Stat 330
Exam III Practice Questions- Key
Spring 2015
1. Given that Leonard’s last cola purchase was Coke, there is a 90% chance that his next cola purchase
will also be Coke. If a Leonard’s last cola purchase was Pepsi, there is an 80% chance that his next cola
purchase will also be Pepsi. Assume that Leonard purchases cola everday and that Coke and Pepsi are
his only choices.
(a.) Give the 1-step transition matrix P.
P =
0.9 0.1
0.2 0.8
(b.) Given that Leonard purchased a Coke on today, what is the probability that he will purchase
Pepsi three days from now?
P
(3)
3
=P =
0.9 0.1
0.9
·
0.2 0.8
0.2
0.1
0.9 0.1
0.781 0.219
·
=
0.8
0.2 0.8
0.438 0.562
So, P (Pepsi 3 days from now|Coke today) = 0.219
(c.) Suppose that on Monday, Leonard is likely to drink Coke 60% of the time and 40% likely to drink
Pepsi, what are the likelihoods of his drinking Coke and Pepsi on Thursday?
P3 = P0 · P 3 = (0.6, 0.4) ·
0.781 0.219
0.438 0.562
= (0.6438, 0.3562)
(d.) Obtain the steady-state distribution of this Markov chain? Explain what it means to you.
(π1 , π2 ) ·
0.9 0.1
= (π1 , π2 )
0.2 0.8
0.9π1 + 0.2π2 = π1
0.9π1 + 0.2(1 − π1 ) = π1
(1 − 0.9 + 0.2)π1 = 0.2
2
3
1
π2 =
3
π1 =
This means that Leonard has a
any given day.
2
3
chance of drinking Coke and a
1
3
chance of drinking Pepsi on
2. The number of hits on a popular Web page is distributed according to a Poisson distribution with
arrival rate = 5 per minute. One begins observation at exactly noon tomorrow (central standard
time).
(a) What is the probability of 4 or less hits in the first minute? Give a distribution and parameter(s).
Let X be the number of hits in one minute. Then X ∼ P oλ with λ = 5 per minute.
P (X ≤ 4) = P o5 (4)
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(b) Compute the probability that the time till the first hit exceeds 10 seconds. Give a distribution
and parameter(s).
Let Y be the time until the first hit. Then Y ∼ Expλ with λ = 5 per minute.
P (Y ≥ 1) = 1 − P (Y < 1) = 1 − (1 − e−λ·1 ) = e−5
(c) What is an appropriate distribution for the time until the 4th hit? State the distribution and
parameter value(s).
Let Z be the time until the fourth hit. Then Z ∼ Erlang(4, λ) with λ = 5 per minute.
(d) Evaluate the probability that the time till the 4th hit exceeds 24 seconds.
P (Z ≥ 24s) = P (Z > 24/60) = 1 − P (Z < 0.4) = 1 − Erl4,5 (0.4) = P o2 (3)
(e) Find the mean and the variance of the time till the 4th hit.
E[Z]
=
V ar[Z]
=
1
= 0.8 min
λ
1
4 · 2 = 0.16 min2
λ
4·
(f) The number of hits in the first hour is Poisson with mean 300. You would like to know the
probability of more than 350 hits. Exact calculation isn’t really feasible. So approximate this
probability and justify your approximation.
Let X be the number of hits in one hour. Then X ∼ P oλ with λ = 300 per hour.
P (X > 350) = 1 − P o300 (350) ≈ 1 − N300,300 (350) = 1 − N0,1 (
350 − 300
√
)
300
The approximation is based on the CLT. The parameters of the normal distribution are given by
expected value and variance of the Poisson distribution.
3. People arrive at a telephone booth at the Fly-by-Nite airline terminal in a random pattern with an
average interarrival time of 12 minutes. The length of phone calls from the booth is exponentially
distributed with an average time of 4 minutes.
(a) What is the probability that an arriving person will have to wait?
this system is an M/M/1 queue with λ = 5 arrivals per hour and µ = 15 departures per hour.
The probability that an incoming person will have to wait for service is
P (X ≥ 1) = 1 − p0 = 1 − (1 − a) = a = 1/3
(b) What is the average length of waiting lines that form, i.e. those lines that are not of zero length?
Lq =
a2
= 1/6
1−a
(c) What is the probability that an arrival will have to wait for more than 10 minutes before the
phone is available?
Let X be the time spent in the queue, then
P (X ≥ 1/6) = 1 − (1 − ae−x/W ) = 1/3 · e−1/6·15(1−1/3) = 0.063
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(d) The telephone company plans to add an additional booth when the traffic increases so much that
Wq ≥ 5 minutes. At what average interarrival time will Wq = 5 minutes occur?
Wq = W − Ws =
1
1
1
− =
− 4 min
µ−λ µ
0.25 − λ
Wq > 5 if
1
>9
0.25 − λ
Therefore
λ ≥ 8/36,
i.e. if more than 13.3 people arrive on average per hour, the airline should put a second booth
into place.
4. A small Telemarketing Company has 4 telephones and 4 members of the staff answering calls. If all
lines are busy, incoming callers are being told to call back later.
Calls come in at a rate of 6 per min. Staff members need on average 30 seconds to answer a call.
Assume that you are dealing with an M/M/x/y queuing system. If you are using any formulas from
your notes, please specify which.
(a) Draw a transition state diagram of the number of busy telephone lines.
6
0
6
1
2
6
2
4
6
3
6
4
8
(b) (Find the (steady state) probability that none of the lines is busy.
p0 = 1/S = (1 + 3 + 4.5 + 4.5 + 0.75 · 4.5)−1 = 1/16.375 = 0.06.
(c) The manager of the telemarketing company is anxious that as few customers as possible should
get a busy signal when calling. What is the rate of customers who are turned away?
λ4 · p4 = λ4 · 0.75 · 4.5 · p0 = 20.25/16.375 = 1.24 customers per min
5. A company has 4 software systems that periodically need the attention of a systems analyst. The
company employs 3 analysts. Suppose that when operating, each system crashes (putting it out of
operation) at rate λ = 2 per hour and that when working on a crashed system, an analyst returns it to
working order (and back into operation) at rate µ = 1 per hour. (An analyst works on a single system
at a time.) We wish to model this situation using a Birth and Death process with ”State i” being the
situation ”i software systems are down.”
(a) Sketch an appropriate state transition diagram (putting appropriate rates on it).
8
0
6
1
1
4
2
2
2
3
3
4
3
(b) Find the (steady-state) probability p0 that none of the system analysts are busy.
p0 = 1/S = (1 + 8 + 24 + 32 + 64/3)−1 = 3/259 = 0.0116.
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6. A toll area on a highway has three toll booths. On the average cars arrive at the rate of one car every
6 seconds and joins a line. It takes 12 seconds to pay toll at any booth (not including waiting in the
single line).
(a) Find the (steady state) probability that there are no cars waiting or paying toll. It is a M/M/3
queuing system with λ = 10/min, µ = 5/min, c = 3, a = λ/µ = 2 and ρ = a/c = 2/3.
p0 =
c−1 k
X
a
k=0
ac 1
+
k!
c! 1 − ρ
!−1
=
22
23
1
1+2+
+
·
2!
3! 1 − 2/3
−1
= 1/9 = 0.11111.
(b) Find the (steady state) average number of cars in the line.
Lq = p0 ·
ac
ρ
= 8/9 = .88889.
·
c! (1 − ρ)2
(c) Find the average time in the toll booths.
Ws =
1
= 1/5min = 12 sec
µ
(d) Find the (steady state) average number of cars in the system (in the line and paying toll).
First,
Wq = Lq /λ = (8/9)/(10) = .088889 min.
Then,
W = Ws + Wq = .2 + .088888 = .288888 min
Thus,
L = W λ = (.288888)(10) = 2.89 cars.
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