practice-exam3-sol

practice-exam3-sol - AMS 311 Joe Mitchell Practice Exam 3...

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AMS 311 Joe Mitchell Practice Exam 3 – Solution Notes Statistics: n =30, μ =64 . 1, median 68.5, σ =22 . 9; score range: 23–99 1. (a). f Y ( y )= ± -∞ f ( x, y ) dx = ² ³ y 0 (2 x +2 y ) dx =3 y 2 if 0 <y< 1 0 otherwise (b). P ( X> 0 . 1 | Y =0 . 5) = ± 0 . 1 f X | Y ( x | 0 . 5) dx = ± 0 . 1 f ( x, 0 . 5) f Y (0 . 5) dx = ± 0 . 5 0 . 1 2 x +2(0 . 5) 3(0 . 5) 2 dx = 64 75 , where we have used the answer to part (a) to obtain f Y (0 . 5) = 3(0 . 5) 2 . (c). E ( X | Y . 5) = ± -∞ x · f X | Y ( x | 0 . 5) dx = ± 0 . 5 0 x · 2 x . 5) 3(0 . 5) 2 dx = 5 18 2. [This is similar to Problem 5, HW7 and example 8.17 of the suggested reading in the text.] We are told that Y is exponential with mean 120 minutes. (Note, since Y is the lifetime in MINUTES, its mean is 120, NOT 2.) Thus, Y is exponential( 1 120 ). Let X be the number of minutes after 5:00pm when Joe enters the room. Then, given the Y = y , X is Uniform(0 ,y ); thus, E ( X | Y = y y/ 2, for any y> 0. (a). We can compute E ( X ) by conditioning on Y : E ( X ± -∞ E ( X | Y = y ) f Y ( y ) dy = ± 0 y 2 · 1 120 e - y 120 dy =60 Thus, I expect Joe to enter at 60 minutes after 5:00pm, i.e., at time 6:00pm. (b). We want to compute P ( 100) (since 6:40pm is 100 minutes after 5:00pm). We condition on the value of Y : P ( 100) = ± -∞ P ( 100 | Y = y ) f Y ( y ) dy = ± 100 0 P ( 100 | Y = y ) 1 120 e - y 120 dy + ± 100 P ( 100 | Y = y ) 1 120 e - y 120 dy = ± 100 0 0 · 1 120 e - y 120 dy + ± 100 y - 100 y · 1 120 e - y 120 dy where we have used the fact that P ( 100 | Y = y ² y - 100 y if 100 0i f y< 100 Alternatively, both parts can be solved directly, using the joint density (see Example 8.17): f ( x, y f X | Y ( x | y ) · f Y ( y ² 1 y · 1 120 e - y 120 0 <y ,0 <x<y 0o . w . Then, we can compute (look at a picture of the support set): P ( 100) = ± 100 ± x 1 y · 1 120 e - y 120 dydx and E ( X ± 0 ± x x · 1 y · 1 120 e - y 120 dydx 1
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3. (a). Let p i = P ( she makes it to the summit | she starts i feet above the cliF). Our goal is to ±nd p 1 .W e condition on the ±rst step she takes, which takes her up by 1 foot with probability 0.3 and down by 1 foot with probability 0.7. Thus, we get a system of 3 equations in 3 unknowns: p 1 =( 0 . 7) · 0+(0 . 3) p 2 p 2 0 . 7) p 1 +(0 . 3) p 3 p 3 0 . 7) p 2 . 3) · 1 Now, solve the system of three equations in three unknowns; the ±nal answer is p 1 . (b). Let t i = E (number of steps she takes before she is done | she starts i feet above the cliF). Condition on the result of the ±rst step: t 1 0 . 7)(1 + 0) + (0 . 3)(1 + t 2 ) t 2 0 . 7)(1 + t 1 )+(0 . 3)(1 + t 3 ) t 3 0 . 7)(1 + t 2 . 3)(1 + 0) Now, solve the system of three equations in three unknowns; the ±nal answer is t 1 . 4. Let X be the number of people that show up at the party. We know that X 0 and that E ( X )=25.Wewantto estimate p = P ( X 61); by Markov’s inequality we get P ( X 61) E ( X ) 61 = 25 61 . Thus, we know that 0 p 25 61 . 5. Let X 1 ,...,X 150 be the 150 random points from (0,1). Let X = X 1 + ··· X 150 .E a c h X i is Uniform(0,1), so E ( X i )=0 . 5 and var ( X i )=1 / 12. Thus, E ( X )=150 · (0 . 5) = 75 and ( X · (1 / 12) = 25 / 2 (since the X i ’s are independent, by assumption, so ( X )= ( X 1 )+ + ( X 150 )).
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This note was uploaded on 06/06/2010 for the course EC 11 taught by Professor All during the Spring '10 term at UCLA.

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practice-exam3-sol - AMS 311 Joe Mitchell Practice Exam 3...

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