CISC 7700X Final Exam

1. b
2. 1    # ( (1 + 0.35) + (1 + 0.35) + (1 -0.70)) / 3 = 1, or 0% net-return. 
3. 0.8177    # (1.35 * 1.35 * 0.30)^(1/3) 
4. c
5. b
6. b, or e if ~10  # sqrt(2*7^2)
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. d
11. b
12. a
13. c
14. 0.3571
   # P(speeding) = 0.1, P(-speeding) = 0.9, P(bmw|speeding) = 0.5, P(bmw|-speeding) = 0.1
   # P(speeding|bmw) = P(bmw|speeding)P(speeding) / ( P(bmw|speeding)P(speeding) + P(bmw|-speeding)P(-speeding))
   #                 = (0.5 * 0.1) / ((0.5 * 0.1) + (0.1 * 0.9))  = 0.3571
15. 0.4255
   # P(speeding) = 0.1, P(-speeding) = 0.9, P(2door|speeding) = 1/3, P(2door|-speeding) = 1/20
   # P(speeding|bmw) = P(2door|speeding)P(speeding) / ( P(2door|speeding)P(speeding) + P(2door|-speeding)P(-speeding))
   #                 = ( (1/3) * 0.1 ) / ( ((1/3) * 0.1) + ((1/20) * 0.9) = 0.4255)   

16. not enough information.
   # P(speeding|two-door-coupe,bmw) = P(2door,bmw|speeding)P(speeding) / ( P(2door,bmw|speeding)P(speeding) + P(2door,bmw|-speeding)P(-speeding))
   # we don't have P(2door,bmw|speeding) nor P(2door,bmw|-speeding)

17. 0.7874
   # P(2door,bmw|speeding) = P(2door|speeding)*P(bmw|speeding)
   # P(speeding|2door,bmw) = P(2door,bmw|speeding)P(speeding) / ( P(2door,bmw|speeding)P(speeding) + P(2door,bmw|-speeding)P(-speeding))
   # P(speeding|2door,bmw) = P(2door|speeding)*P(bmw|speeding)*P(speeding) / 
   #   ( P(2door|speeding)*P(bmw|speeding)*P(speeding) + P(2door|-speeding)*P(bmw|-speeding)*P(-speeding))
   #                      = ( (1/3) * 0.5 * 0.1 ) / ( ( (1/3) * 0.5 * 0.1 ) + ( (1/20) * 0.1 * 0.9 )) = 0.7874 

18. d
19. c
20. 0.64952;   # exp( 3*( 0.5*log(1+0.5)+0.5*log(1-0.5)) )