ISYE 6644 FINAL ALL EXAMS
COMBINED TEST 1 WITH COMPLETE
SOLUTION
DECISION MAKING (GEORGIA INSTITUTE OF TECHNOLOGY
, 1
ISyE 6644 — Sp/Su 2026 — Test #1 Solutions
This test is 120 minutes. You’re allowed one cheat sheet (both sides). I also give you
permission to use your computer (if you want) for Questions 30 and 34.
In any case, this test requires a proctor. All questions are 3 points, except 34, which is
1 point — and I’ll probably give you that 1 point as long as you give me any interesting answer.
Good luck! I want you to make this test wish that it had never been born!!!
1. TRUE or FALSE? Simulation can be used to analyze queueing models that are
too complicated to solve analytically.
Solution: TRUE (of course!) □
2. Use bisection (or any other method) to find x such that e−x = x.
(a) x = 0
(b) x = e
(c) x = e/2
(d) .
x = 0.567
(e) None of the above.
Solution: Let’s use bisection to find the zero of g(x) = e−x − x.
x g(x) comments
0 1
1 —0.6321 look in [0, 1]
0.5 0.1065 look in [0.5, 1]
0.75 —0.2776 look in [0.5, 0.75]
0.625 —0.0897 look in [0.5, 0.625]
0.5625 0.0073 look in [0.5625, 0.625]
0.59375 −0.0415 look in [0.5625, 0.59375]
0.578125 −0.0172 look in [0.5625, 0.578125]
0.5703125 −0.0050 OK, stop here.
, 2
.
If we keep going, we get closer and closer to x = 0.5671; so the answer is (d). □
3. Suppose that X is a continuous random variable with p.d.f. f (x) = 2x for 0 < x < 1.
Find Pr(X < 1/2 | X > 1/4).
(a) 0
(b) 0.2
(c) 0.5
(d) 0.8
(e) 1/16
Solution: We have
Pr(X < 1/2 ∩ X > 1/4)
Pr(X < 1/2 | X > 1/4) =
Pr(X > 1/4)
Pr(1/4 < X < 1/2)
=
Pr(X > 1/4)
∫ 1/2
2xdx
= ∫1/4
1
1/4
2xdx
= 0.2,
after the smoke clears. So the answer is (b). □
4. Suppose I conduct a series of 4 independent experiments, each of which has a 20%
chance of success. What’s the probability that I’ll see at least 3 successes?
(a) 0.027
(b) 0.181
(c) 0.819
(d) 0.973
(e) 1
, 3
Solution: The number of successes X ∼ Bin(4, 0.2). Thus,
Σ4
4
Pr(X ≥ 3) = (0.2)x(0.8)4−x = 0.0272.
x=3
x
So the answer is (a). □
5. If X ∼ Bern(0.5), find E[ln(X + 1)].
(a) 1
(b) e/2
(c) 0.347
(d) 1.38
(e) None of the above.
Solution: By the Unconscious Statistician and the fact that X ~ Bern(0.5), we
have
Σ1 1 1
E[ln(X + 1)] = ln(x + 1) Pr(X = x) = ln(1) + ln(2) = 0.347,
2 2
x=0
so the answer is (c). □
6. If X has a mean of −2 and a variance of 3, find Var(−2X + 1).
(a) −5
(b) −6
(c) −12
(d) 12
(e) 13
Solution: Var(−2X + 1) = 4Var(X) = 12, so the answer is (d). □
COMBINED TEST 1 WITH COMPLETE
SOLUTION
DECISION MAKING (GEORGIA INSTITUTE OF TECHNOLOGY
, 1
ISyE 6644 — Sp/Su 2026 — Test #1 Solutions
This test is 120 minutes. You’re allowed one cheat sheet (both sides). I also give you
permission to use your computer (if you want) for Questions 30 and 34.
In any case, this test requires a proctor. All questions are 3 points, except 34, which is
1 point — and I’ll probably give you that 1 point as long as you give me any interesting answer.
Good luck! I want you to make this test wish that it had never been born!!!
1. TRUE or FALSE? Simulation can be used to analyze queueing models that are
too complicated to solve analytically.
Solution: TRUE (of course!) □
2. Use bisection (or any other method) to find x such that e−x = x.
(a) x = 0
(b) x = e
(c) x = e/2
(d) .
x = 0.567
(e) None of the above.
Solution: Let’s use bisection to find the zero of g(x) = e−x − x.
x g(x) comments
0 1
1 —0.6321 look in [0, 1]
0.5 0.1065 look in [0.5, 1]
0.75 —0.2776 look in [0.5, 0.75]
0.625 —0.0897 look in [0.5, 0.625]
0.5625 0.0073 look in [0.5625, 0.625]
0.59375 −0.0415 look in [0.5625, 0.59375]
0.578125 −0.0172 look in [0.5625, 0.578125]
0.5703125 −0.0050 OK, stop here.
, 2
.
If we keep going, we get closer and closer to x = 0.5671; so the answer is (d). □
3. Suppose that X is a continuous random variable with p.d.f. f (x) = 2x for 0 < x < 1.
Find Pr(X < 1/2 | X > 1/4).
(a) 0
(b) 0.2
(c) 0.5
(d) 0.8
(e) 1/16
Solution: We have
Pr(X < 1/2 ∩ X > 1/4)
Pr(X < 1/2 | X > 1/4) =
Pr(X > 1/4)
Pr(1/4 < X < 1/2)
=
Pr(X > 1/4)
∫ 1/2
2xdx
= ∫1/4
1
1/4
2xdx
= 0.2,
after the smoke clears. So the answer is (b). □
4. Suppose I conduct a series of 4 independent experiments, each of which has a 20%
chance of success. What’s the probability that I’ll see at least 3 successes?
(a) 0.027
(b) 0.181
(c) 0.819
(d) 0.973
(e) 1
, 3
Solution: The number of successes X ∼ Bin(4, 0.2). Thus,
Σ4
4
Pr(X ≥ 3) = (0.2)x(0.8)4−x = 0.0272.
x=3
x
So the answer is (a). □
5. If X ∼ Bern(0.5), find E[ln(X + 1)].
(a) 1
(b) e/2
(c) 0.347
(d) 1.38
(e) None of the above.
Solution: By the Unconscious Statistician and the fact that X ~ Bern(0.5), we
have
Σ1 1 1
E[ln(X + 1)] = ln(x + 1) Pr(X = x) = ln(1) + ln(2) = 0.347,
2 2
x=0
so the answer is (c). □
6. If X has a mean of −2 and a variance of 3, find Var(−2X + 1).
(a) −5
(b) −6
(c) −12
(d) 12
(e) 13
Solution: Var(−2X + 1) = 4Var(X) = 12, so the answer is (d). □
