STAT_333_Test_1_Solutions

STAT_333_Test_1_Solutions - STAT 333 Spring 2010 Test 1...

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STAT 333 Spring 2010 Test 1 SOLUTIONS Thurs, June 10, 2:30 – 4:00 pm First (given) name:__________________ Last (family) name:____________________ Student ID #:__________________ UW userid:____________________ Instructions: 1. Please fill in the above information 2. This test has 7 pages, including this cover page 3. Answer all questions in the space provided 4. You have 90 minutes for the test 5. Show all your work and justify your steps 6. Good luck! Question Marks available Marks obtained 1 9 2 13 3 10 4 8 5 8 6 12 Total 60
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1. Harry Potter is trapped in the Department of Mysteries. Three doors lead out from the centre room, one returning back to the centre after 2 minutes, one returning back to the centre after 4 minutes, and one leading to his goal (the Hall of Prophecies) after 10 minutes. Unfortunately, every time he enters the centre room, the doors randomly spin around so he cannot remember which doors have been tried already, so he chooses randomly. [3] a) Let X = the number of minutes to reach his goal. Find E[X]. Let D be the door Harry chooses. E[X|D = 1] = E[2 + X] = 2 + E[X] since once he gets back, the time has the same dist as X E[X|D = 2] = E[4 + X] = 4 + E[X] similarly E[X|D = 3] = 10 So E[X] = E[E[X|D] = E[X|D = 1]P(D = 1) + E[X|D = 2]P(D = 2) + E[X|D = 3]P(D = 3) = (2 + E[X]) (1/3) + (4 + E[X]) (1/3) + 10 (1/3) = 16/3 + (2/3)E[X] So E[X] = 16 minutes [4] b) Find Var(X). E[X 2 |D = 1] = E[(2 + X 1 ) 2 ] = 4 + 4E[X] + E[X 2 ] E[X 2 |D = 2] = E[(4 + X 1 ) 2 ] = 16 + 8E[X] + E[X 2 ] E[X 2 |D = 3] = 10 2 = 100 So E[X 2 ] = E[E[X 2 |D] = E[X 2 |D = 1]P(D = 1) + E[X 2 |D = 2]P(D = 2) + E[X 2 |D = 3]P(D = 3) = (4 + 4E[X] + E[X 2 ])(1/3) + (16 + 8E[X] + E[X 2 ])(1/3) + 100 (1/3) = 312/3 + (2/3)E[X 2 ] since E[X] = 16 So E[X 2 ] = 312 And Var(X) = E[X 2 ] – E[X] 2 = 312 – 16 2 = 56 [2] c) Harry’s friend Hermione figures out a way to label the doors so they will not try the same door twice. Describe how you would find E[X] and Var(X) in this case. (No calculations are necessary.) There are a number of ways to do this. The easiest way would be to list out all possible paths Harry could take. There are only 5 cases if he can never choose the same door twice: 1, 2, 3 (prob 1/6), 1, 3 (prob 1/6), 2, 1, 3 (prob 1/6), 2, 3 (prob 1/6), and 2 (prob 1/3). Then the expected value and variance can be calculated from first principles.
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STAT_333_Test_1_Solutions - STAT 333 Spring 2010 Test 1...

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