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.dsy:it. : Powered by vBulletin version 2.3.1 .dsy:it. > Didattica > Corsi A - F > Calcolo delle probabilità e statistica matematica > Appello di Gennaio 2010 - Zanaboni
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middu
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HAI STANDARIZZATO???

26-01-2010 10:38
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garfa84
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dovrebbe essere perchè la media campionaria standardizzata quando n tende all'infinito tende alla funzione di ripartizione di gauss quindi
P(|Xn*| < 1) = P(|G|<1) = P(-1 < G < 1) =
diciamo F(1) = funz. ripartizione di gauss in 1 allora
F(1) - F(-1) = = F(1) - (1 - F(1)) = 0.8413- ( 1 -0.8413) = 0.6826

F(1) - F(-1) = 2F(1)-1

26-01-2010 10:42
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middu
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ALLORA p(|XN*| < 1 ) = P(-1<XN*<1) = P(XN*<1) - P(XN >-1) = P(XN*<1) - P(XN* < - 1) = P(XN*<1) - 1 + P(XN* <1) = 2P(XN*<1) - 1 = 2F(2)- 1

26-01-2010 10:42
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garfa84
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ma per l'esercizio 3....

26-01-2010 10:44
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R1cky`
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Ma non è che qualcuno di voi è in comelico che ne parliamo di persona? :)

26-01-2010 10:45
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middu
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SI

26-01-2010 10:45
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R1cky`
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Io sono in biblioteca, se passi di qui vedi due tizi con il tavolo pieno di robi di statistica e il pc con la schermata sul dsy :D

26-01-2010 10:47
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middu
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NON SONO IN UNI

26-01-2010 10:53
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Metteus
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edit

Last edited by Metteus on 26-01-2010 at 11:38

26-01-2010 11:36
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Teju
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Originally posted by garfa84
ma per l'esercizio 3....

Devi usare il teorema del limite centrale, sul Mood è a pg 204 in fondo.

Domattina chi di voi c'è? E se ci trovassimo intorno alle 9.20 in Comelico? Giusto per vedere le ultime cose al volo insieme??

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26-01-2010 11:41
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Metteus
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quindi nel 2.5 la prob è (1-p)^x, quindi in 0 la probabilità vale 1 ... ma ha senso ? non dovrebbe essere 0.9 ?


e poi il grafico come sarebbe ?

Last edited by Metteus on 26-01-2010 at 11:55

26-01-2010 11:53
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middu
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nel caso nostro 1- P(G<=X) = 1- (1-p)^X

26-01-2010 12:18
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middu
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1- p(1-p)^x

26-01-2010 12:19
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middu
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ma per chi ha gia fatto l'orale

26-01-2010 12:20
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middu
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era giusto il precedente. Ma per chi ha fatto l'orale posso sapere cosa chiede???

26-01-2010 12:28
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