Download HONR 399 November 5, 2010 Chapter 29 Answers 1. At a certain

HONR 399
November 5, 2010
Chapter 29 Answers
1. At a certain university, each student who tries to make an appointment for registration
is successful with probability .85 each time that he or she makes a request. If unsuccessful
in making an appointment, he or she makes another request, and another, etc., until finally
successfully getting an appointment. The attempts are independent. Also, the various
students’ attempts are independent among students. Within a college of 300 students, find
the probability that strictly more than 360 attempts are necessary (altogether) for all students
to successfully get appointments.
If X1 , . . . , X300 denote the number of attempts that each student makes until successfully
getting an appointment, we are asked to compute
P (360 < X1 + · · · + X300 ),
or equivalently,
P (361 ≤ X1 + · · · + X300 ).
Since we are approximating an integer-valued sum of random variables, we again need to use
continuity correction. So we compute
P (360.5 ≤ X1 + · · · + X300 ).
We notice that each Xj is geometric with p = .85, so E[Xj ] = 1/.85 and Var(Xj ) =
q/p = .15/.852 .
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So we compute:
P (360.5 ≤ X1 + · · · + X300 )
=P
X1 + · · · + X300 − (300)(1/.85)
360.5 − (300)(1/.85)
p
p
≤
(300)(.15/.852 )
(300)(.15/.852 )
!
≈ P (.96 ≤ Z)
= 1 − P (Z ≤ .96)
≈ 1 − .8315
= .1685
So the probability is approximately 16.85% that strictly more than 360 attempts are needed
altogether.
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2. At an auction, exactly 282 people place requests for an item. The bids are placed “blindly”,
which means that they are placed independently, without knowledge of the actions of any
other bidders. Assume that each bid is a continuous uniform random variable on the interval
[10.50, 19.30]. Find the probability that the sum of all the bids exceeds 4150.
We let X1 , . . . , X282 denote the auction bids. We are making an approximation of a sum
of continuous random variables, so no continuity correction is needed.
Since each Xj is uniform on [10.50, 19.30], then E[Xj ] = (10.50 + 19.30)/2 = 14.9 and
Var(Xj ) = (19.30 − 10.50)2 /12 ≈ 6.453.
So we compute:
P (4150 ≤ X1 + · · · + X282 )
=P
X1 + · · · + X282 − (282)(14.9)
4150 − (282)(14.9)
p
p
≤
(282)(6.453)
(282)(6.453)
!
≈ P (−1.21 ≤ Z)
= P (1.21 ≥ Z)
= P (Z ≤ 1.21)
≈ .8869
So the probability is approximately 88.69% that the sum of all the bids exceeds 4150.
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3. A certain DJ takes requests for songs at a party. Assume that there are 120 people the
a party, each of whom makes exactly one request for a song. All of their requests are made
independently. Assume that each person asks for a pop song with probability .37, a rock
song with probability .20, or a rap song with probability .43. What is the probability that
50 or more requests are made for pop songs?
Let X1 , . . . , X120 be independent Bernoulli random variables, with
Xj = 1
if the jth song request is for a “pop” song,
and Xj = 0 otherwise. Then X1 + · · · + X120 is the total number of requests that are made
for pop songs. We need to compute
P (50 ≤ X1 + · · · + X120 ),
or equivalently,
P (49 < X1 + · · · + X120 ).
Here we are approximating an integer-valued sum of random variables, so we need to use
continuity correction. So we compute
P (49.5 ≤ X1 + · · · + X120 ).
Since each Xj is Bernoulli with p = .37, then E[Xj ] = p = .37 and Var(Xj ) = p(1 − p) =
(.37)(.63) = .2331.
So we compute:
P (49.5 ≤ X1 + · · · + X120 )
=P
49.5 − (120)(.37)
X1 + · · · + X120 − (120)(.37)
p
p
≤
(120)(.37)(.63)
(120)(.37)(.63))
!
≈ P (.96 ≤ Z)
= 1 − P (Z ≤ .96)
≈ 1 − .8315
= .1685
So the probability is approximately 16.85% that 50 or more requests are made for pop songs.
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4. Now consider the 300 students who (finally!) all got appointments for registration. When
each students arrives at the registration office, even though he or she has an appointment,
a short wait is always required. Suppose that the waiting times are independent, and each
waiting time (in hours) has density fX (x) = 6e−6x for x > 0, and fX (x) = 0 otherwise. Find
the probability that the 300 students collectively (i.e., altogether) spend between 45 and 52
hours waiting for their appointments.
We let X1 , . . . , X300 denote the waiting times for the 300 students, respectively. We
are making an approximation of a sum of continuous random variables, so no continuity
correction is needed.
Each Xj has expected value
Z ∞
∞
E[Xj ] =
(x)(6e−6x ) dx = (x)(−e−6x ) − (1)(1/6)(e−6x ) x=0 = 1/6
0
and also the expected value of Xj2 is
Z ∞
∞
2
E[Xj ] =
(x2 )(6e−6x )dx = (x2 )(−e−6x ) − (2x)(1/6)(e−6x ) − (2)(1/36)(e−6x ) x=0 = 2/36
0
and thus
Var(Xj ) = 2/36 − (1/6)2 = 1/36.
So we compute:
P (45 ≤ X1 + · · · + X300 ≤ 52)
=P
X1 + · · · + X300 − (300)(1/6)
52 − (300)(1/6)
45 − (300)(1/6)
p
p
≤
≤ p
(300)(1/36)
(300)(1/36)
(300)(1/36)
!
≈ P (−1.73 ≤ Z ≤ .69)
= P (Z ≤ .69) − P (Z ≤ −1.73)
= P (Z ≤ .69) − P (Z ≥ 1.73)
= P (Z ≤ .69) − 1 + P (Z ≤ 1.73)
≈ .7549 − 1 + .9582
= .7131
So the probability is approximately 71.31% that the students spend between 45 and 52 hours
waiting for their appointments.
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