Georgia Institute Of Technology ISYE 6644 EXTRA OFFICE HOUR β MIDTERM 1 SOLUTIONS
ISYE 6644
EXTRA OFFICE HOUR β MIDTERM 1 SOLUTIONS
(SORRY IN
... [Show More] ADVANCE FOR MY LACK OF POWERPOINT SKILLS!)With only one server, customer arrivals will overwhelm the server and a queue WILL form.10AM 11AM 1PM 2PMπΆππ£ π, π = πΆππ£ π, π2 = πΈ π3 β πΈ π πΈ[π2]
This leaves us with 0 β 0(πΈ π2] = 0
*Extra example here; not needed to derive/calculate
this to answer this question.The question was supposed to state βsum is less than or equal to 2β (or alternatively βsum is less than 3β). This would have made
the correct answer to be FALSE because
P(sum is less than or equal to 2)= 1
P(sum is greater than 11)= 1
P(sum is less than or equal to 2 AND sum is greater than 11)= 0 and then 0 β 1
which means they are dependent.
However, the way the question is worded, the P(sum is less than 2)= 0, so we would have 0 = 0 1
36
which means they are
independent.P(Blue)=4
P(Blue, Green) = P(Blue)P(Green) = 4
First Draw (8 pens) Second Draw (8 pens)P(Blue)=4
First Draw (8 pens) Second Draw (7 pens)Only possibility of being square is when π = 4.7 and π = 4.7, so we want to find π(π = 4.7, π = 4.7)1.000.20 1.00
8 9Let π=number of successes in the 6 experiments
π~π΅πππππππ(π = 6, π = 0.25)
= π π β€ 0 = 0.5Time between events (interarrival times) in a Poisson process follow an exponential distribution with π = 1
Let π = time between the 5th and 6th arrivals; π~πΈπ₯π(π = 1)
Note: The cumulative distribution function for πΈπ₯π(π) is πΉ π₯ = π π β€ π₯ = 1 β πβππ₯
Could also try integrating the probability density function
= πβ1 β 0.368If ππ has mean π = π and variance π2 = π, then the Central Limit Theorem implies that
πWould generate sum of two dice
Would include 1.5, 2.5, etc.
ππ would be creating ππππ(0, π). Using the ceiling function would then create a discrete distribution with the
possible outcomes of 1, 2, 3, 4, β¦ , π which is a realization of an π-sided die toss.Using properties of logarithms, β2 ln π4 = β8ln(π)
This has the same form as β 1
π ln(π) which is ~πΈπ₯π(π), so β8 ln π ~πΈπ₯π 1 8
Note, this is not Erlang because we only have one π here, so we are not adding iid
exponentials. We would be adding up the same exponential four times.
Erlang would look like
β2 ln π1π2π3π4 = β2 ln π1 β 2ln(π2) β 2ln(π3) β 2ln(π4)This is an application of inverse transform since Ξ¦(π₯) is the cumulative distribution function for π(0,1) distribution.
2 β 0.345Integral of π 0, 1 distribution, so this is the same as π 0 β€ π β€ 1 = Ξ¦ 1 β Ξ¦ 0 = 0.8413 β 0.5000 = 0.3413πΰ· = 4πΖΈ = 4
168
200 = 3.36π1 = 5π0 + 1 πππ 8 = 16πππ 8 = 0
Has a cycle length of 8, so one way is to take 100πππ 8 = 4, so we want π4 = 7
Or π100 = π96 = π92 = β― = π4Using inverse transform, π = β 1
π ln π = β 0.125 ln(0.5) β 2.77Arrive Start Service Leave Wait [Show Less]