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Unformatted text preview: μ we obtain μ = 2 . 5 + 2 . 33 σ = 2 . 73 . 3. a) No – it takes negative values. b) This can be a pdf – everywhere nonzero and R ∞∞ f ( x ) dx = 1. c) This one is ﬁne – for the same reasons as b) d) No since R ∞∞ f ( x ) dx = 1 . 1 6 = 1. 1 e) This one is ﬁne. f) No – it can take negative values (for large x ). g) No since R ∞∞ f ( x ) dx = 1 √ 2 6 = 1. 4. We need to choose C such that R ∞∞ f ( x ) dx = 1. a) Z ∞∞ f ( x ) dx = C Z ∞ exp(λx ) dx = C 1 λ . So we have C 1 λ = 1, and, therefore, C = λ . b) Because of symmetry of the function we can write Z ∞∞ f ( x ) dx = 2 C Z ∞ exp(λx ) dx = 2 C 1 λ . So C = λ/ 2. c) Z ∞∞ f ( x ) dx = C Z 3 xdx = C 9 2 . So C = 2 9 . 2...
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 Spring '08
 Perevalov
 dx, vial, larger company, smaller company

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