Binomial taqsimotga oid misol savollar va muhokama
Binomial taqsimot eng ko'p qo'llaniladigan diskret ehtimollik taqsimotlaridan biridir. U bir qator bir xil, mustaqil sinovlarda muvaffaqiyatlar sonini modellashtirish uchun foydalidir, ularning har biri muvaffaqiyat yoki muvaffaqiyatsizlikka olib keladi. Ushbu maqolada biz bir nechta misollar va batafsil muhokamalar orqali binomial taqsimotni chuqurroq o'rganamiz.
Binomial taqsimotga kirish
Binomial taqsimotning asosiy xususiyatlari:
1. n: Sinovlar yoki takrorlashlar soni.
2. p: Har bir sinovda muvaffaqiyat qozonish ehtimoli.
3. q = 1-p: Har bir sinovda muvaffaqiyatsizlik ehtimoli.
Binomial taqsimotning ehtimollik massasi funksiyasi quyidagicha:
\[ P(X = k) = {n \choose k} p^k (1-p)^{nk} \]
Qayerda:
– \( {n \choose k} = \frac{n!}{k!(nk)!} \)
– \( X \): Muvaffaqiyatlar sonini ifodalovchi tasodifiy o'zgaruvchi.
– \( k \): Izlangan muvaffaqiyatlar soni.
Namunaviy savollar va muhokama
Binomial taqsimot tushunchasini batafsilroq tushunish uchun ba'zi misol masalalardan boshlaylik.
1-misol: Talabalar guruhidan tanlash
Masalan, bizda 10 talabadan iborat guruh bor deylik va har bir talabaning musobaqada ishtirok etish uchun tanlanish ehtimoli 0,3 ga teng. Biz aniq 4 talabaning tanlanish ehtimolini bilmoqchimiz.
1-qadam: Binomial taqsimot parametrlarini aniqlang.
– \( n = 10 \)
– \( p = 0.3 \)
2-qadam: Binomial taqsimotdan foydalanib, \(X = 4 \) ehtimolini hisoblang.
\[ P(X = 4) = {10 \choose 4} (0.3)^4 (0.7)^6 \]
\( {10 \choose 4} \ ni hisoblash ):
\[ {10 \choose 4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = 210 \]
Endi \( (0.3)^4 \) va \( (0.7)^6 \) ni hisoblang:
\[ (0.3)^4 = 0.0081 \]
\[ (0.7)^6 = 0.117649 \]
Shunday qilib,
\[ P(X = 4) = 210 \cdot 0.0081 \cdot 0.117649 \taxminan 0.20012 \]
Demak, aynan 4 talabaning tanlanish ehtimoli taxminan 0.20012 yoki 20.012% ni tashkil qiladi.
2-misol: Ehtimollik 2 dan kichik yoki teng
Endi, masalan, bizdan 2 tadan kam yoki teng talaba tanlanish ehtimoli haqida so'raladi.
1-qadam: Biz \(P(X = 0) \), \(P(X = 1) \) va \(P(X = 2) \) ni hisoblashimiz kerak.
– \(P(X = 0) \) uchun:
\[ P(X = 0) = {10 \choose 0} (0.3)^0 (0.7)^{10} \]
\[ {10 \choose 0} = 1 \]
\[ (0.7)^{10} = 0.0282475 \]
\[ P(X = 0) = 1 \cdot 1 \cdot 0.0282475 = 0.0282475 \]
– \(P(X = 1) \) uchun:
\[ P(X = 1) = {10 \choose 1} (0.3)^1 (0.7)^9 \]
\[ {10 \choose 1} = 10 \]
\[ (0.3) \cdot (0.7)^9 = 0.1210608 \]
\[ P(X = 1) = 10 \cdot 0.3 \cdot 0.1210608 = 0.3631824 \]
– \(P(X = 2) \) uchun:
\[ P(X = 2) = {10 \choose 2} (0.3)^2 (0.7)^8 \]
\[ {10 \choose 2} = 45 \]
\[ (0.3)^2 \cdot (0.7)^8 = 0.2334744 \]
\[ P(X = 2) = 45 \cdot 0.09 \cdot 0.2334744 = 0.2334744 \]
2-qadam: Ehtimolliklarni qo'shing.
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X \leq 2) = 0.0282475 + 0.3631824 + 0.3826372 = 0.7740671 \]
Demak, 2 tadan kam yoki teng talabaning tanlanish ehtimoli taxminan 0.7740671 yoki 77.41% ni tashkil qiladi.
3-misol: Kamida 8 ehtimollik
Agar tajriba 12 marta o'tkazilsa va har bir sinovda muvaffaqiyat qozonish ehtimoli 0.5 ga teng bo'lsa, kamida 8 ta muvaffaqiyat qozonish ehtimoli qanday?
1-qadam: Binomial parametrlarni o'rnating: \( n = 12, p = 0.5 \).
2-qadam: \(X \geq 8 \) uchun ehtimollikni toping.
Bu bir nechta individual ehtimolliklarni hisoblash va ularni qo'shishni talab qiladi:
\[ P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) \]
Birma-bir sanang:
– \(P(X = 8) \) uchun:
\[ P(X = 8) = {12 \choose 8} (0.5)^8 (0.5)^4 \]
\[ {12 \choose 8} = 495 \]
\[ (0.5)^{12} = 0.0002441406 \]
\[ P(X = 8) = 495 \cdot 0.0002441406 = 0.1208496 \]
– \(P(X = 9) \) uchun:
\[ P(X = 9) = {12 \choose 9} (0.5)^9 (0.5)^3 \]
\[ {12 \choose 9} = 220 \]
\[ P(X = 9) = 220 \cdot 0.0002441406 = 0.05371094 \]
– \(P(X = 10) \) uchun:
\[ P(X = 10) = {12 \choose 10} (0.5)^{10} (0.5)^2 \]
\[ {12 \choose 10} = 66 \]
\[ P(X = 10) = 66 \cdot 0.0002441406 = 0.01611328 \]
– \(P(X = 11) \) uchun:
\[ P(X = 11) = {12 \choose 11} (0.5)^{11} (0.5)^1 \]
\[ {12 \choose 11} = 12 \]
\[ P(X = 11) = 12 \cdot 0.0002441406 = 0.002929688 \]
– \(P(X = 12) \) uchun:
\[ P(X = 12) = {12 \choose 12} (0.5)^{12} \]
\[ {12 \choose 12} = 1 \]
\[ P(X = 12) = 1 \cdot 0.0002441406 = 0.0002441406 \]
3-qadam: Barcha ehtimolliklarni qo'shing.
\[ P(X \geq 8) = 0.1208496 + 0.05371094 + 0.01611328 + 0.002929688 + 0.0002441406 \taxminan 0.1938477 \]
Demak, 12 ta sinovda kamida 8 ta muvaffaqiyatga erishish ehtimoli taxminan 0.1938477 yoki 19.38% ni tashkil qiladi.
Xulosa
Binomial taqsimot statistikada ko'plab amaliy qo'llanmalarda juda muhim bo'lgan asosiy tushunchadir. Yuqoridagi misollarda ko'rsatilgandek, binomial taqsimotning turli holatlari uchun ehtimolliklarni qanday hisoblashni tushunish orqali biz ushbu tushunchani real hayotdagi vaziyatlarda qo'llashimiz mumkin. Ushbu mashq, shuningdek, ehtimollik tuzilmalari aniq va tartibli kontekstda qanday ishlashini tushunishimizni mustahkamlaydi.
