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516hw5

516hw5 - STAT516 Section 3.2 FALL 2005 Solution to Homework...

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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) & d) Can’t 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 j-th roll, j=1,2,3. So, E(#of sixes in 3 rolls) = 1/6 + 1/6 + 1/6 = ½ 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: Markov’s 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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