Imibuzo Eyisibonelo Exoxa Ngomsebenzi Wokusabalalisa Okubili
Ukusatshalaliswa kwe-binomial kuwukusatshalaliswa kwamathuba okuhlukile okuchaza inani lempumelelo ekuhlolweni okuhlanganisa inani lezilingo ezizimele ezinemiphumela emibili engaba khona: impumelelo nokwehluleka. Isivivinyo ngasinye sibizwa ngokuthi isivivinyo, futhi ukusatshalaliswa kwe-binomial kuvame ukusetshenziswa ezimweni lapho inani lempumelelo kuzo zonke izivivinyo ezizimele lithakazelisa khona. Kulesi sihloko, sizoxoxa ngemiqondo eyisisekelo yokusatshalaliswa kwe-binomial futhi sinikeze izibonelo nezixazululo.
Imiqondo Eyisisekelo Yomsebenzi Wokusabalalisa Okubili
Ngaphambi kokuba singene emibuzweni eyisibonelo kanye nengxoxo, ake sixoxe ngemiqondo eyisisekelo ehlobene nokusatshalaliswa kwe-binomial.
1. Incazelo: Ukusatshalaliswa kwe-binomial kuchazwa njengesamba sempumelelo ezivivinyweni ezizimele ze-'n', lapho isilingo ngasinye sinemiphumela emibili engaba khona: impumelelo (enamathuba angu-p) noma ukwehluleka (enamathuba angu-q = 1 – p).
2. Umsebenzi Wokungenzeka: Umsebenzi wokungenzeka wokusabalalisa kwe-binomial uthi:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{nk}
\]
Kuphi:
– \( P(X = k) \) amathuba okuba nempumelelo ka-k ezivivinyweni zika-n.
– \( \binom{n}{k} \) iyinhlanganisela ka-n take k, echazwa ngokuthi \( \frac{n!}{k!(nk)!} \).
– \( p \) amathuba okuphumelela esivivinyweni ngasinye.
– \( (1-p) \) amathuba okwehluleka esivivinyweni ngasinye.
3. Inani Elilindelekile kanye Nokwehluka:
– Inani elilindelekile (isilinganiso) sokusatshalaliswa kwe-binomial lingu-\( \mu = np \).
– Ukwehluka kokusatshalaliswa kwe-binomial kungu-\( \sigma^2 = np(1-p) \).
Manje, ake sisebenzise le mibono enkingeni yesibonelo ukuze sithole ukuqonda okujulile.
Isibonelo Umbuzo 1: Izibalo Eziyisisekelo Zokusabalalisa Okubili
Umbuzo:
Inkampani ikhiqiza izingxenye ze-elekthronikhi ezinethuba elingu-0.95 lokuthi ingxenye ngayinye iphumelele ukuhlolwa kwekhwalithi. Uma kukhiqizwa izingxenye eziyi-10, bala amathuba okuthi izingxenye eziyi-8 ziphumelele ukuhlolwa kwekhwalithi.
Ingxoxo:
Singasebenzisa ifomula yokusabalalisa ye-binomial ukuxazulula le nkinga. Okokuqala, sithola amapharamitha alandelayo:
– \( n \) (inani eliphelele lezilingo) = 10
– \( k \) (inani lempumelelo) = 8
– \( p \) (amathuba empumelelo) = 0.95
– \( q \) (amathuba okwehluleka) = 1 – 0.95 = 0.05
Bese ufaka la manani esikhundleni sefomula yokusabalalisa ye-binomial:
\[
P(X = 8) = \binom{10}{8} (0.95)^8 (0.05)^2
\]
Okokuqala, bala inhlanganisela \( \binom{10}{8} \):
\[
\binom{10}{8} = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!} = \frac{10 \izikhathi ezingu-9 \izikhathi ezingu-8!}{8! \izikhathi ezingu-2!} = \frac{10 \izikhathi ezingu-9}{2 \izikhathi ezingu-1} = 45
\]
Bese, bala amathuba \( (0.95)^8 \) kanye \( (0.05)^2 \):
\[
(0.95)^8 \cishe 0.6634
\]
\[
(0.05)^2 = 0.0025
\]
Ekugcineni, phinda wonke lawo manani ukuze uthole:
\[
P(X = 8) = 45 \izikhathi 0.6634 \izikhathi 0.0025 \cishe 0.0744
\]
Ngakho-ke, amathuba okuthi izingxenye ezingu-8 kwezingu-10 ziphumelele ukuhlolwa kwekhwalithi cishe angu-0.0744 noma u-7.44%.
Isibonelo Umbuzo 2: Amathuba Ahlanganisiwe
Umbuzo:
Noma kunjalo ngenkampani efanayo, bala amathuba okuthi okungenani izingxenye ezingu-9 kwezingu-10 ziphumelele ukuhlolwa kwekhwalithi.
Ingxoxo:
Ukuze sixazulule le nkinga, sidinga ukubala amathuba ahlanganisiwe. Amathuba okuthi okungenani izingxenye ezingu-9 kwezingu-10 ziphumelele ukuhlolwa asho ukuthi sibala \( P(X \geq 9) \), okungabhalwa kanje:
\[
P(X \geq 9) = P(X = 9) + P(X = 10)
\]
Ukusebenzisa ifomula yokusabalalisa ye-binomial:
\[
P(X = 9) = \binom{10}{9} (0.95)^9 (0.05)^1
\]
\[
I-P(X = 10) = \binom{10}{10} (0.95)^{10} (0.05)^0
\]
Okokuqala, bala inhlanganisela yecala ngalinye:
\[
\binom{10}{9} = \frac{10!}{9!(10-9)!} = 10
\]
\[
\binom{10}{10} = 1
\]
Bese, bala amathuba e- \( P(X = 9) \) kanye ne- \( P(X = 10) \):
\[
P(X = 9) = 10 \izikhathi (0.95)^9 \izikhathi 0.05
\]
\[
(0.95)^9 \cishe 0.6302
\]
\[
P(X = 9) = 10 \izikhathi 0.6302 \izikhathi 0.05 \cishe 0.3151
\]
\[
P(X = 10) = 1 \izikhathi (0.95)^{10} \izikhathi 1
\]
\[
(0.95)^{10} \cishe 0.5987
\]
\[
I-P(X = 10) = 0.5987
\]
Amathuba aphelele e- \( P(X \geq 9) \):
\[
I-P(X \geq 9) = 0.3151 + 0.5987 \cishe kube ngu-0.9138
\]
Ngakho-ke, amathuba okuthi okungenani izingxenye ezingu-9 kwezingu-10 ziphumelele ukuhlolwa kwekhwalithi cishe angu-0.9138 noma angu-91.38%.
Isibonelo Umbuzo 3: Inani Elilindelekile kanye Nokwehluka
Umbuzo:
Bala inani elilindelekile kanye nokwehluka kwenani lezingxenye eziphumelela ukuhlolwa kwekhwalithi kwezingxenye eziyi-10 ezikhiqizwe, kanye nethuba lokudlula elingu-0.95.
Ingxoxo:
Sebenzisa ifomula elandelayo:
– Inani elilindelekile (isilinganiso) \( \mu = np \)
– Ukwehluka \( \sigma^2 = np(1-p) \)
Nge-\( n = 10 \) kanye ne-\( p = 0.95 \):
\[
\mu = 10 \izikhathi 0.95 = 9.5
\]
\[
\sigma^2 = 10 \izikhathi 0.95 \izikhathi 0.05 = 0.475
\]
Ngakho-ke, inani elilindelekile lenani lezingxenye ezidlula ukuhlolwa kwekhwalithi lingu-9.5, kanti umehluko ungu-0.475.
Isiphetho
Ngezinkinga ezintathu zezibonelo ezingenhla, sixoxe ngendlela yokubala amathuba sisebenzisa ukusatshalaliswa kwe-binomial ezimweni ezahlukahlukene: ukubala amathuba aqondile, amathuba aqongelelekayo, kanye nokubala inani elilindelekile kanye nokwehluka. Ulwazi ngokusatshalaliswa kwe-binomial luwusizo emikhakheni ehlukahlukene, njengokukhiqiza, ucwaningo lwezokwelapha, kanye nezibalo zomphakathi, lapho imiphumela yokuhlolwa okuphindaphindiwe enemiphumela emibili engaba khona ingahlaziywa ukusiza ekwenzeni izinqumo. Ngethemba ukuthi izinkinga zezibonelo kanye nezingxoxo ezinikeziwe kuzokusiza ukuthi uqonde kangcono ukusatshalaliswa kwe-binomial.
