CEE3770_Sum09_HW2_solns

CEE3770_Sum09_HW2_solns - CEE 3770 - Summer 2009 Homework 2...

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CEE 3770 - Summer 2009 Homework 2 (due on 06/02) 1. (10 points) Suppose that X is continuous with p.d.f. f ( x ) = cx 2 , 0 x 1. (a) Find the value of c that will make the p.d.f. integrate to 1. Solution: 1 = Z 1 0 cx 2 dx = c/ 3 This implies that c = 3. (b) Find the c.d.f. F ( x ) for all x . Solution: F ( x ) = Pr ( X x ) = 0 if x < 0 x 3 if 0 x 1 1 if x 1 (c) Calculate E [ X ]. Solution: E [ X ] = Z 1 0 xf ( x ) dx = Z 1 0 x 3 x 3 dx = 3 / 4 (d) Calculate V ar ( X ). Solution: E [ X 2 ] = Z 1 0 x 2 f ( x ) dx = Z 1 0 3 x 4 dx = 3 / 5 This implies that V ar ( X ) = E [ X 2 ] - ( E [ X ] ) 2 = 3 / 80. (e) Calculate Pr (0 X 1 / 2). Solution: Pr (0 X 1 / 2) = F (1 / 2) - F (0) = 1 / 8 . 2. (14 points) Jake is chosen as a semifinalist in GT Sweepstakes. This entitled him to enter the grand prize drawing for $1 million dollars. Being skeptical, Jake sat down to figure out how much money he could expect to win if he played. The prize breakdown is following: Number of winners Prize Amount 1 Grand prize $1 million 5 1st prize $1000 25 2nd prize $100 50 3rd prize $10 1
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(a) If 250,000 people have been selected as semifinalists, how much money could Jake ap- proximately expect to win if he entered. Solution: Let X denote the the amount of money you may win. X = $1 million w.p. 1 / 250 , 000 $1 , 000 w.p. 5 / 250 , 000 $100 w.p. 25 / 250 , 000 $10 w.p. 50 / 250 , 000 $0 w.p. 249 , 919 / 250 , 000 Thus, E [ X ] = $1 million × 1 / 250 , 000 + $1 , 000 × 5 / 250 , 000 + $100 × 25 / 250 , 000 + $10 × 50 / 250 , 000 = 4 . 032 (b) What is the variance of his possible gain? Solution: E [ X 2 ] = $1 million 2 × 1 / 250 , 000 + $1 , 000 2 × 5 / 250 , 000 +$100 2 × 25 / 250 , 000 + $10 2 × 50 / 250 , 000 = 4000021 . 02 V ar ( X ) = E [ X 2 ] - ( E [ X ]) 2 = 4000021 . 02 - 4 . 032 2 = 4000004 . 763 3. (15 points) Car arriving at an intersection may either turn or go straight. The probability of
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CEE3770_Sum09_HW2_solns - CEE 3770 - Summer 2009 Homework 2...

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