CISC 7700X Final Exam

1. b
2. 1    # ( (1 + 0.25) + (1 + 0.25) + (1 -0.50)) / 3 = 1, or 0% net-return. 
3. 0.9210    # (1.25 * 1.25 * 0.50)^(1/3) 
4. b
5. c
6. b
7. b   # we found a random widget, there is a 50% chance that the serial number 1234 is within the interquartile range of all serial numbers.
8. d
9. e  #invalid; p(x,y)=p(x|y)p(y)=p(y|x)p(x)
10. c
11. b
12. a
13. c

14. 0.6400
    Given: P(Leave) = 0.1, P(-Leave) = 0.9; 
    P(Leave|Promote) = P(Promote|Leave) P(Leave) / P(Promote)
    P(Leave|Promote) = P(Promote|Leave) P(Leave) / (P(Promote|Leave) P(Leave) + P(Promote|-Leave) P(-Leave) )
    = (0.8 * 0.1) / (0.8 * 0.1 + 0.05 * 0.9) = 0.6400

15. 0.068966
   Given: P(Leave) = 0.1, P(-Leave) = 0.9; 
   P(Leave|Lt5years) = P(Lt5years|Leave) P(Leave) / P(Lt5years)
   P(Leave|Lt5years) = P(Lt5years|Leave) P(Leave) / ( P(Lt5years|Leave) P(Leave) + P(Lt5years|-Leave) P(-Leave) ) 
      = (0.6 * 0.1) / ( 0.6 * 0.1 + 0.9 * 0.9) = 0.068966

16. no answer: 
    we need P(Leave|Lt5years,Promote) = P(Lt5years,Promote|Leave) P(Leave) / P(Lt5years,Promote) 
    we don't know P(Lt5years,Promote|Leave) or P(Lt5years,Promote)

17. 0.5424
 
  P(Leave|Lt5years,Promote) = P(Lt5years,Promote|Leave) P(Leave) / P(Lt5years,Promote)

  Naive bayes: 
   P(Leave|Lt5years,Promote) = P(Lt5years|Leave)P(Promote|Leave) P(Leave) / P(Lt5years)P(Promote)

  marginalize: 
    P(Lt5years|Leave)P(Promote|Leave) P(Leave) / 
      (P(Lt5years|Leave)P(Promote|Leave) P(Leave) + 
        P(Lt5years|-Leave)P(Promote|-Leave) P(-Leave)))
    = 0.6* 0.8 * 0.1 / ( 0.6* 0.8 * 0.1 + 0.9*0.05*0.9) = 0.5424

18. d
19. c

20. 0.64952 # exp( 3*( 0.5*log(1+0.5)+0.5*log(1-0.5)) ) =0.64952