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Projekt binomialfordeling

Teori binomialfordeling Et forsøg er et binomialforsøg hvis: der ved hvert forsøg er to udfald: succes eller fiasko resultaterne er uafhængige af hinanden når forsøget bliver udført flere gange sandsynligheden for succes og fiasko er den samme ved hver gentagelse af forsøget Formlen for binomialfordeling er:

P = sandsynligheden et bestemt antal succeser (r) n = antalsparameteren, hvor mange gange udføres forsøget r = antal succeser p = sandsynlighedsparameteren, dvs. sandsynligheden for succes K(n,r) = binomialkoefficienten og udregnes på denne måde: K(n,r) = K(n,r) beskriver antallet af kombinationer med r antal objekter ud af n mulige. n! læses "n falkultet" og udregnes: n!= Eksempel på binomialfordelingsforsøg: En terning kastes 15 gange - derfor er er n=15 Terningen er 6-sidet og sandsynligheden for at få f.eks. en 6-er er 1/6, og sandsynligheden for at f.eks. at få 3 eller derunder er derfor 1/2 Vi vil udregne chancen for at få netop fire 6'ere, dvs. r = 4

(1) Sandsynligheden for at få fire 6-ere er altså 14 % Middelværdien

Middelværdien er det gennemsnit af en række tal man forventer hvis man udfører et forsøg mange gange. Middelværdien udregnes med formlen:

Eksempel på udregning af middelværdi I et stort vareparti er 8 % af bananerne defekte. Der udtages en stikprøve på 38 bananer. Bestem middelværdien: Udregning af middelværdi p = 0,08 n = 38 (2) Der vil i gennemsnit være ca. 3 defekte bananer i hver stikprøve. Kumuleret sandsynlighed Sandsynligheden for op til og med "r" succeser i "n" forsøg Eksempel på kumuleret sandsynlighed Vi har det samme vareparti som i foregående eksempel. Denne gang ønsker vi at bestemme sandsynligheden for at der højst er 3 defekte bananer i stikprøven.

(3) Først regner vi sandsynligheden for at der er 0 defekte bananer: (4) Udregning 1 defekt banan: (5) Udregning 2 defekte bananer: (6) Udregning 3 defekte bananer: (7) Udregning af kumuleret sandsynlighed - altså sandsynligheden for at der er op til tre defekte bananer: 0.6380176310 (8) Den kumulerede sandsynlighed er 0,64 - dvs. sandsynligheden for at der er højst 3 defekte bananer i stikprøven. Dette kan også regnes med Hennings pakke:

(9) Binomialtest Nulhypotesen

Den hypotese som vi vil undersøge Fejl af 1. og 2. art: Sand Forkastes Forkastes ikke Fejl af 1. art Korrekt Falsk Korrekt Fejl af 2. art

Fejl af 1. art er altså når man forkaster en sand hypotese Fejl af 2. art er når man ikke forkaster en falsk hypotese Signifikansniveau: Man fastlægger en procentvis risiko for at begå fejl af 1. art. Hvis man vælger et signifikansniveau på 5% skal man i hver ende af prøvens resultatmængde forkaste 2,5 %. Kritisk mængde: Den mængde af testen som hører til det valgte signifikansniveau. Acceptmængden: Resten prøvens resultater - de resultater man accepterer. For eksempel af signifikansniveau og kritisk mængde se vores egen binomialfordelingstest: Marabou vs. billig chokolade - MUMS!!!!! ;-) Opgave 952 Formel: a) Vi angiver at succes ikke er at få nogle 6'ere - fiasko er at få 6'ere n=5 p = 5/6 r=0


b) Vi angiver at succes ikke er at få nogle 1'ere eller 6'ere - fiasko er at få 1'ere eller 6'ere n=5 p = 2/3


(11) c) Vi angiver at succes er at få et ulige antal øjental - fiasko er at få et lige antal øjental n=5 p = 1/2 r=3


Opgave 957 Vi kan bruge binomialfordeling fordi at vi har at gøre med to muligheder: pige eller dreng. Desuden er udfaldene ved hver fødsel uafhængige af hinanden og chancen for at en dreng/pige bliver født er ved hver fødsel den samme. Bestem sandsynligheden for at alle de fødte er drenge: p = 0,513 n = 13 r = 13

(13) Sandsynligheden for at alle de fødte er drenge er 0.0001704220348 Bestem sandsynligheden for at 8 eller 9 af børnene er piger Udregning af sandsynligheden for at få 8 piger: p = 1-0.513 n = 13 r=8


Sandsynligheden for at få 8 piger ud af de 13 børn er: 0.1446761897 Udregning af sandsynligheden for at få 9 piger: p = 1-0.513 n = 13 r=9

(15) Sandsynligheden for at få 9 piger ud af de 13 børn er: 0.07630204067 Udregning af sandsynligheden for at få enten 8 eller 9 piger: (16) Sandsynligheden for at få enten 8 eller 9 piger ud af de 13 børn er: 0.2209782304 Bestem det mest sandsynlige antal drenge blandt de 13 nyfødte: Det mest sandsynlige antal drenge vil være tæt på middelværdien, da middelværdien er det gennemsnitlige antal fødte drenge: Vi udregner derfor middelværdien ved hjælp af formlen:

(17) Middelværdien giver ikke et helt tal. Det mest sandsynlige antal drenge vil derfor være 6 eller 7:


(19) Det mest sandsynlige antal drenge blandt de 13 nyfødte er altså 7. Opgave 961 I en klassisk tipskupon er der 3 kolonner og 13 rækker Der er altså 13 kampe hvor man skal tippe resultatet i hver kamp, hvor der er følgende muligheder: hold 1 vinder, uafgjort, hold 2 vinder. Udregn sandsynligheden for at Yrsa får 13 rigtige: n=13 r=13 p = 1/3

(20) Sandsynligheden for at Yrsa for 13 rigtige ud af 13 mulige er 1

Udregn sandsynligheden for at Yrsa får 12 rigtige n=13 r=12 p = 1/3

(21) Sandsynligheden for at Yrsa får 12 ud af 13 rigtige er 26

Bestem det antal rigtige som der er størst sandsynlighed for at få: Vi udregner igen middelværdien:

(22) Det antal rigtige som der er størst sandsynlighed for at få er altså enten 4 eller 5


(24) Der er altså størst sandsynlighed for at få 4 rigtige ud af 13 mulige

Test marabou vs. billig chokolade Hypotese: man kan ikke smage forskel! 11 gange har vi kunnet kende forskel på chokoladen og 4 gange kunne vi ikke kende forskel n = 15

p = 0,5 r=4 Signifikansniveau = 5 % derfor må den kumulerede sandsynlighed i hver ende ikke overstige 2,5 %



0.0175781250 Udregning af sandsynligheden for at begå fejl af 1. art: 0.03515625000 Sandsynligheden for at begå fejl af 1. art er derfor: Den kritiske mængde er derfor for r = 0-3 og r = 11-15 Altså: K={0,1,2,3,11,12,13,14,15}



Da vi i vores forsøg fik 11 succeser, lå vores resultater derfor inde for den kritiske mængde. Hypotesen skal derfor forkastes, men der er 3,52 procent chance for at der er fejl af 1. art, og at forsøget ikke burde forkastes. Fejl af 1. art ville betyde at hypotesen er rigtig, selvom vores resultater lå i det kritiske niveau.…...

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