Ngā Tauira Pātai me te Kōrero mō te Tohatoha Binomial
Ko te tohatoha rua-ira tētahi o ngā tohatoha tūponotanga motuhake e whakamahia whānuitia ana. He mea whai hua mō te whakatauira i te maha o ngā angitu puta noa i te maha o ngā whakamātautau ōrite, motuhake, ā, ka puta he angitu, he kore rānei i ia whakamātautau. I roto i tēnei tuhinga, ka ruku hohonu ake mātou ki te tohatoha rua-ira mā te whakarato i ētahi tauira me tētahi matapakinga taipitopito.
He Kupu Whakataki ki te Tohatoha Binomial
Ngā āhuatanga matua o te tohatoha rua-ira:
1. n : Te maha o ngā whakamātautau, ngā tāruarua rānei.
2. p : Te tūponotanga o te angitu i ia whakamātautau.
3. q = 1-p : Te tūponotanga o te korenga i roto i ia whakamātautau.
Ko te mahi papatipu tūponotanga o te tohatoha rua ko:
\[ P(X = k) = {n \choose k} p^k (1-p)^{nk} \]
kāore i te mana:
– \( {n \choose k} = \frac{n!}{k!(nk)!} \)
– \( X \): He taurangi matapōkere e tohu ana i te maha o ngā angitu.
– \( k \): Te maha o ngā angitu i rapua.
Ngā Pātai Tauira me te Kōrero
Me tīmata tātou me ētahi tauira rapanga hei mārama ake i te ariā o te tohatoha rua-ira.
Tauira 1: Te Kōwhiri mai i tētahi Rōpū Ākonga
Hei tauira, me kī he rōpū ākonga tekau tā tātou, ā, ko te tūponotanga ka tohua ia ākonga ki te whai wāhi ki tētahi whakataetae he 0,3. E hiahia ana tātou ki te mōhio ki te tūponotanga ka tohua kia whā tonu ngā ākonga.
Hipanga 1: Tāutuhia ngā tawhā o te tohatoha rua-ira.
– \( n = 10 \)
– \( p = 0.3 \)
Hipanga 2: Whakamahia te tohatoha rua-ira hei tatau i te tūponotanga o \( X = 4 \).
\[ P(X = 4) = {10 \kōwhiria 4} (0.3)^4 (0.7)^6 \]
Te tatau \( {10 \choose 4} \):
\[ {10 \kōwhiri 4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = 210 \]
Tātaihia inaianei \( (0.3)^4 \) me \( (0.7)^6 \):
\[ (0.3)^4 = 0.0081 \]
\[ (0.7)^6 = 0.117649 \]
Nō reira,
\[ P(X = 4) = 210 \cdot 0.0081 \cdot 0.117649 \approx 0.20012 \]
Nō reira, ko te tūponotanga ka tīpakohia kia 4 ngā ākonga he tata ki te 0.20012, arā, 20.012%.
Tauira 2: Tūponotanga Iti Iho, Ōrite rānei ki te 2
Hei tauira, ka pātaihia mai ki a tātou te tūponotanga ka tīpakohia he iti iho i te 2 ngā ākonga, ka ōrite rānei ki te 2.
Hipanga 1: Me tatau tātou i \( P(X = 0) \), \( P(X = 1) \), me \( P(X = 2) \).
– Mō \( P(X = 0) \):
\[ P(X = 0) = {10 \kōwhiria 0} (0.3)^0 (0.7)^{10} \]
\[ {10 \kōwhiri 0} = 1 \]
\[ (0.7)^{10} = 0.0282475 \]
\[ P(X = 0) = 1 \cdot 1 \cdot 0.0282475 = 0.0282475 \]
– Mō \( P(X = 1) \):
\[ P(X = 1) = {10 \kōwhiria 1} (0.3)^1 (0.7)^9 \]
\[ {10 \kōwhiri 1} = 10 \]
\[ (0.3) \cdot (0.7)^9 = 0.1210608 \]
\[ P(X = 1) = 10 \cdot 0.3 \cdot 0.1210608 = 0.3631824 \]
– Mō \( P(X = 2) \):
\[ P(X = 2) = {10 \kōwhiria 2} (0.3)^2 (0.7)^8 \]
\[ {10 \kōwhiri 2} = 45 \]
\[ (0.3)^2 \cdot (0.7)^8 = 0.2334744 \]
\[ P(X = 2) = 45 \cdot 0.09 \cdot 0.2334744 = 0.2334744 \]
Hipanga 2: Tāpirihia ngā tūponotanga.
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X \leq 2) = 0.0282475 + 0.3631824 + 0.3826372 = 0.7740671 \]
Nō reira, ko te tūponotanga ka tīpakohia he iti iho i te 2 ngā ākonga, he ōrite rānei ki te 0.7740671, arā, 77.41%.
Tauira 3: Te tūponotanga o te 8 neke atu rānei
Mena ka mahia he whakamātautau kia 12 ngā wā, ā, ko te tūponotanga o te angitu i ia whakamātautau he 0.5, he aha te tūponotanga kia 8 pea ngā angitu, neke atu rānei?
Hipanga 1: Tautuhia ngā tawhā rua-ira: \( n = 12, p = 0.5 \).
Hipanga 2: Kimihia te tūponotanga mō \( X \geq 8 \).
Me tatau i ētahi tūponotanga takitahi mō tēnei, me te tāpiri i aua tūponotanga:
\[ P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) \]
Tatauhia tētahi i muri i tētahi:
– Mō \( P(X = 8) \):
\[ P(X = 8) = {12 \kōwhiria 8} (0.5)^8 (0.5)^4 \]
\[ {12 \kōwhiri 8} = 495 \]
\[ (0.5)^{12} = 0.0002441406 \]
\[ P(X = 8) = 495 \cdot 0.0002441406 = 0.1208496 \]
– Mō \( P(X = 9) \):
\[ P(X = 9) = {12 \kōwhiria 9} (0.5)^9 (0.5)^3 \]
\[ {12 \kōwhiri 9} = 220 \]
\[ P(X = 9) = 220 \cdot 0.0002441406 = 0.05371094 \]
– Mō \( P(X = 10) \):
\[ P(X = 10) = {12 \kōwhiria 10} (0.5)^{10} (0.5)^2 \]
\[ {12 \kōwhiri 10} = 66 \]
\[ P(X = 10) = 66 \cdot 0.0002441406 = 0.01611328 \]
– Mō \( P(X = 11) \):
\[ P(X = 11) = {12 \kōwhiria 11} (0.5)^{11} (0.5)^1 \]
\[ {12 \kōwhiri 11} = 12 \]
\[ P(X = 11) = 12 \cdot 0.0002441406 = 0.002929688 \]
– Mō \( P(X = 12) \):
\[ P(X = 12) = {12 \kōwhiria 12} (0.5)^{12} \]
\[ {12 \kōwhiri 12} = 1 \]
\[ P(X = 12) = 1 \cdot 0.0002441406 = 0.0002441406 \]
Hipanga 3: Tāpirihia ngā tūponotanga katoa.
\[ P(X \geq 8) = 0.1208496 + 0.05371094 + 0.01611328 + 0.002929688 + 0.0002441406 \tata 0.1938477 \]
Nō reira, ko te tūponotanga kia puta te angitu o te 8, neke atu rānei, i roto i ngā whakamātautau 12, tata ki te 0.1938477, arā, 19.38%.
Whakamutunga
He ariā taketake te tohatoha rua-ira i roto i ngā tatauranga, he mea nui hoki i roto i ngā mahi maha. Mā te mārama ki te tatau i ngā tūponotanga mō ngā momo take o te tohatoha rua-ira, e whakaaturia ana i ngā tauira i runga ake nei, ka taea e tātou te whakamahi i tēnei ariā i roto i ngā āhuatanga o te ao tūturu. Ka whakapakari anō hoki tēnei mahi i tō tātou māramatanga ki te mahi a ngā hanganga tūponotanga i roto i tētahi horopaki mārama, whakarite hoki.
