26 7.43. P (X = 33) ≈ 0, 030, P (20 ≤ X ≤ 40) ≈ 0, 905. 7.44. n ≥ 127, n ≥ 12608 (suureneb ligikaudu 100 korda). 7.45. ≈ 0, 411. 7.46. F (x, y) = 0, 3 · 1(x)1(y − 2) + 0, 1 · 1(x)1(y − 3)+ +0, 2 · 1(x − 1)1(y − 3) + 0, 2 · 1(x − 2)1(y − 2) + 0, 2 · 1(x − 2)1(y − 3), f(x, y) = 0, 3·δ(x)·δ(y −2)+0, 1·δ(x)·δ(y −3)+0, 2·δ(x−1)·δ(y −3)+ +0, 2 · δ(x − 2) · δ(y − 2) + 0, 2 · δ(x − 2) · δ(y − 3), f 1 (x) = 0, 4 · δ(x) + 0, 2 · δ(x − 1) + 0, 4 · δ(x − 2), f 2 (y) = 0, 5 · δ(x − 2) + 0, 5 · δ(x − 3), EX = 1, EY = 2, 5, DX = 0, 8, DY = 0, 25, σ x ≈ 0, 9, σ y = 0, 5, cov(X, Y ) = 0, 1, K = ( ) 0, 8 0, 1 , R = 0, 25 ( ) 1 0, (2) , 1 r(X, Y ) ≈ 0, (2), korreleeruvad. 7.47. y k \ x i 0 1 2 2 6/20 0 0 1 0 12/20 0 0 0 0 2/20 F (x, y) = 6 20 12 2 · 1(x)1(y − 2) + · 1(x − 1)1(y − 1) + · 1(x − 2)1(y), 20 20 f(x, y) = 6 12 2 · δ(x)δ(y − 2) + · δ(x − 1)δ(y − 1) + · δ(x − 2)δ(y), 20 20 20 f 1 (x) = 6 12 2 · 1(x) + · 1(x − 1) + · 1(x − 2), 20 20 20
27 f 2 (y) = 2 12 · 1(y) + 20 20 · 1(y − 1) + 6 · 1(y − 2), 20 EX = 0, 8, EY = 1, 2, DX = DY = 0, 36, σ x = σ y = 0, 6, cov(X, Y ) = −0, 36, r(X, Y ) = −1, sõltuvad, korreleeruvad, ⎧ ⎪⎨ 2, x = 0, y = 1, x = 1, ⎪⎩ 0, x = 2, ⎧ ⎪⎨ 2, y = 0, x = 1, y = 1, ⎪⎩ 0, y = 2. 7.48. f 1 (x) = f(x, y) = { 1 , (x, y) ∈ [0, 2] × [−1, 2], 6 0, (x, y) ∉ [0, 2] × [−1, 2], { 0, 5, x ∈ [0, 2], 0, x ∉ [0, 2], f 2 (y) = { 1 , y ∈ [−1, 2], 3 0, y ∉ [−1, 2], EX = 1, EY = 0, 5, DX = 0, 25, DY = 0, 75, σ x = 0, 5, σ y = √ 3 2 ≈ 0, 87, cov(X, Y ) = 0, sõltumatud, mittekorreleeruvad. 7.49. 0n, f 1 (x) = { 0, 25(x + 1), x ∈ [0, 2], 0, x ∉ [0, 2], f(y/x) = f 2 (y) = { 1, y ∈ [0, 1], { 2(x − xy + y)/(x + 1), x ∈ [0, 2], 0, x ∉ [0, 2], 0, y ∉ [0, 1], EX = 7 11 , EY = 0, 5, DX = 6 36 , DY = 1 12 , σ x = √ 11 6 ≈ 0, 55, σ y = √ 3 6 ≈ 0, 29, cov(X, Y ) = − 1 36 , r(X, Y ) = − 1 √ 33 ≈ −0, 174,
- Page 1 and 2: TALLINNA TEHNIKAÜLIKOOL Matemaatik
- Page 3 and 4: 3 5. Täiendav kirjandus 1. Jõgi A
- Page 5 and 6: 6.9. Juhusliku suuruse momendid. Ju
- Page 7 and 8: 7 6.20. Juhusliku funktsiooni karak
- Page 9 and 10: 7.2. Arvude moodustamiseks saab kas
- Page 11 and 12: Signaali tabas üks radar. Leidke t
- Page 13 and 14: 13 7.35. Juhusliku suuruse X jaotus
- Page 15 and 16: 15 ja x = E(X/y). (J 8.24, T 3.2.3,
- Page 17 and 18: 17 = 2, DU = 4, DV = 9 ja cov(U, V
- Page 19 and 20: 7.77. Leidke ülesande 7.76. andmet
- Page 21 and 22: 21 F (x) ✻ 1 ✛ ✛ 0, 6 ✛ ✛
- Page 23 and 24: 23 7.25. 8. 7.26. x k 0 1 2 3 4 p k
- Page 25: 25 f(x) ✻ 1 1 F (x) ✻ −2 −1
- Page 29 and 30: 29 EX = π 8 , EY = 4 π · ln √
- Page 31 and 32: 31 7.65. E y (t) = 1, K y (t 1 , t
- Page 33 and 34: 7.80. H 0 : EX = EY, t arvutuslik =
- Page 35 and 36: 35 y ✻ y = 2x 1 ✁ ❅ ❅❅
- Page 37 and 38: 37 Arvestades sündmuste A 1 , A 2
- Page 39 and 40: 39 Ka pärast ümberhindamist P (H
- Page 41 and 42: p 5 = P (X = 5) = P (AAAAA) = ... =
- Page 43 and 44: 43 ühega, saame silmade arvu X jao
- Page 45 and 46: Kuna σ = √ DX, leiame dispersioo
- Page 47 and 48: kus t ≥ 0 ja parameeter λ - sün
- Page 49 and 50: DX = 1 i 2 · 0, 4i2 e 0 (1 + 0, 6e
- Page 51 and 52: Lahendus. Mõõdetava suuruse mista
- Page 53 and 54: muutumistendentsiga korrelatiivne s
- Page 55 and 56: cov(X, Y ) = E(XY )−EX ·EY = ∫
- Page 57 and 58: 57 z ✻ B ❆❆ ❆ ❆ ✁ ✁
- Page 59 and 60: mis ongi hajuvusellipsite kanoonili
- Page 61 and 62: = E1 − 2EX 2 + EX 4 − 4 9 = 1
- Page 63 and 64: 63 = cov(X,X) σ X σ X = DX σ 2 X
- Page 65 and 66: 65 = ∫ 2 0 u 3 2 du − 16 9 = u4
- Page 67 and 68: 67 K y (t 1 , t 2 ) = t 1 t 2 ∫ t
- Page 69 and 70: Punkt koordinaatidega (3, 70; 9, 04
- Page 71 and 72: Vabaliikme b 0 määrame tingimuses
- Page 73 and 74: 73 9. Teadmiste kontroll ja hindami