HW1_and_2_sol

HW1_and_2_sol - ECON 414 SOLUTION TO HOMEWORK 1 & 2...

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ECON 414 FALL 2010 B.2 (i) P(X 6) = P[(X – 5)/2 (6 – 5)/2] = P(Z .5) .692, where Z denotes a Normal (0,1) random variable. [We obtain P(Z .5) from Table G.1.] (ii) P(X > 4) = P[(X – 5)/2 > (4 – 5)/2] = P(Z > - .5) = P(Z .5) .692. (iii) P(|X – 5| > 1) = P(X – 5 > 1) + P(X – 5 < –1) = P(X > 6) + P(X < 4) (1 – .692) + (1 – . 692) = .616, where we have used answers from parts (i) and (ii). B.4 We want P(X .6). Because X is continuous, this is the same as P(X > .6) = 1 – P(X . 6) = F(.6) = 3(.6)2 – 2(.6)3 = .648. One way to interpret this is that almost 65% of all counties have an elderly employment rate of .6 or higher. B.8 The weights for the two-, three-, and four-credit courses are 2/9, 3/9, and 4/9, respectively. Let Y j be the grade in the jth course, j = 1, 2, and 3, and let X be the overall grade point average. Then X = (2/9)Y1 + (3/9)Y2 + (4/9)Y3 and the expected value is E(X) = (2/9)E(Y1) + (3/9)E(Y2) + (4/9)E(Y3) = (2/9)(3.5) + (3/9)(3.0) + (4/9)(3.0) = (7 + 9 + 12)/9 3.11. B.10 (i) E(GPA|SAT = 800) = .70 + .002(800) = 2.3. Similarly, E(GPA|SAT =1,400) = .70 + . 002(1400) = 3.5. The difference in expected GPAs is substantial, but the difference in SAT scores is also rather large. (ii) Following the hint, we use the law of iterated expectations. Since E(GPA|SAT) = .70 + .002 SAT, the (unconditional) expected value of GPA is .70 + .002 E(SAT) = .70 + .002(1100) = 2.9. C.2 (i) E(Wa) = a 1 E(Y 1 ) + a 2 E(Y 2 ) + K + anE(Y n ) = (a 1 + a 2 + K + a n )µ. Therefore, we must have a 1 + a 2 + K + a n = 1 for unbiasedness. (ii) Var(Wa) =
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HW1_and_2_sol - ECON 414 SOLUTION TO HOMEWORK 1 &amp; 2...

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