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Unformatted text preview: Final Practice Problems  Answer Key 1. The following is a fictional joint probability distribution for years of education an individual possesses and their income. Notice that 12 years of education corresponds to graduating from high school, 16 to graduating from college and 20 to having an advanced degree. Years Education/Income $ 30,000 $ 60,000 $ 90,000 12 years 0.05 16 years 0.2 0.4 20 years 0.35 (a) What is the probability of having 12 years of education? What is the probability of having 16 years of education? What is the probability of having 20 years of education? Letting E denote years of education and I denote income, we can then obtain: P ( E = 12) = P ( E = 12 ,I = 30 , 000)+ P ( E = 12 ,I = 60 , 000)+ P ( E = 12 ,I = 90 , 000) = 0 . 05 . In a similar way we obtain P ( E = 16) = 0 . 6 and P ( E = 20) = 0 . 35. (b) What is the probability of earning $ 30,000 conditional on having 12 years of education? Using the formula for conditional probability we obtain: P ( I = 30 , 000  E = 12) = P ( I = 30 , 000 ,E = 12) /P ( E = 12) = 0 . 05 / . 05 = 1 . (c) What is the probability of earning $ 60,000 conditional on having 16 years of education? What is the probability of earning $ 90,000 conditional on having 16 years of education? As in part (b), P ( I = 60 , 000  E = 16) = P ( I = 60 , 000 ,E = 16) /P ( E = 16) = 0 . 2 / . 6 = 1 / 3 and similarly P ( I = 90 , 000  E = 16) = P ( I = 90 , 000 ,E = 16) /P ( E = 16) = 0 . 4 / . 6 = 2 / 3. (d) Compare your answers to (b) and (c). Does it increase your earnings to obtain 16 years of education instead of 12? Yes, we see that people with 12 years of education earn 30,000 with probability 1, while people with 16 years of experience earn 60,000 with probability 1/3 and 90,000 with probability 2/3. (e) What is the covariance between years of education and income? What does your answer suggest about how income relates to years of education? Using the formula for the mean of a discrete random variable and using the answer to (a): E [ E ] = 12 × . 05 + 16 × . 6 + 20 × . 35 = 17 . 2 , while in a similar way we may find E [ I ] = 70 , 500. To find E [ EI ] we note: E [ EI ] = 12 × (30 , 000 × . 05+60 , 000 × 0+90 , 000 × 0)+16 × (30 , 000 × 0+60 , 000 × . 2+90 , 000 × . 4) + 20 × (30 , 000 × 0 + 60 , 000 × . 35 + 90 , 000 × 0) = 1 ,...
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 Spring '10
 Elliot
 Normal Distribution, $100, $ 60,000, $ 30,000, $ 90,000, $ 90,000 0 0.4 0

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