Imibuzo Yezibonelo kanye Nengxoxo Ngokusatshalaliswa Kwamathuba
Ukusatshalaliswa kwamathuba kungenye yemiqondo eyisisekelo kuzibalo kanye namathuba. Kusetshenziselwa ukuqonda amathuba ezinhlobo ezahlukene zenombolo engahleliwe eyenzeka. Ukusatshalaliswa kwamathuba kungathatha izinhlobo eziningi kuye ngohlobo lwedatha ehlaziywayo. Izinhlobo ezimbili ezivame kakhulu zokusatshalaliswa kwamathuba zihlukile futhi ziyaqhubeka. Kulesi sihloko, sizobuyekeza izinkinga eziningana zezibonelo futhi sixoxe ngokusatshalaliswa kwamathuba ukuze kusisize siqonde kangcono lesi sihloko.
Ukusatshalaliswa Okuhlukile
Ukusabalalisa okuhlukile kuwukusabalalisa okubala amathuba okuguquguquka okungahleliwe okuhlukile, okungukuthi, ukuguquguquka okungathatha amanani athile kuphela. Izibonelo ezaziwayo zokusabalalisa okuhlukile yi-Binomial Distribution kanye ne-Poisson Distribution.
Isibonelo 1: Ukusatshalaliswa Kwe-Binomial
Ukusatshalaliswa kwe-binomial kuchaza inani lempumelelo ochungechungeni lwezilingo zaseBernoulli. Isivivinyo ngasinye saseBernoulli sinemiphumela emibili: impumelelo noma ukwehluleka. Amathuba empumelelo ahlala engaguquki kulo lonke isivivinyo.
Umbuzo:
Inkampani yemithi ihlola umuthi omusha ezigulini eziyi-10. Amathuba okuthi umuthi usebenze kunoma yisiphi isiguli esisodwa angu-0.7. Bala amathuba okuthi umuthi usebenza ezigulini eziyi-7 kweziyi-10 ngqo.
Ingxoxo:
I-variable engahleliwe \(X\) ilandela ukusatshalaliswa kwe-binomial nge-\(n = 10\) kanye ne-\(p = 0.7\). Umsebenzi we-binomial probability uthi:
\[ P(X = k) = \binom{n}{k} p^k (1 – p)^{n – k} \]
Ukuze \(k = 7\):
\[ P(X = 7) = \binom{10}{7} (0.7)^7 (0.3)^3 \]
Ukubala i-binomial coefficient \(\binom{10}{7}\):
\[ \binom{10}{7} = \frac{10!}{7!(10-7)!} = \frac{10!}{7!3!} = 120 \]
Ukubala amanani okungenzeka:
\[ P(X = 7) = 120 \izikhathi (0.7)^7 \izikhathi (0.3)^3 \]
\[ P(X = 7) \cishe kube ngu-120 \izikhathi 0.0823543 \izikhathi 0.027 \]
\[ P(X = 7) \cishe 0.231 \]
Ngakho-ke, amathuba okuthi umuthi usebenza ezigulini ezingu-7 kwezingu-10 cishe angama-0.231 noma ama-23.1%.
Isibonelo 2: Ukusatshalaliswa kwe-Poisson
Ukusatshalaliswa kwe-Poisson kusetshenziselwa ukukhombisa inani lezehlakalo zesehlakalo esingavamile ngaphakathi kwesikhathi esithile noma isikhala esithile.
Umbuzo:
Isitolo sithola isilinganiso samakhasimende angu-4 ngehora. Angakanani amathuba okuba isitolo sithole amakhasimende angu-5 ncamashi ngehora elilodwa?
Ingxoxo:
I-variable engahleliwe \(X\) ilandela ukusatshalaliswa kwe-Poisson ngepharamitha \(\lambda = 4\). Umsebenzi we-Poisson probability mass uthi:
\[ P(X = k) = \frac{\lambda^ke^{-\lambda}}{k!} \]
Ukuze \(k = 5\):
\[ P(X = 5) = \frac{4^5 e^{-4}}{5!} \]
Isibalo:
\[ P(X = 5) = \frac{1024 \cdot e^{-4}}{120} \]
\[ P(X = 5) \cishe \frac{1024 \cdot 0.0183}{120} \]
\[ P(X = 5) \cishe 0.156 \]
Ngakho-ke, amathuba okuthi isitolo sithole amakhasimende angu-5 ngqo ngehora elilodwa cishe angu-0.156, noma u-15.6%.
Ukusatshalaliswa Okuqhubekayo
Ukusabalalisa okuqhubekayo kusetshenziswa lapho i-random variable elinganiswayo ingathatha noma yiliphi inani ngaphakathi kobubanzi obuthile. Izibonelo ezaziwayo zokusabalalisa okuqhubekayo yi-Normal Distribution kanye ne-Exponential Distribution.
Isibonelo 3: Ukusatshalaliswa Okuvamile
Ukusatshalaliswa Okujwayelekile, okuvame ukubizwa ngokuthi Ukusatshalaliswa Kwe-Gaussian, ukusatshalaliswa okuvame ukusetshenziswa emikhakheni ehlukahlukene, okuhlanganisa isayensi, ubunjiniyela, kanye nezomnotho.
Umbuzo:
Ukuphakama kwamadoda amadala edolobheni kuvame ukusatshalaliswa ngesilinganiso esingu-170 cm kanye nokuphambuka okujwayelekile okungu-10 cm. Yimaphi amathuba okuthi indoda ekhethwe ngokungahleliwe ibe phakathi kuka-160 cm no-180 cm ubude?
Ingxoxo:
Sidinga ukubala i-z-score engu-160 cm kanye no-180 cm. I-z-score ichazwa kanje:
\[ Z = \frac{X – \mu}{\sigma} \]
Ukuze \(X = 160\):
\[ Z_{160} = \frac{160 – 170}{10} = -1 \]
Ukuze \(X = 180\):
\[ Z_{180} = \frac{180 – 170}{10} = 1 \]
Manje sidinga ukubheka amanani okungenzeka kusukela ku--1 kuya ku-1 kuthebula lika-z. Inani kusukela ku-z = -1 kuya ku-z = 1 cishe liyi-0.6826.
Ngakho-ke, amathuba okuthi indoda ekhethwe ngokungahleliwe ibe phakathi kuka-160 cm no-180 cm ubude cishe angu-0.6826 noma angu-68.26%.
Isibonelo 4: Ukusabalalisa Okubanzi
I-Exponential Distribution isetshenziselwa ukulinganisa isikhathi phakathi kwemicimbi enkambisweni ye-Poisson.
Umbuzo:
Isikhathi esimaphakathi phakathi kokufika kwamakhasimende amabili esitolo yimizuzu eyi-15. Yimaphi amathuba okuthi isikhathi esiphakathi kokufika kwamakhasimende amabili singaphansi kwemizuzu eyi-10?
Ingxoxo:
I-Exponential Distribution inepharamitha \(\lambda\) okuyi-inverse ye-mean (\(\mu\)). Nge-average yemizuzu eyi-15:
\[ \lambda = \frac{1}{\mu} = \frac{1}{15} = 0.0667 \]
Umsebenzi wokusabalalisa okuhlanganisiwe kwe-exponential uthi:
\[ P(X \leq x) = 1 – e^{-\lambda x} \]
Ukuze \(x = 10\):
\[ P(X \leq 10) = 1 – e^{-0.0667 \izikhathi 10} \]
\[ P(X \leq 10) = 1 – e^{-0.667} \]
\[ P(X \leq 10) \cishe 1 – 0.5134 \]
\[ P(X \leq 10) \cishe 0.4866 \]
Ngakho-ke, amathuba okuthi isikhathi esiphakathi kokufika kwamakhasimende amabili singaphansi kwemizuzu eyi-10 cishe angama-0.4866 noma ama-48.66%.
Isiphetho
Ukusatshalaliswa kwamathuba, kokubili okuhlukile nokuqhubekayo, kuyimiqondo ewusizo kakhulu yokwenza amamodeli nokuqonda ukuziphatha kweziguquguquko ezingahleliwe. Ukusatshalaliswa kwe-binomial kanye ne-Poisson kuvame ukusetshenziselwa iziguquguquko ezihlukile, kuyilapho ukusatshalaliswa okujwayelekile kanye nokubonakalayo kuyizibonelo zokusatshalaliswa okuqhubekayo.
Ngezibonelo ezingenhla, sithemba ukuthi usuthole ukuqonda okungcono kokuthi ungabala kanjani futhi uhumushe kanjani amathuba okusatshalaliswa kwamathuba. Ngokuzijwayeza njalo, ikhono lakho lokuqonda ukusatshalaliswa kwamathuba lizothuthuka futhi lingasetshenziswa kuzo zonke izigaba ezahlukene.
