MATHHX B
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15.5 Fordelingsfunktioner
Vi har lært at en kontinuert fordeling er bestemt ved dens tæthedsfunktioner. Der findes imidlertid en anden måde man kan beskrive en kontinuert fordeling, nemlig ved dens fordelingsfunktion. En fordelingsfunktion for
en kontinuert sandsynlighedsfordeling er en funktion \(F\) der til ethvert \(x_0\) knytter sandsynligheden \(P(X\leq x_0)\):
Sagt på en anden måde: Fordelingsfunktionen \(F\) er defineret ved
\[F(x)=P(X\leq x)\]
I det følgende vil vi begrænse os til fordelingsfunktioner for normalfordelinger.
Øvelse 15.5.1
Lad \(X\sim N(2,4)\).
-
a) Bestem \(F(3)\) ved hjælp af tæthedsfunktionen i Geogebra. Ja, jeg har ikke vist et eksempel på hvordan man gør, men øvelsen er en test, om man har
forstået, hvad en fordelingsfunktion er. Den kan regnes med kendt teknik i Geogebra.
Man kan finde fordelingsfunktionen i Geogebra. Vi åbner sandsynlighedslommeregneren som vi plejer. Her er tæthedsfunktionen for \(N(0,1)\):
Klikker man på knappen til venstre for der hvor der står ”Normal” i screenshottet (knappen har en blå graf på sig), får man følgende:
Dette er fordelingsfunktionen, og sådan som den er indstillet i screenshottet, kan vi se at \(P(X\leq 1)=0{,}8413\). Vi ser at sandsynligheden er markeret med en vandret rød linje. Det er fordi at sandsynlighederne her er bestemt
ved funktionsværdier, i stedet for arealer som ved tæthedsfunktionen.
-
Eksempel 15.5.1
Vi vil bestemme sandsynligheden \(P(2\leq X \leq 8 )\) for normalfordelingen \(N(5,10)\) ved hjælp af fordelingsfunktionen. Da fordelingsfunktionen kun kan give os sandsynligheder på
formen \(P(X \leq x)\) må vi først overveje, hvordan vi kan udtrykke den ønskede sandsynlighed ved hjælp af sandsynligheder på formen \(P(X \leq x)\). Det er ikke svært:
\[P(2\leq X \leq 8 )=P(X\leq 8) - P(X\leq 2).\]
Ved hjælp af fordelingsfunktionen bestemmer vi \(P(X\leq 8)\) og \(P(X\leq 2)\). Her er vist \(P(X\leq 8)\):
Vi kan se at \(P(X\leq 8)=0{,}6179\). Tilsvarende findes \(P(X\leq 2)= 0{,}3821\). Så vi får
\[P(2\leq X \leq 8 )=P(X\leq 8) - P(X\leq 2) = 0{,}6179 - 0{,}3821=0{,}2358.\]
Altså har vi at
\[P(2\leq X \leq 8 )=24\%\]
Øvelse 15.5.2
Regn følgende sandsynligheder for fordelingen \(N(20,10)\) ved hjælp af fordelingsfunktionen.
For en fordelingsfunktion for en normalfordeling gælder der altid at \(F(\mu )=0{,}5\). Det er fordi at en normalfordeling altid er symmetrisk omkring middelværdien så \(P(X\leq \mu )=0{,}5\).
En fordelingsfunktion med en lille standardafvigelse er stejl i nærheden af middelværdien.
Mens en fordelingsfunktion med stor standardafvigelse er mindre stejl omkring middelværdien:
Ekstra
-
Eksempel 15.5.2
Vi vil komme med vores bedste bud for middelværdien og standardafvigelsen for normalfordelingen givet ved følgende tæthedsfunktion:
Middelværdien \(\mu \) er nem. Den må være ca. 3. Det kan vi se fordi at det er ved ca. 3 at funktionsværdien er \(\frac {1}{2}\).
Standardafvigelsen er lidt sværere. Vi ved fra 68-95-99,7-reglen at sandsynligheden
\[P(\mu - 2 \sigma \leq X \leq \mu + 2\sigma )\approx 95\%.\]
Dvs. at \(P(X\leq \mu - 2 \sigma )\) må være ca. \(2{,}5\%\). Det svært at aflæse \(2{,}5\%\) på y-aksen ,men det må være ca. ved \(x=2\). Altså er \(\mu - 2 \sigma \approx 2\) og da \(\mu \approx 3\), må det
betyde at \(\sigma \approx \frac {1}{2}\)
Øvelse 15.5.4
Betragt fordelingsfunktionen for en normalfordeling:
Løsning 15.5.4
-
a) \(\mu =5\)
-
b) \(\sigma =2\)
Hvis du er startet på HHX i 2023 eller tidligere skal du huske, at gå tilbage til binomialfordeling og regn afsnittet om normalfordelingsapproksimationen (afsnit 14.2)