Questions
BU.232.630.W6.SP25 Quiz 1 solutions
Single choice
Consider two random variables: X can take values (4,5), and Y can take values (1,2,3). The joint distribution of X and Y is shown in the table below Y 1 2 3 X 4 0.1 0.30 0.20 5 0.1 0.05 0.25 Using this information, please compute the unconditional expected value E(X) and the conditional expected value E(X|Y=2). (Please round your results to the 4th decimal place.)
Options
A.E(X)=4.4;E(X|Y=2)=4.5
B.E(X)=4.4;E(X|Y=2)=4.1429
C.E(X)=4.5;E(X|Y=2)=4.5
D.E(X)=2.25;E(X|Y=2)=2.375
E.E(X)=2.25;E(X|Y=2)=4.1429
F.E(X)=4.4;E(X|Y=2)=2.375
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Step-by-Step Analysis
We start by parsing the joint distribution and computing the requested expectations step by step.
First, calculate E(X) using the marginal distribution of X. The probability that X=4 is the sum over Y: 0.1 + 0.30 + 0.20 = 0.60. The probability that X=5 is the sum over Y: 0.1 + 0.05 + 0.25 = 0.40. Therefore, E(X) = 4*(0.60) + 5*(0.40) = 2.40 + 2.00 = 4.40. When rounded to four decimals, E(X) = 4.4000, which is consistent with a value of 4.4 reported in several optio......Login to view full explanationLog in for full answers
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