Download IE 4521 Midterm #1

IE 4521 Midterm #1
Prof. John Gunnar Carlsson
March 2, 2010
This exam has 9 pages (including the normal distribution table) and a total of 8 problems.
Make sure that all pages are present. To obtain credit for a problem, you must show all your work. if you use
a formula to answer a problem, write the formula down. Do not open this exam until instructed to do so.
Before you begin:
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1. (10 points) Let E , F , and G be three events in a state space S . Write expressions for the events that of E ,
F , and G
(a) only E occurs
Solution
E ∩ F c ∩ Gc
(b) both E and G but not F occur
Solution
E ∩ Fc ∩ G
(c) at least one of the events occurs
Solution
E∪F ∪G
(d) at least two of the events occur
Solution
(E ∩ F ) ∪ (E ∩ G) ∪ (F ∩ G)
(e) all three occur
Solution
E∩F ∩G
(f) none of the events occurs
Solution
E c ∩ F c ∩ Gc
(g) at most one of the events occurs
Solution
(E ∩ F c ∩ Gc ) ∪ (E c ∩ F ∩ Gc ) ∪ (E c ∩ F c ∩ G) ∪ (E c ∩ F c ∩ Gc )
(h) at most two of the events occur
Solution
(E ∩ F ∩ G)
c
(i) exactly two of the events occur
Solution
(E ∩ F ∩ Gc ) ∪ (E ∩ F c ∩ G) ∪ (E c ∩ F ∩ G)
(j) Simplify the expression
(E ∪ F ) ∩ (F ∪ G)
(hint: draw a Venn diagram)
Solution
F ∪ (E ∩ G)
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2. (15 points) You ask your neighbor to water a sickly plant while you are on vacation. Without water it will
die with probability 0.8; with water it will die with probability 0.15. You are 90 percent certain that your
neighbor will remember to water the plant.
(a) What is the probability that the plant will be alive when you return?
Solution Solution Let A be the event that your neighbor waters your plant and B the event that your
plant is alive. By the law of total probability,
Pr (B) = Pr ( B| A) Pr (A) + Pr ( B| Ac ) Pr (Ac ) = 0.85 · 0.9 + 0.2 · 0.1 = 0.785
so there is a 78.5% probability that your plant will remain alive.
(b) If the plant is dead, what is the probability that your neighbor forgot to water it?
Solution With A and B as before, we have by Bayes' Rule
Pr ( Ac | B c ) =
0.8 · 0.1
Pr ( B c | Ac ) Pr (Ac )
=
≈ 0.37
c
Pr (B )
1 − 0.785
so there is a 37% probability that your neighbor forgot to water it.
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3. (12 points) Let X represent the dierence between the number of heads and the number of tails obtained
when a coin is tossed n times.
(a) What are the possible values of X ?
Solution If n is even, X can take the values n + 2k , for k ∈ {− bn/2c , . . . , bn/2c}.
Comment If students only write positive values (i.e. they take the absolute value of the dierence),
this is OK too.
(b) If the coin is assumed to be fair, what are the probabilities associated with the values that X can take,
for n = 3?
Solution By symmetry of the distribution, Pr (X = 3) = Pr (X = −3) = 1/8. The only other possible
values are X = 1 and X = −1, and the distribution is symmetric, so Pr (X = 1) = Pr (X = −1) =
3/8.
4. (13 points) Suppose we ip a fair coin 5 times. Find the probability that:
(a) the rst three ips are the same
Solution
(1/2) + (1/2) = 1/4 (all heads or all tails)
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(b) either the rst three ips are the same, or the last three ips are the same
Solution Let A denote the event that the rst three ips are the same and B the event that the last three
ips are the same. Then Pr (A) = Pr (B) = 1/4, and Pr (A ∪ B) = Pr (A) + Pr (B) − Pr (A ∩ B).
The event A∩B happens if all ve ips are heads or all ve ips are tails, and therefore Pr (A ∩ B) =
5
5
(1/2) + (1/5) = 1/16. Therefore, we nd that
Pr (A ∪ B) = 1/4 + 1/4 − 1/16 = 7/16
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5. (15 points) A random variable X , which represents the weight (in ounces) of an article, has a density function
given by f (x) with


x − 8 if 8 ≤ x ≤ 9
f (x) = 10 − x if 9 ≤ x ≤ 10


(a) Calculate the mean and variance of X .
Solution We have
E (X)
0
ˆ
otherwise
∞
=
xf (x) dx
−∞
ˆ 9
ˆ
x (10 − x) dx
8
=
E X
2
9
ˆ
9
ˆ
9
x2 (10 − x) dx
x (x − 8) dx +
9
8
and therefore
10
2
=
=
10
x (x − 8) dx +
=
487/6
487
2
Var (X) = E X 2 − [E (X)] =
− 92 = 1/6
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(b) If an article has weight X , then the distributor will sell it for $ X2 . He guarantees to refund the purchase
money to any customer who nds the weight of his article to be less than 9 oz. An article costs $2.00 to
produce. Find the expected prot per article.
Solution Let g (X) denote the prot associated with an article of weight X . If X ≤ 9, then the
distributor's prot is −2 dollars, since the customer returns the article. If X > 9, then the distributor
earns a prot of X2 − 2 dollars, and therefore
(
−2
if X ≤ 9
g (X) = X
−
2
if
X>9
2
We nd that the expected prot is
ˆ
E (g (X))
∞
=
g (x) f (x) dx
−∞
ˆ 9
ˆ
(−2) (x − 8) dx +
=
8
=
10
(x/2 − 2) (10 − x) dx
9
1/3 ≈ $0.33
6. (15 points) In a typical sports playo series, two teams play a sequence of games until one team, the winner,
has won four games. Suppose that in each game, team A beats team B with a probability of 0.55, and that
the results of dierent games are independent.
(a) Explain how the binomial distribution can be used to analyze this problem.
Solution The outcome of each game can be regarded as a Bernoulli random variable Xi , with Xi = 1
indicating a win for team A and Xi = 0 a win for team B . Hence, if X is a binomial random variable
with p = 0.55 and n = 7, then a win for A corresponds to X > 3 and a win for B corresponds to
X ≤ 3.
(b) Explain how the negative binomial distribution can be used to analyze this problem.
Solution Let X be a negative binomial random variable with p = 0.55 and r = 4. Then a win for A
corresponds to X ≤ 7 and a win for B corresponds to X > 7. Furthermore, if X ≤ 7, then the value
of X tells us exactly how many games it took team A to win the tournament.
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(c) What is the probability that team A wins the series in game seven?
By part (b) this is just Pr (X = 7) = C (7 − 1, 4 − 1) (0.45)3 (0.55)4 ≈ 0.167
(d) What is the probability that team A wins the series in game six?
Solution
By part (b) this is just Pr (X = 6) = C (6 − 1, 4 − 1) (0.45)2 (0.55)4 ≈ 0.185
(e) What is the probability that team A wins the series?
Solution This is just Pr (X = 4) + Pr (X = 5) + Pr (X = 6) + Pr (X = 7) ≈ 0.608
Solution
7. (15 points) Suppose that X ∼ N (64, 25).
(a) What is Pr (59.9 ≤ X ≤ 70.0)?
Solution We have
X − 64
6.0
−4.1
≤
≤
Pr (59.9 ≤ X ≤ 70.0) = Pr
5
5
5
= Pr (−0.82 ≤ Z ≤ 1.2)
=
Φ (1.2) − Φ (−0.82) ≈ 0.679
(b) What is Pr (|X − 64| ≤ 3)?
Solution We have
X − 64
≤ 67
= Pr 61 ≤
X − 64
= Pr −0.6 ≤
≤ 0.6
5
= Φ (0.6) − Φ (−0.6) ≈ 0.4515
Pr (|X − 64| ≤ 3)
(c) Find y so that Pr (X ≤ y) ≈ 0.30.
Solution We have
Pr
Pr (X ≤ y) ≈ 0.30
y − 64
X − 64
≤
≈ 0.30
5
5
Pr (Z ≤ z) ≈ 0.30
We nd that z ≈ −0.52 and therefore y = 5z + 64 ≈ 61.4.
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8. (10 points) True/false. No justication is needed.
(a) If the conditional probability Pr ( A| B) is equal to Pr (A), then the events A and B are independent.
True.
(b) If Pr ( A| B) = Pr (A), then Pr ( B| A) = Pr (B). True.
(c) The event A and its complement Ac are independent. False.
(d) SAT-style analogy: Exponential distribution:Gamma distribution :: Geometric distribution:Negative
binomial distribution. True.
(e) If X is a memoryless random variable, then Pr ( X ≥ x + y| X ≥ x) = Pr (X ≥ y). True.
(f) If Z ∼ N (0, 1), then E Z 2 = 1. True.
(g) If X ∼ N (0, 1) and Y ∼ N (0, 1), then X + Y ∼ N (0, 1). False.
(h) If X ∼ N (0, 4) and Y ∼ N (0, 1), then X − Y ∼ N (0, 3). False.
(i) If Pr (A) = 0.55 and Pr (B) = 0.65, then A and B cannot be mutually exclusive. True.
(j) If X1 , . . . , X100 are independent random variables with mean 0 and variance 10, then
follows a normal distribution with mean 0 and variance 0.1. True.
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X1 +···+X100
100
roughly