O`zbekiston respublikasi oliy va o`rta maxsus ta`lim vazirligi
Download 1.11 Mb. Pdf ko'rish
|
- Bu sahifa navigatsiya:
- 6- §. Muavr — Laplasning lokal va integral teoremalari
- 23.7- teorema.
- 23.8-teorema.
- Misol.
- 23.6-teorema. (Puasson teoremasi).
|
1-misol. Texnik nazorat bo`limi 24 ta detaldan iborat guruhni tekshirmoqda. Detalning yaroqli standartga muvofiq bo`lish ehtimoli 0,6 ga teng. Yaroqli deb tan olinadigan detalning eng katta ehtimolli soni topilsin.
( )
np − − 1 = 24 ⋅0,6 - (1 - 0,6)= 14,4 - 0,4 = 14, pr + r = 24 ⋅0,6 + 0,6= 14,4 + 0 , 6 = 15 bo`lib, eng katta ehtimolli k
≤ k 0
≤ 15 tengsizliklarni qanoatlantirishi kerak. Demak, bu munosabatdan ko`rinadiki, eng katta ehtimolli son ikkita bo`ladi: k 0
= 14, k 0 + 1 = 15. 6- §. Muavr — Laplasning lokal va integral teoremalari Bernulli sxemasida R p (k) extimolni topish uchun ushbu ( )
( )
n k k n n р р C k Р − − = 1
formulaga ega bo`lgan edik. Bu formula sodda bo`lsa ham undan, ayniqsa, tajribalar soni 15 katta bo`lganda foydalanish ancha qiyin bo`ladi. Natijada bu ifodani o`ziga qaraganda soddaroq va ayni paytda hisoblash uchun oson bo`lgan ifoda bilan taqribiy ifodalash masalasi tug`iladi. Bu masala ba`zi hollar uchun Muavr — Laplasning l o k a l v a i n t e g r a l teoremasi yordamida hal etiladi. Quyida ularni isbotsiz keltiramiz.
(Muavr — Laplasning lokal teoremasi). Agar Bernulli sxemasida p etarlicha katta bo`lib, har bir sinashda A hodisaning ro`y berish ehtimoli r (0 o`zgarmas bo`lsa, u holda R ( )
)
( ( ) p np np k n e p np k Р − − − − ≈ 1 2 2 1 2 1 π (23.21) taqribiy formula o`rinli bo`ladi. Agar
( )
np np k х − − = 1
deyilsa, u holda ( ) ( ) 2 1 2 2 2 х p np np k = − −
bo`lib, yuqoridagi (23.20) formula quyidagi ko`rinishga keladi. ( ) ( ) ( ) 2 2 2 2 2 1 1 1 1 2 1 х х n e p np e p np k Р − − ⋅ − = − ≈ π π
Ushbu ( )
2 2 2 1 х e х − = π ϕ belgilash kiritsak, u holda ( ) ( ) ( ) . 1 1 х p np k Р n ϕ ⋅ − ≈ (23.21) Bu erda ( )
х ϕ juft funktsiya bo`lib, uning qiymatlari uchun jadvallar tuzilgan Misol. Har bir ekilgan chigitni unib chiqish (A hodisa) ehtimoli o`zgarmas bo`lib, R(A) = r = 0,8 ga teng bo`lsa, ekilgan 100 ta chigitdan unib chiqqanlar soni 85 ta bo`lish ehtimolini toping.
echish. Masala shartiga ko`ra p = 100, r = R(A) = 0,8, q = 1 - r = 0,2, k = 85. Ravshanki, talab qilingan
100
(85) ehtimolni Bernulli formulasi bilan ( )
( ) ( ) 15 85 100 2 , 0 8 , 0 ! 85 ! 85 ! 100 85 ⋅ = Р aniq hisoblash (n – katta bo`lgan holda) juda qiyin, bunday holda r - o`zgarmas (0 hisoblaymiz: . 25 , 1 4 5 16 80 85 2 , 0 8 , 0 100
8 , 0 100 85 = = − = ⋅ ⋅ ⋅ − = − = npq np k х
Muavr — Laplasning lokal teoremasiga asosan ( ) ( )
( ) . 25 , 1 4 1 25 , 1 2 , 0 8 , 0 100
1 85 100 ϕ ϕ = ⋅ ⋅ ⋅ ≈ Р
Ilovadagi jadvaldan ( ) 1826
, 0 25 , 1 ≈ ϕ ekanligidan, talab qilingan ehtimollik ( ) 0456
, 0 1826 , 0 4 1 85 100 = ⋅ ≈ Р bo`ladi. Mazkur bobning 7-§ da p ta erkli tajribada A hodisaning kami bilan k
ko`pi bilan k 2 marta ro`y berish hodisasi { } 2 1
k ≤ ≤ µ ning ehtimoli bo`lishini ko`rgan edik. { }
∑ = = ≤ ≤ 2 1 2 1 k k m n n m P k k Р µ
16 Muavr — Laplasning integral teoremasn p etarlicha katta bo`lganda { } 2 1 k k Р n ≤ ≤ µ
ehtimolni taqribiy ifodalovchi formulani beradi. 23.8-teorema. Bernulli sxemasida lokal teorema shartlari { }
1 k k Р n ≤ ≤ µ ehtimol uchun ushbu { } ( ) ( ) ∫ − − = − − ≈ ≤ ≤ −
np np k dx e p np np k k k Р x n 1 1 2 1 1 2 2 2 1 2 π µ
taqribiy formula o`rinli bo`ladi, bu erda 0 < r < 1. Ushbu ( )
∫ − = Φ x t dt e x 0 2 2 2 1 π (23.23) Laplas funktsiyasi toq funktsiya bo`lib, x ning turli qiymatlariga integralning mos qiymatlari jadvali tuzilgan Misol. Tavakkaliga olingan pillaning yaroqsiz chiqish ehtimoli 0,2 ga teng. Tasodifan olingan 400 ta pilladan 70 tadan 130 tagacha yaroqsiz bo`lish ehtimoli topilsin.
Shartga ko`ra p = 400, k 1 = 70, k 2 = 130, r = 0,2, q = 1 - r = 0,8 bo`ladi. Ravshanki, ( ) ( )( ) . 25 , 6 ; 25 , 1 8 10 2 , 0 1 2 , 0 1 2 , 0 400 2 , 0 400
70 1 1 = ′′ − = − = − − ⋅ ⋅ − = − − = ′ х p np np k x
Yuqorida keltirilgan (23.24) formulaga muvofiq izlanayotgan ehtimol taxminan { } { } ( ) ( ) 25 , 1 25 , 6 130
70 400
2 1 − Φ − Φ ≈ ≤ ≤ = ≤ ≤ µ µ
k k Р n
bo`ladi. Jadvaldan hamda ( ) x Φ
ning toq funktsiyaligini e`tiborga olib quyidagilarni topamiz: F(-1,25) = -0,39435, F(6,25) = 0,5 (2-ilovaga qaralsin), u holda { } ( ) 89435 , 0 39435 , 0 5 , 0 130 70 400
= − − ≈ ≤ ≤ µ Р
bo`ladi. Demak, izlanayotgan ehtimollik { } . 89435 , 0 130 70 400 ≈ ≤ ≤ µ Р
Faraz qilaylik, p ta erkli tajribada A hodisa µ marta ro`y bersin. Har bir tajribada A hodisaning ro`y berish ehtimoli r (0 n µ miqdor A hodisaning nisbiy chastotasi bo`ladi. Yuqorida keltirilgan Muavr—Laplasning integral teoremasidan foydalanib, nisbiy chastota n µ ning o`zgarmas ehtimol r dan chetla nish ehtimolini topish mumkin: 0 > ∀ ε olinganda ham ushbu ε µ
− p
tengsizlik orqali ifodalanadigan hodisaning ehtimoli uchun ( ) − ⋅ Φ ≈ < −
p n p n Р 1 2 ε ε µ taqribiy formula o`rinli bo`lishini ko`rsatamiz. Ravshanki, ( ) ( ) ( ) ( ) ( ) ( ) . 1 1 1 1 1 1
p n p np np p p n p p n p p n n np p p n p n p n −
− −
− − ⇔ − < < − − < − − ⇔ < −
− ⇔
− ε µ ε ε µ ε ε µ ε ε µ Demak,
17 ( ) ( ) ( ) . 1 1 1 − < − − < − − = < −
p n p np np p p n P p n P ε µ ε ε µ (23.25) Muavr—Laplasning integral teoremasidan foydalanib topamiz: ( )
) ( ) ( ) ( ) ( ) ( ) . 1 2 2 1 2 1 1 1 1 1 0 2 1 1 2 2 2 − Φ = = = ≈ −
− −
− − ∫ ∫ − − − − − − p p n dx e dx e p p n p np np p p n P p p n x p p n p p n x ε π π ε µ ε ε ε ε
Bu holda (23.25) munosabatdan ( ) − ⋅ Φ ≈ < −
p n p n P 1 2 ε ε µ (23.26) bo`lishi kelib chiqadi. Bu formuladai ε, p va ehtimol qiymatlarini topish mumkin.
bo`lsin. Tangani 400 marta tashlanganda A xodisa nisbny chastotasi 400 µ
absolyut kiymat bo`yicha chetlanishi 0,08 dan kichik bo`lish ehtimolini toping.
ε = 0,08. U holda (23.26) formulaga asosan: ( )
. 2 , 3 2 5 , 0 5 , 0 400 08 , 0 2 8 , 0 5 , 0 400
Φ = ⋅ ⋅ Φ ≈ < − µ P
Jadvaldan F(3,2) = 0,49931 ni olsak, 99862
, 0 8 , 0 5 , 0 400 ≈ < − µ P
bo`ladi. 7- §. Puasson teoremasi Biz yuqorida o`rgangan Bernulli sxemasida p ta erkli tajribada A hodisaning k marta ro`y berishi ehtimoli Bernulli formulasi bilan hisoblanishini ko`rdik. Bernulli formulasini keltirib chiqarishda A hodisaning har bir tajribada ro`y berish ehtimoli o`zgarmas va u r ga teng bo`lsin deb olindi (0 < r < 1].
ortib borishi bilan r ning kamayib borishiga bog`langan bo`ladi. Bunday holda Bernulli sxemasi uchun kuyidagi teorema o`rinli bo`ladi.
→∞ da r→ 0 va pr→λ (λ>0) bo`lsa, u holda n →∞ da ushbu munosabat o`rinli bo`ladi: ( ) λ
− →
k k Р k n ! yoki ( ) . ! λ λ − ≈ e k k Р k n (23.16)
Bu taqribiy formulani Puasson formulasi deyiladi. Isbot. Ma`lumki, p ta o`zaro erkli tajribada A hodisaning k marta ro`y berish ehtimoli ( )
( )
n k k n n р р C k Р − − = 1 bo`ladi, bunda ( ) ! ! !
n k n С k n − = . Keyingi tenglikni quyidagicha yozib olamiz: 18 ( ) ( )( ) ( ) ( ) (
) . 1 1 ...
2 1 1 1 ! ! 1 1 ... 2 1 1 1 ! 1 ... 1 ... 3 2 1 ! ...
1 ...
3 2 1 ! ! ! − − −
− = − − − − ⋅ = = + − − = − ⋅ ⋅ + − − ⋅ ⋅ = − = n k n n k n k n k n n n n n n k k n n n k n k n k n k n k n k n С k k n
Bu tenglikdan esa − − −
− = n k n n С n k k n k 1 1 ... 2 1 1 1 ! (23.17)
bo`lishi kelib chiqadi. Endi ( ) 1 , 1 1 0 ≥ ≤ ≤ ≤ ≤ n n k a k sonlar uchun o`rinli bo`lgan ushbu ( )(
) ( ) n n a a a a a a + + + − ≥ − − − ... 1 1 ... 1 1 2 1 2 1
sodda tengsizlikdan foydalanib topamiz. . 1 ... 2 1 1 1 1 ... 2 1 1 1 − + + + − ≥ − − − − n k n n n k n n
Ravshanki, ( ) ( ) ( ) ( ) . 2 1 2 1 1 1 ... 2 1 1 1 ... 2 1
k k k k n k n n k n n − = − = − + + + = − + + +
Demak, ( ) . 2 1 1 1 1 ... 2 1 1 1
k k n k n n − − ≥ − − − − (23.18)
Natijada (23.17), (23.18) munosabatlardan ( )
k k С n k k n k 2 1 1 ! − − ≥
bo`lishi kelib chiqadi. Agar 1 ! ≤ k n k С n k ekanligini e`tiborga olsak, u holda ( )
! 2 1 1 ≤ ≤ − −
n k С n k n k k
bo`lishini, ya`ni ( ) ! 2 1 1 ! k n С n k k k n k k n k ≤ ≤ − −
bo`lishini topamiz. Bu tengsizlikni ( ) k n k p p − − 1 ga ko`paytirsak, unda quyidagi tengsizliklar hosil bo`ladi: ( )
) ( ) ( ) . 1 ! 1 1 ! 2 1 1
n k k k n k k n k n k k p р k n p р С p р k n n k k − − − − ≤ − ≤ − − −
Demak ( ) ( ) ( ) ( ) . 1 ! 1 ! 2 1 1 k n k k n k n k k p р k n k P p р k n n k k − − − ≤ ≤ − − − (23.19)
Endi shu tengsizlikda qatnashuvchi ( )
n k k p p k n − − 1 ! ifodani quyidagicha yozamiz: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
. 1 1 ! 1 1 ! 1 1 ! 1 ! np np n k k n k k n k k k n k k n np p k np n np p k np p p k np p p k n − − − − − − −
− = −
− = = − − = −
19 Agar n →∞ da pr →λ, ( ) ( ) ( ) 0 1 1 , 1 2 1 1 → → − → − − − p p n k k k va
( ) ( ) λ λ − − − − → −
− e k n np p k np k np np n k k ! 1 1 !
bo`lishini e`tiborga olsak, unda (23.19) munosabatdan ( )
( ) λ λ − − → − = e k р р C k Р k k n k k n n ! 1 bo`lishini topamiz. Teorema isbotlandi. Puasson formulasi tajribalar soni etarlicha katta bo`lib, har bir tajribada hodisaning ro`y berish ehtimoli r etarlicha kichik bo`lganda ( )
Download 1.11 Mb. Do'stlaringiz bilan baham: 
|