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# hw4sol - 36-225 Homework 4 SOLUTIONS 1 4 points 2 for E(Y 2...

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36-225 - Homework 4 SOLUTIONS 1. 4 points: 2 for E(Y), 2 for V(Y) The random variable of interest is Y , the number of successful explorations in n = 10 explo- rations. Then Y has a binomial distribution with p = . 1. Hence E ( Y ) = np = 10( . 1) = 1 and V ( Y ) = np (1 - p ) = 10( . 1)( . 9) = . 9 2. 4 points: 2 for realizing that it is Bin(32,p=.9), 2 for probability calculation Let Y be the number of passengers who arrive for the flight. Y Bin ( n = 32 , p = . 9). P (more people show up than seats) = P ( Y = 31) + P ( Y = 32) = ( 32 31 ) ( . 9) 31 ( . 1) 1 + ( 32 32 ) ( . 9) 32 ( . 1) 0 = 32 × . 9 31 × . 1 + . 9 32 = . 1564. 3. 6 points: 2 for set-up, 3 for work, 1 for final answer E (4 Y ) = 5 X y =0 4 y · P Y ( y ) = 5 X y =0 4 y 5 y ! 1 3 y 2 3 5 - y = 5 X y =0 5 y ! 4 3 y 2 3 5 - y By hint z}|{ = 4 3 + 2 3 5 = 2 5 = 32 . 4. 8 points: 4 for each part (a) P ( Y > a ) = P ( Y a + 1) = X y = a +1 (1 - p ) y - 1 p = p 1 - p X y = a +1 (1 - p ) y geometric series z}|{ = p 1 - p · (1 - p ) a +1 1 - (1 - p ) = (1 - p ) a [ = q a ] (b) P ( Y > a + b | Y > a ) = P ( { Y > a + b } ∩ { Y > a } ) P ( Y > a ) = P ( Y > a + b ) P ( Y > a ) part (a) z}|{ = (1 - p ) a + b (1 - p ) a = (1 - p ) b = P ( Y > b ) 5. 8 points: (a) 2 points, (b) 3 points (c) 3 points Let Y be the number of calls until first “pro-choice” person.= Y Geometric ( p = . 62). (a) P ( Y = 5) = (1 - . 62) (5 - 1) · . 62 = . 013 (b) P ( Y > 2) = ( problem 4( a )) = (1 - . 62) 2 = . 1444 (c) P ( Y 8 | Y > 5) = P ( Y > 7 | Y > 5) = ( memoryless ) = P ( Y > 2) = (1 - . 62) 2 = . 1444 1

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6. 12 points: (a) 3 points [1 for NB, 1 for r 1 for p] (b) 3 points (c) 6 points [2 for X=Y-6, 2 for E(Y-6), 2 for V(Y-6)] Let Y be the number of people that the telemarketer needs to contact until 6 subscriptions are sold. (a) Y NB ( r = 6 , p = 1 / 3) P Y ( y ) = y - 1 6 - 1 ! 1 3 6 2 3 y - 6 y = 6 , 7 , 8 , . . . (b) P ( Y 8) = 1 - P ( Y 7) = 1 - [ P ( Y = 6) + P ( Y = 7)] = 1 - " 6 - 1 6 - 1 !
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