4703-10-Fall-hreview-solu

# 4703-10-Fall-hreview-solu - IEOR 4701 Solutions to review...

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Unformatted text preview: IEOR 4701: Solutions to review of probability problems 1. (a) f ( x ) = λe- λx , if x ≥ 0; , if x < 0. F ( x ) = Z x f ( y ) dy = 1- e- λx , x ≥ E ( X ) = R ∞ xλe- λx dx = 1 /λ via integration by parts ( U = x , dV = e- λx dx ). Via integrating the tail, P ( X > x ) = 1- F ( x ) = e- λx , Z ∞ e- λx dx = 1 λ . (b) f ( x ) = 1 b- a ; if x ∈ ( a,b ) , otherwise. F ( x ) = Z x f ( y ) dy = , if x ≤ a ; x- a b- a , if x ∈ ( a,b ); 1 if x ≥ b . E ( X ) = 1 b- a Z b a xdx = 1 b- a 1 2 ( b 2- a 2 ) = 1 2 ( b + a ) , because ( a- b )( a + b ) = a 2- b 2 . P ( X > x ) = 1- F ( x ) = 1 , if x ≤ a ; b- x b- a , if x ∈ ( a,b ); if x ≥ b , and so Z ∞ P ( X > x ) dx = Z a 1 dx + 1 b- a Z b a ( b- x ) dx = 1 2 ( b + a ) . For a = 1 ,b = 3 we obtain E ( X ) = 2. (c) F ( x ) = 0; if x < 1 1- 1 x 2 , if x ≥ 1 . f ( x ) = d dx F ( x ) = , if x < 1; 2 x 3 , if x ≥ 1....
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4703-10-Fall-hreview-solu - IEOR 4701 Solutions to review...

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