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SOLUTION FOR HOMEWORK 8, ACTS 4306
Welcome to your 8th homework.
The main task of this homework is to recognize an underlying distribution and then solve
the problem as fast as possible using known facts about the distribution. I will use notation
of the text.
1. Solution: Here X, the number of hurricanes, is a classical Binomial(θ = .05, n = 20)
random variable. Then
P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2) =
20! 1
20! 2
20! 0
θ (1 − θ)20 +
θ (1 − θ)19 +
θ (1 − θ)18
0!20!
1!19!
2!18!
= (.95)20 + (20)(.05)(.95)19 + (190)(.05)2(.95)18 = .9245.
=
Note that you need a good calculator for these calculations!
Answer: E
2. Solution: Let X and Y be the number of participants who completed the study in the
first and second groups. Note that these random variables are independent Binomial(θ =
.8, n = 10). Then
P (X ≥ 9) = P (Y ≥ 9) = P (X = 9) + P (X = 10)
10! 9
10! 10
θ (1 − θ) +
θ (1 − θ)0 = (10)(.8)9(.2) + (.8)10 = .376.
9!1!
10!0!
The probability in question is
=
P ({(X ≥ 9) ∩ (Y < 9)} ∪ {(X < 9) ∩ (Y ≥ 9)})
(because X and Y are identically distributed)
= 2P ((X ≥ 9) ∩ (Y < 9))
(because X and Y are independent)
= 2P (X ≥ 9)P (Y < 9) = 2(.376)(1 − .376) = .469.
Answer: E
3. Solution: This is a classical Bayes Theorem’s application problem with the following
twist: we need to calculate probability of the event that among 30 vials 1 is ineffective. This
is the probability for a Binomial(θ, n = 30) random variable X to take on value 1, that is
P (X = 1) =
30! 1
θ (1 − θ)29 = (30)θ(1 − θ)29 .
1!(29)!
1
It is given that for shipments by company A we have θ = .1, and then
P (X = 1|A) = 30(.1)(.9)29 = .1413
and for shipments by other companies θ = .02 and then
P (X = 1|A′ ) = (30)(.02)(.98)29 = .3340.
Here A denotes the vent that a shipment is from company A.
It is given that P (A) = .2 and then the Bayes Theorem yields
P (A|X = 1) =
P (A)P (X = 1|A)
P (A)P (X = 1|A) + P (A′ )P (X = 1|A′ )
(plug in the numbers)
=
(.2)(.1413)
= .096.
(.2)(.1413) + (.8)(.3340)
Answer: A
4. Notation: X is the number of high-performing employees.
Find: The largest C that P (CX > 120) < .01.
Solution: X is Binomial(θ = .02, n = 20). Thus
P (CX > 120) = P (X > 120/C) < .01.
(1)
The simplest was to solve this problem is the normal approximation which will be discussed
in several weeks in class. For now a method of trials and errors will do. Direct calculations
show that
20!
P (X = 0) =
(.02)0(.98)20 = .6676,
0!20!
20!
(.02)1(.98)19 = .2725,
P (X = 1) =
1!19!
20!
P (X = 2) =
(.02)2(.98)18 = .0528.
2!18!
Here we stop because
P (X > 2) = 1 − [P (X = 0) + P (X = 1) + P (X = 2)] = .0071 < .01.
As a result, if we set 120/C = 2, that is C = 60, the requirement of the problem is satisfied.
Remark: Another approach is to try the given Cs and see if they are satisfied.
Answer: D
5. Notation: X is the number of days until an accident.
Given: fX (x) = ce−cx I(x > 0), c is an unknown constant, and P (X ≤ 50) = .3.
Find: P (X ≤ 80).
2
Solution: The underlying distribution is Exponential which we will study in two weeks.
For any a > 0 we have
P (X ≤ a) =
Z
0
a
ce−cx dx = [−e−cx ]a0 = 1 − e−ca .
This, from P (X ≤ 50) = .3 we get that c = − ln(.7)/50. Then
P (X ≤ 80) = 1 − e[ln(.7)/50](80) = 1 − (.7)80/50 = .43
Answer: C
6. Notation: X is the number of claims.
Given: X is P oisson(λ), λ is an unknown positive constant, P (X = 2) = 3P (X = 4).
Find: Var(X).
Solution: We know that for the Poisson random variable X
Var(X) = λ.
Thus we need to find λ from the given information. Write
P (X = k) =
e−λ λk
k!
and then using the given information about probabilities we get
e−λ λ4
e−λ λ2
=3
.
2!
4!
Solving this equation yields λ = 2.
Answer: D.
7. Solution: Let X be the year when the device fails. Then X is Geometric(θ = .4) and
P (X = k) = θ(1 − θ)k−1 I(k = 1, 2, . . .).
By its definition, the benefit B is equal to
B = 4000I(X = 1) + 3000I(X = 2) + 2000I(X = 3) + 1000I(X = 4).
Because E(I(X = k)) = P (X = k) we get
E(B) = 1000[(4)(.4) + (3)(.4)(.6) + (2)(.4)(.6)2 + (1)(.4)(.6)3 ]
= 1000[1.6 + .72 + .288 + .086] = 2694.4.
Answer: E
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8. Solution: Let X be the number of tourists that show up. Note that X is Binomial(θ =
.98, n = 21). The event that a seat is not available occurs if and only if X = 21, and
P (X = 21) = (.98)21 = .6543.
The expected loss of the operator due to overbooking is
(0)P (X < 21) + (100)P (X = 21) = (100)(.6543) = 65.43,
and then the expected revenue is (remember that E(X) = θn) the average sale minus the
expected loss due to overbooking,
E{(50)X} − 65.43 = (50)(.98)(21) − 65.43 = 1029 − 65.43 = 963.57
Remark: Is the strategy of overbooking profitable? Think about this.
Answer: C (the closest one).
9. Notation: L, M and H are the types of drivers.
Given: P (L) = .5, P (M) = .3, P (H) = .2.
Solution: Here the formulation of “what to find” is tricky. Let XL , XM and XH be the
numbers of low-, moderate and high-risk drivers in a random sample of 4 drivers. Then we
need to find the probability
P (XH − HL ≥ 2) = P ((Xl = 0) ∩ (XH ≥ 2)) + P ((XL = 1) ∩ (XH ≥ 3)).
Let us evaluate two probabilities in the right side.
P ((XL = 0) ∩ (XH ≥ 2)
= P (XL = 0, XH = 2, XM = 2)+P (XL = 0, XH = 3, XM = 1)+P (XL = 0, XH = 4, XM = 0)
(note that the underlying distribution is multinomial)
=
4!
4!
4!
(.5)0 (.2)2 (.3)2 +
(.5)0 (.2)3 (.3)1 +
(.5)0 (.2)4 (.3)0 = .0216+.0096+.0016 = .0328.
0!2!2!
0!3!1!
0!4!0!
Further, the event oin the second probability is elementary one,
P ((XL = 1) ∩ (XH ≥ 3)) = P (Xl = 1, XH = 3, XM = 0) =
4!
(.5)1 (.2)3 (.3)0 = .016.
1!3!0!
Combining the results we get
P (XH − XL ≥ 2) = .0328 + .016 = .0488.
Answer: D
10. Solution: Here X is Hypergeometric(N = 25, M = 10, n = 10), I use notation of the
text, p. 557. Then we know that
E(X) =
(10)(10)
nM
=
= 4,
N
25
4
and
Var(X) =
(10)(10)(15)(15)
nM(N − M)(N − n)
=
= 3/2.
2
N (N − 1)
(25)2 (24)
Thus
Var(X)/E(X) = 3/8.
Answer: E
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