ch4 review

ch4 review - -For a continuous probability distribution,...

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Unformatted text preview: -For a continuous probability distribution, pdf: ∫ = < < b a dx x f b x a P ) ( ) ( A cdf is ∫ ∞- = < a dx x f a x P ) ( ) ( Ex) given pdf for 0<x<2, take integral from 0-2 and get c. Then to get p(1<x<2) use pdf integral from 1-2 with c ∫ ∞ ∞- = = dx x xf x E ) ( ) ( μ and 2 2 )] ( [ ) ( ) ( x E x E x V- = (you have to do two separate integrals for V(x) )-Ex) f(t)= , 1000 1000 /- t for e t so take ∫ ∞- 1000 / 1000 dt e t t using value in integral table [given on test]-Uniform f(x;a,b)=1/(b-a) Ex) between 0 and 1 P(x<0.5) is ∫- 5 . 1 1 dx =0.5, P(x=0.2)=0 Ex) min=3, max=10, P(t>5)= ∫- 10 5 3 10 1 dx and the E(X)= ∫- 10 5 3 10 1 dx x- Normal π σ σ μ σ μ 2 ) , ; ( 2 2 2 / ) (-- = x e x f , 1sd=68%, 2sd=98.5%,3sd=99.73%, P(a<x<b)= ∫ b a dx x f ) , ; ( σ μ-A normal has bell, symmetrical about mean, unimodal, mean=median=mode, 1sd=inflection points Ex) mean=0.9, sd=0.002, min=.895, max=.905, DEFECT= 1-(P(.895<x<.905)->P(z<(.905-.9)/.002)=P(z<2.5)=.9938[chart], P(z<-2.5)=0.0062[chart]. Subtract->.9876 good ->P(z<(....
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This note was uploaded on 09/01/2008 for the course ISE 261 taught by Professor Koon during the Spring '08 term at Binghamton.

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