Ceistean Eisimpleir agus Deasbad mu Sgaoileadh Binomial
’S e an sgaoileadh dà-thaobhach aon de na sgaoilidhean coltachd sgarach as cumanta a thathas a’ cleachdadh. Tha e feumail airson modaladh a dhèanamh air an àireamh de shoirbheasan thar grunn dheuchainnean co-ionann, neo-eisimeileach, agus gach fear dhiubh a’ toirt a-mach soirbheachas no fàilligeadh. San artaigil seo, nì sinn sgrùdadh nas doimhne air an sgaoileadh dà-thaobhach le bhith a’ toirt seachad grunn eisimpleirean agus deasbad mionaideach.
Ro-ràdh do Sgaoileadh Binomial
Prìomh fheartan an t-sgaoilidh dà-thomaich:
1. n: Àireamh dheuchainnean no ath-aithris.
2. p : Cothrom soirbheachais anns gach deuchainn.
3. q = 1-p: Cothrom fàilligeadh anns gach deuchainn.
Is e gnìomh mais coltachd an t-sgaoilidh dà-thaobhach:
[P(X = k) = {n \tagh k} p^k (1-p)^{nk} \]
Càite:
– \( {n \tagh k} = \frac{n!}{k!(nk)!} \)
– \( X \): Caochladair air thuaiream a’ riochdachadh an àireamh de shoirbheasan.
– \(k \): An àireamh de shoirbheasan a thathar a’ sireadh.
Ceistean Eisimpleir agus Deasbad
Tòisichidh sinn le beagan eisimpleirean de dhuilgheadasan gus tuigse nas mionaidiche fhaighinn air bun-bheachd an t-sgaoilidh binomial.
Eisimpleir 1: A’ taghadh bho Bhuidheann Oileanach
Mar eisimpleir, can gu bheil buidheann de 10 oileanaich againn, agus gu bheil coltachd 0,3 ann gun tèid gach oileanach a thaghadh airson pàirt a ghabhail ann am farpais. Tha sinn airson faighinn a-mach dè cho coltach ‘s a tha e gun tèid dìreach 4 oileanaich a thaghadh.
Ceum 1: Comharraich paramadairean an t-sgaoilidh dà-thaobhach.
– \(n = 10 \)
– \(p = 0.3 \)
Ceum 2: Cleachd an sgaoileadh dà-thaobhach gus coltachd \( X = 4 \) obrachadh a-mach.
[P(X = 4) = {10 \tagh 4} (0.3)^4 (0.7)^6 \]
A’ tomhas \( {10 \tagh 4} \):
[{10 \tagh 4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = 210 \]
A-nis obraich a-mach \( (0.3)^4 \) agus \( (0.7)^6 \):
\[ (0.3)^4 = 0.0081 \]
\[ (0.7)^6 = 0.117649 \]
Mar sin,
[P(X = 4) = 210 ≥ 0.0081 ≥ 0.117649 timcheall air 0.20012]
Mar sin, tha coltachd timcheall air 0.20012 no 20.012% gun tèid dìreach 4 oileanaich a thaghadh.
Eisimpleir 2: Cothrom nas lugha na no co-ionann ri 2
A-nis, mar eisimpleir, thèid faighneachd dhuinn mun chothrom gum bi nas lugha na no co-ionann ri 2 oileanach air an taghadh.
Ceum 1: Feumaidh sinn obrachadh a-mach \( P(X = 0) \), \( P(X = 1) \), agus \( P(X = 2) \).
– Airson \(P(X = 0) \):
[P(X = 0) = {10 \tagh 0} (0.3)^0 (0.7)^{10} \]
\[ {10 \tagh 0} = 1 \]
\[ (0.7)^{10} = 0.0282475 \]
[P(X = 0) = 1 ⋅ 1 0.0282475 = 0.0282475]
– Airson \(P(X = 1) \):
[P(X = 1) = {10 \tagh 1} (0.3)^1 (0.7)^9 \]
\[ {10 \tagh 1} = 10 \]
[(0.3) ≤ (0.7)^9 = 0.1210608]
[P(X = 1) = 10 ⋅ 0.3 0.1210608 = 0.3631824]
– Airson \(P(X = 2) \):
[P(X = 2) = {10 \tagh 2} (0.3)^2 (0.7)^8 \]
\[ {10 \tagh 2} = 45 \]
[(0.3)^2 ⋅ (0.7)^8 = 0.2334744]
[P(X = 2) = 45 ⋅ 0.09 0.2334744 = 0.2334744]
Ceum 2: Cuir na cothroman ri chèile.
[P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)]
[P(X ≤ 2) = 0.0282475 + 0.3631824 + 0.3826372 = 0.7740671]
Mar sin, tha coltachd gum bi nas lugha na no co-ionann ri 2 oileanach air an taghadh timcheall air 0.7740671 no 77.41%.
Eisimpleir 3: Cothrom co-dhiù 8
Ma thèid deuchainn a dhèanamh 12 uair, agus ma tha coltachd soirbheachais anns gach deuchainn 0.5, dè an coltachd gum bi co-dhiù 8 soirbheasan ann?
Ceum 1: Suidhich na paramadairean dà-thaobhach: \(n = 12, p = 0.5 \).
Ceum 2: Lorg an coltachd airson (X ≥ 8).
Feumaidh seo grunn chothroman fa leth obrachadh a-mach agus an cur ri chèile:
\[ P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) \]
Cunnt aon às dèidh a chèile:
– Airson \(P(X = 8) \):
[P(X = 8) = {12 \tagh 8} (0.5)^8 (0.5)^4 \]
\[ {12 \tagh 8} = 495 \]
\[ (0.5)^{12} = 0.0002441406 \]
[P(X = 8) = 495 ≥ 0.0002441406 = 0.1208496]
– Airson \(P(X = 9) \):
[P(X = 9) = {12 \tagh 9} (0.5)^9 (0.5)^3 \]
\[ {12 \tagh 9} = 220 \]
[P(X = 9) = 220 ≥ 0.0002441406 = 0.05371094]
– Airson \(P(X = 10) \):
[P(X = 10) = {12 \tagh 10} (0.5)^{10} (0.5)^2 \]
\[ {12 \tagh 10} = 66 \]
[P(X = 10) = 66 ≥ 0.0002441406 = 0.01611328]
– Airson \(P(X = 11) \):
[P(X = 11) = {12 \tagh 11} (0.5)^{11} (0.5)^1 \]
\[ {12 \tagh 11} = 12 \]
[P(X = 11) = 12 ≥ 0.0002441406 = 0.002929688]
– Airson \(P(X = 12) \):
[P(X = 12) = {12 \tagh 12} (0.5)^{12} \]
\[ {12 \tagh 12} = 1 \]
[P(X = 12) = 1 ≥ 0.0002441406 = 0.0002441406]
Ceum 3: Cuir na cothroman uile ri chèile.
[P(X ≤ 8) = 0.1208496 + 0.05371094 + 0.01611328 + 0.002929688 + 0.0002441406 timcheall air 0.1938477]
Mar sin, tha coltachd gum bi co-dhiù 8 soirbheasan ann an 12 deuchainnean timcheall air 0.1938477 no 19.38%.
Co-dhùnadh
’S e bun-bheachd bunaiteach ann an staitistig a th’ anns an sgaoileadh dà-thaobhach a tha deatamach ann am mòran thagraidhean practaigeach. Le bhith a’ tuigsinn mar a nì sinn obrachadh a-mach coltachdan airson diofar chùisean den sgaoileadh dà-thaobhach, mar a chithear anns na h-eisimpleirean gu h-àrd, is urrainn dhuinn a’ bhun-bheachd seo a chur an sàs ann an suidheachaidhean fìor. Bidh an eacarsaich seo cuideachd a’ neartachadh ar tuigse air mar a bhios structaran coltachd ag obair ann an co-theacsa soilleir agus eagraichte.
