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# chg3 - T ien hanh n phep th oc l(A = p oi vi moi phep th ap...

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ª Tieán haønh n pheùp thöû ñoäc laäp. ª X laø soá laàn A xaûy ra trong n pheùp thöû, thì X laø ñ.l.n.n rôøi raïc coù theå nhaän caùc giaù trò: 0, 1, 2. . . . , n X phaân phoái theo qui luaät nhò thöùc vôùi caùc tham soá : P(A) = p ñoái vôùi moïi pheùp thöû
Ñaïi löôïng ngaãu nhieân X phaân phoái theo qui luaät nhò thöùc vôùi caùc tham soá n vaø p ñöôïc kyù hieäu laø: X B(n, p)

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) n , .... , 2 , 1 , 0 x ( q p C ) x X ( P P x n x x n x = 2200 = = = - (3.1) b- Coâng thöùc tính xaùc b- Coâng thöùc tính xaùc suaát suaát Neáu X B(n, p)
Neáu X B(n, p) vaø ta caàn tính P(X = x) hoaëc P(X x) thì coù theå duøng haøm BINOMDIST trong Excel. P(X = x) =BINOMDIST(x, n, p,0) P(X x) =BINOMDIST(x,n,p,1)

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Thí duï: X B(50; 0,3) Tính P(X = 16) vaø P(12 X 18) P(X = 16) = BINOMDIST(16,50,0.3,0) = 0,1147 P(12 X 18) = P(X 18) - P(X 11) = BINOMDIST(18,50,0.3,1) - BINOMDIST(11,50,0.3,1) = 0,7204
P(x P(x X X x+h) = P(X = x) + x+h) = P(X = x) + P(X = x+ 1) + . . . . + P(X P(X = x+ 1) + . . . . + P(X = x+h) (3.2) Neáu X B(n, p), thì: Trong ñoù: P(X=x), P(X=x+1),. . . , P(X=x+h) ñöôïc tính theo coâng thöùc (3.1)

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c- Caùc tham soá ñaëc c- Caùc tham soá ñaëc tröng: tröng: Kyø voïng toaùn: Kyø voïng toaùn: Neáu X B(n , p) thì: E(X) = np Phöông sai: Phöông sai: Neáu X Neáu X B(n , p) B(n , p) thì: thì: Var(X) = npq Var(X) = npq
Giaù trò tin chaéc nhaát: Giaù trò tin chaéc nhaát: Neáu X B(n , p) thì: np + p - 1 Mod(X) np + p

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X B(n, p) nhöng n lôùn, p nhoû (p < 0,1), np = λ khoâng ñoåi thì ta coù theå coi X phaân phoái theo qui luaät Poisson vôùi tham soá λ

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X coù phaân phoái Poisson vôùi tham soá λ ñöôïc kyù hieäu laø: X P( λ )
e - haèng soá neâpe: e = ; e 2,71828 n n n 1 1 Lim + Neáu X P ( λ ) thì: P k = P(X = k) = e - λ ! k k λ (k = 0, 1, 2, . . .)

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Neáu X P ( λ ) thì: P(k X k+h) = P k + P k+1 +. . . +P k+h (3.9)
Chuù yù: Neáu X P( λ ), ñeå tính P(X = k) hoaëc P(X ≤ k) ta coù theå duøng haøm POISSON trong EXCEL P(X = k) = POISSON(k, λ ,0) P(X ≤ k) = POISSON(k, λ ,1)

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Thí duï 1: Cho X P(1,5), Tính P (X = 5) vaø P(X ≤ 3) P (X = 5) = POISSON(5,1.5,0) = 0,01412 P(X ≤ 3) = POISSON(3,1.5,1) = 0,934358
Thí duï 2: Moät maùy deät coù 500 oáng sôïi. Xaùc suaát ñeå moät oáng sôïi bò ñöùt trong khoaûng thôøi gian 1 giôø maùy hoaït ñoäng laø 0,004. Tìm xaùc suaát ñeå trong moät giôø coù khoâng quaù 2 oáng sôïi bò ñöùt.

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Giaûi: Neáu coi vieäc quan saùt 1 oáng sôïi xem coù bò ñöùt hay khoâng trong khoaûng thôøi gian 1 giôø laø moät pheùp thöû thì ta coù 500 pheùp thöû ñoäc laäp. Trong moãi pheùp thöû bieán coá A (oáng sôïi bò ñöùt) xaûy ra vôùi xaùc suaát laø p =
Neáu goïi X laø soá oáng sôïi bò ñöùt trong khoaûng thôøi gian 1 giôø thì X ~ B(500; 0,004) Vì n = 500 khaù lôùn, p = 0,004 raát nhoû; np = 500×0,004 = 2 khoâng ñoåi neân ta coù theå coi X ~ P(2)

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Xaùc suaát ñeå coù khoâng quaù 2 oáng sôïi bò ñöùt trong khoaûng thôøi gian 1 giôø laø: P(0 ≤ X ≤ 2) = P 0 + P 1 + P 2 2 2 0 0 e e ! 0
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