# hw2s - MSPE PROGRAM ECON 506- FALL 2011 SOLUTION TO HW2...

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MSPE PROGRAM ECON 506- FALL 2011 SOLUTION TO HW2 (The problem numbers refer to numbers in 7 th edition) 3.11 There is a 1/3 chance a person has O + blood and 2/3 they do not. Similarly, there is a 1/15 chance a person has O blood and 14/15 chance they do not. Assuming the donors are randomly selected, if X = # of O + blood donors and Y = # of O blood donors, the probability distributions are 0 1 2 3 p ( x ) (2/3) 3 = 8/27 3(2/3) 2 (1/3) = 12/27 3(2/3)(1/3) 2 =6/27 (1/3) 3 = 1/27 p ( y ) 2744/3375 3*(196/3375) 3*(14/3375) 1/3375 Note that Z = X + Y = # will type O blood. The probability a donor will have type O blood is 1/3 + 1/15 = 6/15 = 2/5. The probability distribution for Z is 0 1 2 3 p ( z ) (2/5) 3 = 27/125 3(2/5) 2 (3/5) = 54/27 3(2/5)(3/5) 2 =36/125 (3/5) 3 = 27/125 3.12 E ( Y ) = 1(.4) + 2(.3) + 3(.2) + 4(.1) = 2.0 E (1/ Y ) = 1(.4) + 1/2(.3) + 1/3(.2) + 1/4(.1) = 0.6417 E ( Y 2 – 1) = E ( Y 2 ) – 1 = [1(.4) + 2 2 (.3) + 3 2 (.2) + 4 2 (.1)] – 1 = 5 – 1 = 4. V ( Y ) = E ( Y 2 ) = [ E ( Y )] 2 = 5 – 22 = 1. 3.23

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## This note was uploaded on 01/27/2012 for the course ECON 506 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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hw2s - MSPE PROGRAM ECON 506- FALL 2011 SOLUTION TO HW2...

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