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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 fourcredit 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(GPASAT = 800) = .70 + .002(800) = 2.3.
Similarly, E(GPASAT =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(GPASAT) = .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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 Spring '11
 CARRILLO
 Economics

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