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Unformatted text preview: Poisson f ( x ) = Gamma f ( x ) = Weibull 1 e x / = x , =failure rate, =time to failure = = 1/, 2 = 2 = 1/2 e ( r ) ( x ) r 1 e x , =failure rate, r = # failures =r/ 2 = r/2 f ( x) = x 1e x , = avg. time to failure, = shape parameter, x = time to failure = 1/ (1 + 1/ ) 2 = 2 / (1 + 2 / )  2 If X is binomial and np and n(1 p) are both 5, then X is approximately normal with = np and = npq. Hence, z = X  = X np is standard normal. npq Discrete Continuous P(X=18) P(17.5 < X < 18.5) P(X > 18) P(X > 18.5) P(X < 18) P(X < 17.5) P(X >=18) P(X > 17.5) P(X <= 18) P(X < 18.5) Linear Relations 1 r +1. r > 0 indicates a positive (x inc. / y dec.), r < 0 indicates a negative (x inc. / y dec), r = 0 indicates no relationship r= SS ( xy ) SS ( x) SS ( y ) SS ( x) = x 2 ( x) 
n 2 SS ( y ) = y 2 ( y) 
n 2 SS ( xy ) = xy  x y
n Least squares reg. y = bx+a B = Joint density SSxy y x A= b SSx n n f XY 1) f XY (x,y) o 0 2) (x,y)=1 3) P(a f X X Y b and c Y d) = f ( x, y )dydx XY a c
Y X b d Independence
2 f XY (x,y)=f X ( x ) fY ( y )
2 yf xf and cov = 0 E [ X ] = XY ( x, y )dxdy E [Y ] = XY ( x, y )dxdy
Y X Var(x) = E(x ) [E(x)] Cov( X , Y ) = E[ XY ]  E[ X ]E[Y ] Cov ( X , Y ) fxy ( x, y ) fxy ( x, y ) Pearson Correlation = fxy= fyx = XY (VarX )(VarY ) fy ( y ) fx ( x ) x  y = E ( x  y ) = x * f x  y dx y  x = E ( y  x) =
Density y* f
x  y x dx f ( x) = 1 2
x e x 1 2 2 Cumul. Dist. Func.= f (t )dt Expected Value E ( x) = xf ( x)dx, E ( x 2 ) = x 2 f ( x)dx
x ...
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 Spring '07
 MQLemons

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