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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 = 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 = 0 . 5) = -∞ x · f X | Y ( x | 0 . 5) dx = 0 . 5 0 x · 2 x + 2(0 . 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 ( X > 100) (since 6:40pm is 100 minutes after 5:00pm). We condition on the value of Y : P ( X > 100) = -∞ P ( X > 100 | Y = y ) f Y ( y ) dy = 100 0 P ( X > 100 | Y = y ) 1 120 e - y 120 dy + 100 P ( X > 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 ( X > 100 | Y = y ) = y - 100 y if y > 100 0 if 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 ) = 1 y · 1 120 e - y 120 0 < y ,0 < x < y 0 o.w. Then, we can compute (look at a picture of the support set): P ( X > 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 cli ff ). Our goal is to find p 1 . We condition on the first 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 + (0 . 3) · 1 Now, solve the system of three equations in three unknowns; the final answer is p 1 . (b). Let t i = E (number of steps she takes before she is done | she starts i feet above the cli ff ). Condition on the result of the first 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 ) + (0 . 3)(1 + 0) Now, solve the system of three equations in three unknowns; the final 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. We want to 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 . Each 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 var ( X ) = 150 · (1 / 12) = 25 / 2 (since the X i ’s are independent, by assumption, so var ( X ) = var ( X 1 ) + · · · + var ( X 150 )).
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