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gogodr
I post too much
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Joined: 19 Dec 2006
Posts: 2041

PostPosted: Tue Aug 14, 2012 7:27 pm    Post subject: Ignore this please Reply with quote

Tecincas inferenciales:
-estimacion de parametros
-prueba de hipotesis

Distribuciones:
Normal - N
ji^2 - X(n)^2
t - t(n)
Fisher - F

N::
Dominio = Reales
Maximo x en moda
simetrico
suma de dos normales es otra normal

X1~n(5;2^2)
X2~n(20;5^2)

(x1+x2)~n(25,29)
(x1-x2)~n(-5,29)

a) P(x<4) = 0.2266
b) P(x>10) = 0.2266
c) P(5<x<Cool = 0.2902
d) a tal que P(x<a) = 0.9 , a = 12.13
e) a tal que P(x>=a)=0.6 , a = 5.9877
f) a y b tal que P(x<a)= 0.01 y P(a<=x<=b)=0.98 , a= -2.305 b=16.31
g) a y b tal que P(x>b)= 0.03 y P(a<=x<=b)=0.9 , a= 1.097 b=14.52
---------------------------------------------------------------------

X~n(µ;5^2) y P(x<Cool=0.3
a) µ ?
P( (x-µ)/σ < (8-µ)/5 ) = 0.3

P(Z<(8-µ)/5) = 0.3
-0.5244 = (8-µ)/5
µ = 10.62

b) P(x>12) = 0.3913

---------------------
X:Tiempo de vide del escape
X~n(3;0.5^2) = 0.02275
C = Costo total del remplazo
Y: numero de escapes con fallas para reparar
C = 10*Y
E(c)= E(10Y) = 10*E(Y) = 10(500000) * 0.02275 = 113750
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trebor
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PostPosted: Tue Aug 14, 2012 7:29 pm    Post subject: Reply with quote

huehuehuehuehuehue
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Up2Admin
I'm a spammer
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Joined: 17 Oct 2007
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PostPosted: Tue Aug 14, 2012 7:38 pm    Post subject: Reply with quote

jajajajajajajajajajaja
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Cryoma
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Joined: 14 Jan 2009
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PostPosted: Tue Aug 14, 2012 7:56 pm    Post subject: Reply with quote

could you imagine if you looked at your math homework and there were a bunch of smug sunglasses smileys?
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