bernulli formulasi

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11-amaliy. mavzu: bog’liqsiz tajribalar ketma-ketligi. bernulli, puasson va muavr–laplas teoremalari. bernulli formulasi mnmm nn qpcmr )( . (81.1) mnmm nn qpcvpmp  )()( yoki mnm n qp mnm n mp    )!(! ! )( . 1-misol. har bir detalning yaroqli bo’lish ( a hodisa) ehtimolligi 0,8 ga teng. tayyorlangan 5 detaldan 3 tasining yaroqli bo’lish ehtimolligi topilsin. yechilishi. shartga binoan 2,01)(,8,0)(,3,5  pqappapmn bo’lgani uchun (81.1) ga asosan. 2048,02,08,0 )!35(!3 !5 )3( 23233 55    qrsp bo’ladi. mnmm n qpc  ifoda mqpx )(  binom yoyilmasidagi mx qatnashgan hadning koeffitsiyenti bo’lgani uchun )(mpn larni ehtimollikning binominal taqsimot qonuni deyiladi. endi bog’liqmas tajribalar ketma-ketligida hodisaning ro’y berish sonini  bilan belgilab, quyidagi hodisalarning ehtimolliklarini yozamiz: 1) hodisaning m dan kam marta ro’y berish ehtimolligi )()10( 1 0 kpmp n m k n      (81.2) bo’ladi; 2) hodisaning m dan ko’p marta ro’y …

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uhni tekshirmoqda. detalning standartga muvofiq bo’lish ehtimolligi 0,7 ga teng. standart detalning eng katta ehtimolligi soni topilsin. yechilishi. shartga ko’ra n=30, p=0,7. unda 7,20)7,01(7,030)1(  pnp , 7,217,021  pnp bo’lib, eng katta ehtimollik 0m (81.11) ga ko’ra: 7,217,20 0  m , ya’ni 210 m . 4-misol. ikki mergan bir vaqtda nishonga o’q uzmoqda. bitta o’qni uzishda nishonga tekkizish ehtimolligi birinchi mergan uchun 0,8 ga, ikkinchi mergan uchun 0,6 ga teng. agar bir yo’la 15 marta o’q uziladigan bo’lsa, ikkala merganning ham nishonga tekkizishlarining eng katta ehtimolligi sonini toping. yechilishi. birinchi merganni o’qni nishonga tekkizish hodisasini a, ikkinchisinikini b orqali belgilaymiz. u holda shartga ko’ra p(a)=0,8, p(b)=0,6 bo’lib har ikkala merganning bir martadan otishda o’qni nishonga tekkizish ehtimolligi 48,06,08,0)()()(  bpapabp bo’ladi. (81.11)ga n=15, p=0,48 qiymatlarni qo’yib topamiz: ,58,652,02,7)48,01(48,015)1(  pnp 68,748,02,7  pnp va 68,758,6 0  m . demak, 70 m . 5-misol. texnik nazorat bo’limi 24 …

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rmulasini n ning katta qiymatlarida qo’llash qiyin ekan. tajribalar soni yetarlicha katta bo’lib, har bir tajribada hodisaning ro’y berish ehtimolligi p yetarlicha kichik bo’lganda )(mpn ehtimollikni taqribiy hisoblash imkonini beruvchi formula bilan tanishamiz. puasson formulasi agar bernulli sxemasida n da 0p va )0(  np bo’lsa, u holda n da ushbu munosabat urinli bo’ladi:   e m mp m n ! )( . (81.12) shuni aytish joizki puassonning assimptotik formulasi (81.12) tajribalar soni yetarlicha katta bo’lib, har bir tajribada hodisaning ro’y berish ehtimolligi p juda kichik bo’lganda yaxshi natija beradi. bu formulani puasson formulasi deyiladi. endi 3-misolni puasson formulasidan foydalanib yechamiz. shartga ko’ra 1000005,0200000,5,00005,0,200000  npmpn bo’lib, (81.12) formulaga asosan 0375,0 !5 10 ! )( 10 5   ee m mp m n  bo’ladi. demak, izlanayotgan ehtimollik 0375,0)5(200000 p . muavr-laplasning lokal va integral teoremalari muavr-laplasning lokal teoremasi. agar a hodisaning ro’y berish ehtimolligi har bir sinovda …

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p ga teng bo’lsa, ekilgan 100 ta chigitdan unib chiqqanlar soni 85 ta bo’lish ehtimolligini toping. yechilishi. masala shartiga ko’ra .85,2,08,01,8,0)(,100  mqappn masalani muavr–laplasning lokal teoremasidan foydalanib hal etamiz, chunki bernulli formulasidan foydalanish ancha qiyin. 25,1 4 5 16 8085 2,08,0100 8,010085         npq npm x . jadvaldan 1826,0)25,1(  ga ega bo’lamiz. (81.13) formulaga binoan izlanayotgan ehtimollik quyidagiga teng bo’ladi. 0456,01826,0 4 1 )25,1( 2,08,0100 1 )85(100    p 8-misol. tavakkaliga olingan pillaning yaroqsiz chiqish ehtimolligi 0,2 ga teng. tasodifan olingan 400 ta pilladan 70 tadan 130 tagacha yaroqsiz bo’lish ehtimolligi topilsin. yechilishi. shartga ko’ra 8,02,01,2,0,130,70,400 21  qrmmn bo’ladi. ayonki, .25,6 8 80130 ,25,1 8 10 8,02,0400 2,040070 2 2 1 1             npq pnm x npq pnm x (81.14) formulaga binoan izlanayotgan ehtimollik )25,1()25,6()25,1()25,6()13070(400  mp …

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00 n nm p                  bo'lgani uchun,   9962,0071,02  n yoki   4981,0071,0  n bo’ladi. jadvaldan   4981,09,2  ekanini topamiz. demak .1664,8,4084507,40,9,2071,0  nnn 11-misol. hodisaning erkli sinovlarning har birida ro’y berish ehtimolligi 0,8 ga teng. hodisaning kamida 75 marta ro’y berish ehtimolligini 0,9 ehtimollik bilan kutish mumkin bo’lishi uchun nechta sinov o’tkazish lozim? yechilishi. shartga ko’ra 9,0)75(;;75;2,0;8,0 21  npnmmqp n  . laplasning ushbu integral teoremasidan foydalanamiz:                     npq npm npq npm xxnmpn 12 121 )()()(  . bunga masalaning shartidagi ma’lumotlarni qo’yib, quyidagini hosil qilamiz:                      …

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11-amaliy. mavzu: bog’liqsiz tajribalar ketma-ketligi. bernulli, puasson va muavr–laplas teoremalari. bernulli formulasi mnmm nn qpcmr )( . (81.1) mnmm nn qpcvpmp  )()( yoki mnm n qp mnm n mp    )!(! ! )( . 1-misol. har bir detalning yaroqli bo’lish ( a hodisa) ehtimolligi 0,8 ga teng. tayyorlangan 5 detaldan 3 tasining yaroqli bo’lish ehtimolligi topilsin. yechilishi. shartga binoan 2,01)(,8,0)(,3,5  pqappapmn bo’lgani uchun (81.1) ga asosan. 2048,02,08,0 )!35(!3 !5 )3( 23233 55    qrsp bo’ladi. mnmm n qpc  ifoda mqpx )(  binom yoyilmasidagi mx qatnashgan hadning koeffitsiyenti bo’lgani uchun )(mpn larni ehtimollikning binominal taqsimot qonuni deyiladi. endi bog’liqmas tajribalar ketma-ketligida hodisaning ro’y berish sonini  bilan belgilab, quyidagi hodisalarnin...

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