Tambayoyi Misali Game da Aikin Rarraba Binomial
Rarraba binomial wani rarrabawar yiwuwar da ba ta da bambanci wanda ke bayyana adadin nasarorin da aka samu a cikin gwaji wanda ya ƙunshi gwaje-gwaje masu zaman kansu da dama tare da sakamako biyu masu yiwuwa: nasara da gazawa. Kowace gwaji ana kiranta gwaji, kuma ana amfani da rarraba binomial sau da yawa a cikin yanayi inda adadin nasarorin da aka samu a cikin gwaje-gwaje masu zaman kansu da yawa abin sha'awa ne. A cikin wannan labarin, za mu tattauna mahimman ra'ayoyin rarraba binomial kuma mu ba da misalai da mafita.
Ka'idoji na Asali na Aikin Rarraba Binomial
Kafin mu shiga cikin tambayoyin misalai da tattaunawa, bari mu tattauna wasu muhimman ra'ayoyi da suka shafi rarraba binomial.
1. Ma'ana: Rarraba binomial an bayyana shi a matsayin jimlar nasarorin da aka samu a gwaje-gwajen 'n' masu zaman kansu, inda kowace gwaji ke da sakamako biyu masu yiwuwa: nasara (tare da yuwuwar p) ko gazawa (tare da yuwuwar q = 1 – p).
2. Aikin Yiwuwa: Aikin yuwuwar rarraba binomial shine:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{nk}
\]
Ina:
– \( P(X = k) \) shine yuwuwar samun nasarar k a cikin gwaje-gwajen n.
– \( \binom{n}{k} \) haɗin n take k ne, wanda aka bayyana a matsayin \( \frac{n!}{k!(nk)!} \).
– \( p \) shine yuwuwar samun nasara a kowace gwaji.
– \( (1-p) \) shine yuwuwar gazawa a kowace gwaji.
3. Darajar da ake tsammani da Bambancin da ke Tsakaninsu:
– Ƙimar da ake tsammani (matsakaicin) na rarraba binomial shine \( \mu = np \).
– Bambancin rarrabawar binomial shine \( \sigma^2 = np(1-p) \).
Yanzu, bari mu yi amfani da waɗannan ra'ayoyi a cikin matsala misali don samar da fahimta mai zurfi.
Misali Tambaya ta 1: Lissafi na Asali na Rarraba Binomial
Tambaya:
Kamfani yana samar da kayan lantarki tare da yuwuwar 0.95 cewa kowane sashi ya ci gwajin inganci. Idan aka samar da kayan aiki 10, a ƙididdige yiwuwar cewa daidai sassa 8 sun ci gwajin inganci.
Tattaunawa:
Za mu iya amfani da dabarar rarraba binomial don magance wannan matsalar. Da farko, za mu gano waɗannan sigogi:
– \( n \) (jimillar adadin gwaje-gwajen) = 10
– \( k \) (adadin nasarorin) = 8
– \( p \) (yiwuwar nasara) = 0.95
– \( q \) (yiwuwar gazawa) = 1 – 0.95 = 0.05
Sai a maye gurbin waɗannan dabi'u a cikin tsarin rarraba binomial:
\[
P(X = 8) = \binom{10}{8} (0.95)^8 (0.05)^2
\]
Da farko, ƙididdige haɗin \( \binom{10}{8} \):
\[
\binom{10}{8} = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!} = \frac{10 \sau 9 \sau 8!}{8! \sau 2!} = \frac{10 \sau 9}{2 \sau 1} = 45
\]
Sannan, ƙididdige yiwuwar \( (0.95)^8 \) da \( (0.05)^2 \):
\[
(0.95)^8 \kimanin 0.6634
\]
\[
(0.05) ^ 2 = 0.0025
\]
A ƙarshe, ninka duk waɗannan dabi'un don samun:
\[
P(X = 8) = 45 \sau 0.6634 \sau 0.0025 \sau 0.0744
\]
Don haka, yuwuwar cewa daidai sassa 8 cikin 10 sun ci gwajin inganci shine kusan 0.0744 ko 7.44%.
Misali Tambaya ta 2: Tarin Yiwuwar Tarawa
Tambaya:
Duk da haka, idan kana da kamfani ɗaya, ƙididdige yiwuwar aƙalla abubuwa 9 cikin 10 su ci jarrabawar inganci.
Tattaunawa:
Domin magance wannan matsala, muna buƙatar ƙididdige yiwuwar tarin abubuwa. Yiwuwar aƙalla abubuwa 9 cikin 10 sun ci jarabawar yana nufin mun ƙididdige \( P(X \geq 9) \), wanda za a iya rubutawa kamar haka:
\[
P (X \geq 9) = P (X = 9) + P (X = 10)
\]
Amfani da dabarar rarraba binomial:
\[
P(X = 9) = \binom{10}{9} (0.95)^9 (0.05)^1
\]
\[
P(X = 10) = \binom{10}{10} (0.95)^{10} (0.05)^0
\]
Da farko, ƙididdige haɗin kowace shari'a:
\[
\binom{10}{9} = \frac{10!}{9!(10-9)!} = 10
\]
\[
\binom{10}{10} = 1
\]
Sannan, ƙididdige yiwuwar \( P(X = 9) \) da \( P(X = 10) \):
\[
P(X = 9) = 10 \sau (0.95)^9 \sau 0.05
\]
\[
(0.95)^9 \kimanin 0.6302
\]
\[
P(X = 9) = 10 \sau 0.6302 \sau 0.05 \sau 0.3151
\]
\[
P(X = 10) = sau 1 (0.95)^{10} \sau 1
\]
\[
(0.95)^{10} \kimanin 0.5987
\]
\[
P(X = 10) = 0.5987
\]
Jimlar yiwuwar \( P(X \geq 9) \):
\[
P(X \geq 9) = 0.3151 + 0.5987 \kimanin 0.9138
\]
Don haka, yuwuwar aƙalla abubuwa 9 cikin 10 su ci jarrabawar inganci kusan 0.9138 ne ko 91.38%.
Misali Tambaya ta 3: Darajar da ake tsammani da Bambancin da ke Tsakaninsu
Tambaya:
Lissafa ƙimar da ake tsammani da bambancin adadin abubuwan da suka wuce gwajin inganci daga cikin abubuwa 10 da aka samar, tare da yuwuwar wucewar 0.95.
Tattaunawa:
Yi amfani da dabarar da ke ƙasa:
– Ƙimar da ake tsammani (matsakaicin) \( \mu = np \)
– Bambancin \( \sigma^2 = np(1-p) \)
Tare da \( n = 10 \) da \( p = 0.95 \):
\[
\mu = 10 \sau 0.95 = 9.5
\]
\[
\sigma^2 = 10 \sau 0.95 \sau 0.05 = 0.475
\]
Don haka, ƙimar da ake tsammani na adadin abubuwan da suka wuce gwajin inganci shine 9.5, kuma bambancin shine 0.475.
Kammalawa
Ta hanyar misalan matsaloli guda uku da ke sama, mun tattauna yadda ake ƙididdige yiwuwar amfani da rarraba binomial don yanayi daban-daban: ƙididdige yiwuwar daidaito, yuwuwar tarin bayanai, da ƙididdige ƙimar da ake tsammani da bambancin bayanai. Sanin rarraba binomial yana da amfani a fannoni daban-daban, kamar masana'antu, binciken likita, da ƙididdigar zamantakewa, inda za a iya nazarin sakamakon gwaje-gwajen da aka maimaita tare da sakamako biyu masu yiwuwa don taimakawa wajen yanke shawara. Da fatan, matsalolin misalai da tattaunawar da aka bayar za su taimaka wajen ƙara fahimtar rarraba binomial.
