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

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-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)= 0 , 1000 1000 / - t for e t so take - 0 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 . 0 0 0 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 - Standard Normal π 2 ) 1 , 0 ; ( 2 / ) ( 2 z e z f - = with Z= σ μ σ μ - = X N ) , ( 2 , given lookup table with z values ->always looking for P(z<value) in table, if you need z>value do 1-value in chart Ex) find P(x>5) if X=N(3,4)-> Z=(5-3)/2=1 ->chart of -1.0=0.1587
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