516hw5 - 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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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) & 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 j-th 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!...
View Full Document

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.

Page1 / 3

516hw5 - 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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online