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# HW1Sol - μ we obtain μ = 2 5 2 33 σ = 2 73 3 a No – it...

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Homework 1 – solution 1. Either company needs to adjust its premium so that it can make payoﬀs for up to N ships without losing money, where N is the smallest number such that P ( X > N 5 ) < 0 . 05 (where X Bin ( n, 0 . 015) is the number of lost ships). Using excel, it’s easy to ﬁnd that, for the smaller company ( n = 1000), N = 22, and for the larger company ( n = 2000), N = 39. So the minimum premium that allows to satisfy the requirement of risk being no more than 0.05 is: 22 · 100000 1000 = 2200, for the smaller company; and 39 · 100000 2000 = 1950, for the larger company. 2. Let X be the (random) volume to be put in a vial. We know that X N ( μ,σ ) , with σ = 0 . 1 and μ unknown (to be determined). The requirement that the fraction of under-ﬁlled vials be 0.01 is expressed as P ( X < 2 . 5) = 0 . 01 . Standardizing, we obtain P ± Z < 2 . 5 - μ σ = 0 . 01 , from which it follows that 2 . 5 - μ σ = - z 0 . 01 = - 2 . 33 . Solving for

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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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HW1Sol - μ we obtain μ = 2 5 2 33 σ = 2 73 3 a No – it...

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