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Assignment 5 Solutions

# Assignment 5 Solutions - CE93-Engineering Data Analysis...

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CE93-Engineering Data Analysis Joan Walker, Spring 2011 Assignment 5 1. License plates a) 26 2 ( 10 4 ) 4! = 3407040 b) (26 2 25 2 ) 9 3 4 26 2 10 4 = 0.022 2. State name permutation Tennessee: ( 9 4 ) ( 5 2 ) ( 3 2 ) = 9! 4! 2! 2! 1! = 3780 Florida: 7! = 5040 Although Tennessee has more number of letters, Florida has more distinct letters. As a result of this, Florida generates more permutations. 3. Bus driver a) 0.9 12 =0.2824 b) 14 12 0.9 12 0.1 2 = 0.2570 c) 14 14 0.9 14 + 14 13 0.9 13 0.1 + 14 12 0.9 12 0.1 2 = 0.8416 4. Rainstorm a) ? 𝑋 = ? ∗ ? ? ?? = ? 8 2 0 ??? + 2 ? 2 ??? 8 2 −∞ ? 𝑋 = ? 3 24 2 0 + 2 ln ? 8 ? = 1 3 + 2 ln 8 2 ln 2 = 3.106 b) 𝑃 ? ≥ 3| ? ≥ 2 = 𝑃 ?≥ 3 𝑃 ?≥ 2 = 2 ? 2 ??? 8 3 2 ? 2 ??? 8 2 = 2 ( 1 3 1 8 ) 2 ( 1 2 1 8 ) = 5 9 5. Storm runoff a) Find ? and draw PDF. Find CDF. Integrate the PDF to find the CDF. If we set the CDF=1 over 0 ≤ ? ≤ 25 , we can find ? .

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CE93-Engineering Data Analysis Joan Walker, Spring 2011 ? ? = ? ? ?? 25 0 = 1 = ? ∗ ? ? 5 ?? 25 0 = ? ∗ 2 ? 3 2 3 ? 2 10 + 𝐶 0 25 = ? ∗ 2 25 3 2 3 25 2 10 + 𝐶 1 = ? ∗ 2 25 3 2 3 25 2 10 + 𝐶 But we know 𝐶 = 0 because ? 0 must be zero. Therefore, ? = 0.048 b) c)
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