12(33K =2(214K = 2λe− λxx > 00 x ≤ 0x f (x) = (13(1(11− λe− λx(41+ λe− λx(31− e −λx(21+ e −λx(1

123 13 (141Lnλ2(42Lnλ(31 Ln2λ(2λLn2(1 E(X) f (x)=x20 < x < 2 X (150 452(4252(3522(2542(1 P( X1< )312 f (x) = (160−1 < x

12482-8383 . 3 (1P (µ < x < µ + δ) = P( 0 < z < 1)= Q1−0/5 = 0/8413 −0/5 = 0/3413f i ′ = F i ′ − F i ′ − 1 = 0/5 −0/35 = 0/15ff ′ = i i ⇒ f i = f i ′ × n = 0/15×40=6ny = ax + b ⇒ y = ax + b = 2×12/5 −1= 24Var(y) = a2Var(x) =22. 4 (2. . 4 (3× 1/5 = 6⎛ x + r −1⎞⎛4+ 3 −1⎞⎛6⎞⎛6⎞⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟⎝r−1⎠ ⎝3−1⎠ ⎝2⎠⎝4⎠P( ) = 1− P(P( ) :W )=41 − =9 : R59. 2 (4. (5. 2 (6. 1 (7. 2 (8. 4 (95 4 4 3 32= P(R1 ) ⋅ P(R2/ R1)+ P(W 1)⋅ P(W2/ R1)= × + × = =9 8 9 8 7249

125 . 4 (10 37374747××××26463636====17272727611511511411⇒ P( ) =17×6112+ ×7{(5,5)(6,4)(4,6)(5,6)(6,5)(6,)}A =6n(A) = 6 n(S) = 62= 36n(A) 6 1P(A) = = =n(S)36 6aS = 1 − q1 k1∑ f (x) = ⇒ = 1⇒k1= 21−2511+27×5112+ ×7411=3477. 2 (11 . 3 (12. 2 (13x xxF f (x)dx λeλxdx eλxeλx( e ) eλxx (x) = =−= −− ⎤ = −−− −−0= −−∫ ∫ 1−∞ 0⎥⎦ 0. 2 (141 1F 0 5 1λx0 5λx0 5λxx (x) = / ⇒ − e−= / ⇒ e−= / ⇒ = ⇒ e = 2 ⇒ λx = Ln2eλx 21x = Ln2λ

126+∞2E(X) = ∫ xf (x)dx ⇒ E( x ) = ∫−∞0233 ⎤1 2 1 1 + 1⎥= ∫ x 2dx= × x 2 =2 2 3⎥0+ 1 ⎥2 ⎦0P( X1 ⎡11 ⎤==2 ⎢− ( − )3 3 ⎥⎣ ⎦13152xf (x)dx = ∫052⎤x 2 ⎥ =⎥⎦01 11 1 1 3 3 1< ) = P( − < x < ) = ∫ f (x)dx = ∫ (3 3 3 1 1− −2 )dx3 3452. 4 (15xx ( )dx2. 3 (16ρ =Cov(x, y)δ x δ yρ (ax + b,ey + d) =− 2 2= = −2/5×2 5P( ) = 1− P(ρ(x, y) . 2 (17 . 3 (18 )⎡ 3 ⎛10⎞1 0 4i0 6n i ⎤− ⎢ ⎜ ⎟ / /−⎥ = 1−0/3823 = 0/6177⎣i=0⎝i⎠ ⎦= ∑Var (x) = npq = 10 × 0/4×0/6 = 2/4. 1 (19. 2 (2013 −12P (x > 3 ) = P(z > ) = P(z > 0/67)= 1−φ0/66 = 1−0/745 = 0/2551/5

127 84-85 60: ( ) 20: . (1 P ( )= 0/ 2P ( )= 0/1 0/5 (40/3 (30/72 (20/02 (1150 225 240 260 275 290 300 1500 : 8 (2: 267/51250 (4 287/51350 (3 267/51350(2 287/51250(1 2/4979 9/4 15 (3: 30/571 (4 26/573 (3 25/450 (2 20/773 (1: (4 1 2 3 4 5 62/0 -2/42/4 -2/82/8 -3/23/2 -2/63/6-4/04/0 -4/45544930/4 (4 4 (25 (4: P(A / B) 0/3 (3 5 4 (14 5 (3P (A I B) = 0/16 P (B) = 0/0/2 (24 (50/1 (1B A k ,..., A 2 , A 1 (6

128[ ] P(Bc)P U A i(4∑ P(A Bc i I )(3: E(X) f (x): =x20f (x) =1 t2e24 (44 (4: 3 (3∑ P(AiIkP( U A i B)i=1B)(2 ∑ P(A i IP(B)B)(1X (72 (21 (1: X (81 21 − xe 22π(4Cov(2 X −13, Y + 46 (4) 3 (3: 2 (2 : Mx (t)et(3tE(XY) = 8 5 (3e 2t(2E(Y) = 14 (22E(X + )31 (1X (9− ∞

1291f (x) =θ (44X، 0< x < θ(4 (3 (2 ( n −1)s2δ2 (1 X (15: θ 3X X δ 2 µ (16: 0/96 (4(3⎡ δP⎢X−1/96 < µ < X + 1/96⎣ n0/95 (3δ2Xn⎤⎥⎦(20/94 (2 Xn(10/93 (1: (17. H 1 H0 (1. H0 H0 (2. H 1 H0 (3. H 1 H 1 n X: P (X2 22 − X 1 > µ 2 − µ 1 ) δ 2 , δ 1 µ 2 , µ 1Xf (x).-20/15 (40/75 (30/5 (2(42 , X 1 (180/25 (1 (19r > −1/ 5-10/100/5(4R2= r2(3r = ±1(2−1≤r ≤ 1 X (2010/120/15. ( α 3 ) (1 (40/2 (31 (20/3 (1

130P (A′ I B ′)= P(A ′)⋅ P(B ′)= ( 1 −0/1)(1−0/2)= 0/72R = X max − X min = 1500−150= 1350260+275m = M e = = 267 / 52δ 2/4979CV = × 100 = × 100 = 26 / 573x 9/484-8585 . 2 (1. 2 (2. 3 (3 . 1 (4P(A I B) 0/16P (A / B) = = = 0/4P(B) 0/41E(X) = np = 2 × = 12E (ax + b) = aE(x) + b+∞ 2 x2123 ⎤ 8E(x) = ∫ xf (x)dx = ∫ dx = x0 2 6 ⎥ =0 6−∞⎦2 2 4 2E(x + ) = E(x) + = + = 23 3 3 3. 4 .=43N2 . 4 (5. 1 (6. 1 (7. 2 (8. . 4 (9. 4 (10

131Cov(x, y)= E(xy) − E(x)E(y) = 8 − 7×1=1Cov(ax + b,ey + d) =aeCov(x, y)= 2×3×1=6 79 − 75 79 − 75P( 71 < x < 79)= P( < z < ) = P( −0/4 < z < 0/4)10 10= 2φ0/4 −1= 2×0/6554 −1= 0/3108. 1 (11. 3 (12. 1 (13. 3 (14. 2 (15. 3 (16. 2 (17. 2 (18P (X2 − X1> µ 2 − µ 1)= P(Z > 0)= 0/5X f (x)xf (x).−1 ≤ r ≤ +1 . 4 (19. 4 (20- 2- 10-0/150/10/5-0/3-0/10E(X) = µ = ∑ xf (x)= 0mb =3 = 0s310/10/120/150/30

13285 ( )30: . 3 4 (1 . 47 . (437 . 25 40 (2 . 4 10⎛25⎞⎛15⎞⎜ ⎟⎜⎟⎝4⎠⎝6⎠⎛40⎞⎜ ⎟⎝10⎠(4 . ( r

133 711E(X / Y = 1)1520(4(4 f (x, y)1420(3710(3 411(24103 ⎛ 2 ⎞⎜ x= xy + ⎟,0

134( 0,1)θ (2 (4 α ،n, m1θ (1 (3 X (12. ...... ...... Var (X i ) =2 − 3 Y = X4( 1− α),n mα,n m(2(4δ 2 X1,X2,X 12(4X1 + X2+ XT =332X1+ X21T = +3 5(2(41x( 1− α),m nα,m n(1(33 (13E (Xi ) =X 1 + XT = 322X1+ 3X2+ 2XT =36 4 X (1441 . P = 50 5050⎛ 1 ⎞1−⎜ ⎟⎝ 50 ⎠(450⎛ 49 ⎞1−⎜ ⎟⎝ 50 ⎠(3(3250⎛ 49 ⎞⎜ ⎟⎠⎝ 50(2(21µ(1(3(1X (1550⎛ 1 ⎞⎜ ⎟⎠⎝ 50 θ X (16: θ 21021031108(4810(3 (0,θ) X (17: 1435 134(43(334(2(226 / 5(1(1(1

135 . 225 µ X (18%95 . [ 5,9](4 [ 4,10 ](3X = 01 , , 2,...,nX = 7 , n = 25 ( z0/025 ≅ 2)[ 3,11](2: µ [ 1,13 ](1 (19 θ ( 1− α)X − ZαnX ⎛ X ⎞⎜1− ⎟ < θ

136{( X ,..., X ) X > K , X < }1 5 1 K2(4{( X ,..., X ) X ≠ K} . 8 50 (22؟ %20 0/04Z0=1 8×4250 50−0/16Z0=1 8×4250 50: a ∑(X i − X)Yâ =i∑(X X)2 i −â =∑(X i Y i )∑(X X)2 i − n2 1 nSY= ∑ (Yi − Ŷ)n − 2 i=12 1 nSY= ∑ (Yi − Y)n − 2 i=1X-2-1021Y413-1-2( 2(41Yi= aXi + e,i = 1,22 (22 (4( 2(4b,a Y = aX + b +5(3−0/04Z0=1 8×4250 500/16Z0=1 8×4250 50(1(3,...,n (23∑(X i − X)Yâ =i∑ X2i∑(X i Y )â = i∑ X2i(1(3e (24: 2 1 nS)2Y= ∑ (Yi − Ŷn −1i=12 1 nSY= ∑ (Yi − Ŷ)n i=1Y = aX +(12 (3b (25 ( â, bˆ) = ( 1/3,−1)( â, bˆ) = ( 1/3,1)( â, bˆ) = ( −1/3,1)( â, bˆ) = ( −1/3,−1)(1( 2(3(4

137 . 380 34-1 (40/75 (4n 2 n(Yi − Ŷ)= 8,∑ (Xi − Xi=1i=1∑ )Y = 5 − X2 Y, X1 (3: (26(Y, X2)0/4 ( 2 4 (4 2= 2-1 (3⎛240P⎜∑Xi⎝i=1 3 (3βˆ⎞< 192⎟⎠Y = aX + X1,...,X 240 1 (2 0 (10 (1(27b (282 (2 ...... (2 (4ρ nt= 41 (1 (29. X 1,X2,X3 ,....X n (1 (3 ρ E(Xi) = 0 ,Var(Xi ) =147(437(346(2 (30.36(1

13885 P (W2 ) = P(W 1)⋅ P(W2/ W 1)+ P(B1)⋅ P(W2/ B 1 )4 3 3 4 12 + 12 24P(W2 ) = × + × = = =7 6 7 6 42 42P( 4 647. 4 (1 :W : B. . 4 (2⎛25⎞⎛15⎞⎜ ⎟⎜⎟⎝4⎠⎝6) =⎠⎛40⎞⎜ ⎟⎝10⎠. . 2 (3P(A) =SASsπr2=πR2⎛= ⎜⎝rR2⎞⎟⎠. 2 (4P( ) = P( ) + P( )P = P1 ( 1−P2) + P2( 1−P1) = 0/95×0/1+0/5×0/9 = 0/14. . 2 (51 4 1 7P(W) = P(K1 ) ⋅ P(W / K1)+ P(K2)⋅P(W / K2)= × + × =2 10 2 101 4×P(K1)⋅P(W / K1)2 10 4P(K1 / W) == =P(W) 11 1120E(x / y+∞413 11= 1)= ∫ xf (x / y = 1)dx=2∫(3x+ x3) dx =−∞70 2 141120 : K. 1 (6: y= 1

139f (x / y) =fx (x) =fy (y) =3 x23 3(xy + ) 3xy+ x23x+=4 2=2=23 1 37y2+ y2+ 116 4 4423x23 3f (x, y)dy = ∫ (xy + )dy = x2+ x042 4 213x23 1f (x, y)dx = ∫ (xy + )dx = y2+ x042 16 4f (x, y)f (y)+∞∫−∞+∞∫−∞ x2X − µZ =δ40−50 60−50P( 40

140. . 2 (132 1Var(ax + b) = a2Var(x) = ( )2Var(x)= × 4 = 14 4P( ) = 1− P(θˆ = λ = E(x)10=813θˆ = 2 x = 2×= 6 / 54L = X − Z α2U = X − Z α2Z =xP = = nx− P0nx x( 1−)n nnS= 7 − 2×nS= 7 + 2×n850= )15= 12515= 13258 20−50 1008 8( 1−)50 5050−0/04=1 8×4250 5050⎛ 49 ⎞= 1−⎜ ⎟⎝ 50 ⎠. 1 (14. 3 (15. 4 (16. 1 (17. 1 (18. 3 (19. 2 (20. 1 (21. 1 (22

141x-2-10210y413-1-25 . 2 (23. 2 (24. 3 (25∑ xx = i= 0n∑ y 5y = i= = 1n 5S xy = ∑ xy − nx.y = −13−0×1 = −13S2 2xx = ∑ x − nx = 10−5×0 = 10iS xy −13a = = = −1/3S xx 10b = y − ax = 1−0×( −1/3)= 13µ = = 0/7542401920/8 −0/75P( ∑ X i < 192 )P(X < ) = P(X < 0/8)= P(Z < ) = 1i=1240380240P(Z < 4 ) = φ4= 1xy-8-10-2-2-132x4104110. 4 (26. 2 (27. 2 (28Var(Bˆ )=2δe∑ − x)2(xi=1∑2(yi − ŷ)n − 2∑ − x)2(xi=1× 84 − 2= 22. 2 (29. 2 (30

142 : : . : : : : -1 -2 -3 -4( ) -5