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Unformatted text preview: STAT516 FALL 2005 Solution to Homework 5 Section 3.2 2. a) average(3 rd list) = average(1 st list) + average(2 nd list) = 5.8 b) average(3 rd list) = average(1 st list) average(2 nd list) = 2.2 c) &amp; d) Cant do it: need to know the order of the numbers in the two lists 3. #of sixes in 3 rolls = I 1 + I 2 + I 3 , where I j is the indicator of getting a six on the jth roll, j=1,2,3. So, E(#of sixes in 3 rolls) = 1/6 + 1/6 + 1/6 = Similarly, E(# of odds in 3 rolls) = 3/2. 4. If 25 of the numbers are 8, then all the others must be 0, since the average must be 2. If 26 or more of the numbers are 8 or more, there is no way the average can be 2, since some of the other numbers would have to be negative. Or you can use a formal argument like this: Markovs inequality gives ( 8) ( )/8 2/8 .25 P X E X P = = . So number of values greater than 8 is at most (100)(.25) = 25. 5. a) The reasoning is incorrect. Notice that your possible gains are 1, 1, 2 and 3. Getting your stake back is not a gain, but not getting it back is definitely a loss!...
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This note was uploaded on 09/14/2009 for the course STAT 134 taught by Professor Aldous during the Fall '03 term at University of California, Berkeley.
 Fall '03
 aldous

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