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APPENDIX B
SOLUTIONS TO PROBLEMS
B.1 Before the student takes the SAT exam, we do not know – nor can we predict with certainty
B.2 (i) P(X 6) = P[(X – 5)/2 (6 – 5)/2] = P(Z .5)
.692, where Z denotes a Normal (0,1)
.692) = .616, where we have used answers from parts (i) and (ii).
B.3 (i) Let Yit be the binary variable equal to one if fund i outperforms the market in year t. By
assumption, P(Yit = 1) = .5 (a 50-50 chance of outperforming the market for each fund in each
(ii) Let X denote the number of funds out of 4,170 that outperform the market in all 10 years.
(iii) Using the Stata command Binomial(4170,5,1/1024), the answer is about .385. So there
is a nontrivial chance that at least five funds will outperform the market in all 10 years.
B.4 We want P(X .6). Because X is continuous, this is the same as P(X > .6) = 1 – P(X .6) =
264
(ii) Above, we computed P(X = 0) as about .069. We need P(X = 1), which we obtain from
B.6 E(X) =
3
()xf x dx
=
32
[(1/9) ] x x dx
= (1/9)
. But
33
x dx
= (1/4)x4
3
0
|
= 81/4.
B.7 In eight attempts the expected number of free throws is 8(.74) = 5.92, or about six free
throws.
B.8 The weights for the two-, three-, and four-credit courses are 2/9, 3/9, and 4/9, respectively.
B.9 If Y is salary in dollars then Y = 1000
X, and so the expected value of Y is 1,000 times the
B.10 (i) E(GPA|SAT = 800) = .70 + .002(800) = 2.3. Similarly, E(GPA|SAT = 1,400) = .70 +
(ii) Following the hint, we use the law of iterated expectations. Since E(GPA|SAT) = .70 +