final_old_sol - NTHU MATH 2810 1. (16pts, 2pts for each)...

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NTHU MATH 2810 Final Examination Solution Jan 8, 2008 1. ( 16pts, 2pts for each ) (a) True. (b) False. The values of a pdf can be larger than 1. However, the integration of a pdf over any region must have values between 0 and 1. (c) False. It must be a one-to-one transformation. (d) True. (e) True. (f) True. (g) False. If X and Y are independent , then E ( X | Y ) = E ( X ). Zero correlation (i.e., uncorrelated) is a weaker condition than independence. It cannot guarantee this property. (h) False. When X and Y are independent, E ( X/Y ) = E ( X ) E (1 /Y ) 6 = E ( X ) /E ( Y ) in general. 2. ( 15pts, 3pts for each ) (a) Normal( μ,σ 2 ) with μ = 68 and σ 2 = 2. (b) Gamma( α,λ ) with α = 1000 and λ = 5. (An alternative answer that is acceptable is Exponential( λ ) with λ = 1 1000 / 5 = 1 200 . ) (c) Poisson( λ ) with λ = 2 × 2 = 4. (d) Binomial( n,p ) with n = 20 and p = 1 / 8. (e) Uniform( a,b ) with a = 0 and b = 360. 3. ( 6pts ) Let X be the number of “5” that occurs in the 500 rolls, then X Binomial(500 , 1 / 6) since the die is fair. Therefore, E ( X ) = 500 × (1 / 6) = 500 / 6 , and V ar ( X ) = 500 × (5 / 36) = 2500 / 36 . Because n = 500 is large, we can use Normal approximation to evaluate P ( X 100) as follows: P ( X 100) = P ( X 99 . 5) = P X - 500 6 q 2500 36 99 . 5 - 500 6 q 2500 36 P Z 99 . 5 - 500 6 q 2500 36 = 1 - Φ 99 . 5 - 500 6 q 2500 36 = 1 - Φ(1 . 94) , where Z Normal(0, 1). 1
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4. (a) ( 2pts ) Because X Uniform(0 ,L/ 2), Y Uniform( L/ 2 ,L ), and they are indepen- dent, their joint pdf is f X,Y ( x,y ) = f X ( x ) f Y ( y ) = ( 1 L/ 2 1 L/ 2 = 4 L 2 , for 0 < x < L/ 2 and L/ 2 < y < L, 0 , otherwise. (b) (
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This note was uploaded on 02/12/2012 for the course MATH 2810 taught by Professor Shao-weicheng during the Fall '11 term at National Tsing Hua University, China.

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final_old_sol - NTHU MATH 2810 1. (16pts, 2pts for each)...

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