201
(k)
(k)
(k)
(k)
for k > M . The increments I1 , B1 , . . . , In , Bn are all disjoint
and account for the entire interval [0, sn]. Now dene the event
Ak = cfw_I1 = 0
.
cfw_In = 0
cfw_B1 = 1
.
cfw_Bn = 1.
Then Ak implies Ak1 (that is, Ak is contained i
Stat 465/865 Assignment 4
Solutions
J. Pohlkamp-Hartt
March 26, 2015
1
Textbook Problems [102]
6th Edition in brackets
1. Montgomery Ch. 10 P. 1 (same)[6]
We nd the means and standard deviations for each product and then normalize the data for the followi
Stat 465/865 Assignment 5
Due April 8 2015
J. Pohlkamp-Hartt
1
Textbook Problems(6th Edition in brackets)
1. Montgomery Ch. 13 P. 2 (DB)
Corrections in ANOVA table: M SA = 9.999 and SST = 357.653
2. Montgomery Ch. 13 P. 16 (13.14)
3. Montgomery Ch. 14 P.
Stat 465/865 Assignment 1
Due by January 22, 2015
J. Pohlkamp-Hartt
January 5, 2015
1
Textbook Problems
1. Montgomery Ch. 1 P. 23
2. Montgomery Ch. 2 P. 12
3. Montgomery Ch. 2 P. 18
4. Montgomery Ch. 3 P. 8 (parts: a, b, d)
5. Montgomery Ch. 3 P. 13
6. Mo
Stat 465/865 Assignment 3
Due Feb 26 2015
J. Pohlkamp-Hartt
1
Textbook Problems(6th Edition in brackets)
1. Montgomery Ch. 6 P. 21 (6.15)
2. Montgomery Ch. 6 P. 22 (6.16)
3. Montgomery Ch. 6 P. 52 (6.44)
4. Montgomery Ch. 7 P. 5 (copied to dropbox)
5. Mon
Stat 465/865 Assignment 5[49]
Solutions
J. Pohlkamp-Hartt
April 7, 2015
1
Textbook Problems [23]
6th Edition in brackets
1. Montgomery Ch. 13 P. 2 (DB)[3,1,1,1]
DF MS
F
p
Source SS
A
9.999
1
9.999 .504 .498
180.378
3
60
3.02 .093
B
a)
AB
8.479
3
2.8
.142
library(qcc)
# Chapter 12
#Integral Control
#Make Data
xt<-rnorm(100,0,5)
yt<-200+(1/4)*xt+ .13*1:100 - .002*(1:100)^2
#Set Target Value
target<-200
# find g
g<-1/4 #if given
#g<-lm(yt-target)~xt-1)$coef #if need to estimate
#Set Lambda (.2-.4)
lambda<-0.
Stat 465/865 Assignment 3
Solutions
J. Pohlkamp-Hartt
March 11, 2015
1
Textbook Problems [79]
6th Edition in brackets
1. Montgomery Ch. 6 P. 21 (6.15)[5,5]
a) Yes, the process is under control although a cyclic pattern does exist in the x chart.
1
b) Stro
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(\Drc A‘Oq ha ’7 'Dwtﬁmx {Gram “ah/U:
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+73 “’3; 333 3 W “‘3
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2)
-Cavx (‘10 Sam gnaC/Srmys Lz) ki'xcfbAQdS “gLL 3,0 \‘hijl
¥\~°~ W‘Qrk TARA
* Fer lawman; M3443 ; OWL: *0?)
(9a5\r w ML/gjc),% m can at was CM maids.
,Cw @mmma gm
Stat 465/865 Assignment 1
Solutions
J. Pohlkamp-Hartt
February 3, 2015
1
Textbook Problems
6th Edition in brackets
1. Montgomery Ch. 1 P. 23 (1.23)
P cfw_Single Meal Good = .99910 = .99
P cfw_All Meals Good = .994 = .9608[1]
P cfw_All Visits Good = .96081
208
24. FURTHER PROPERTIES OF THE POISSON PROCESS
Superposition of Poisson Processes:
Suppose that cfw_N1(t) : t 0 and cfw_N2(t) : t 0 are two independent Poisson processes with rates 1 and 2, respectively. The sum of
N1(t) and N2(t),
cfw_N (t) = N1(t) +
205
So for n a positive integer, the Gamma(n,) density can be written
as
fX (x|n, ) =
n
n1 x
e
(n1)! x
0
for x 0
.
for x < 0
An important special case of the Gamma(, ) distribution is the
Exponential() distribution, which is obtained by setting = 1. Getti
211
Thinning a Poisson Process:
Let cfw_N (t) : t 0 be a Poisson process with rate . Suppose we mark
each event with probability p, independently from event to event, and
let cfw_N1(t) : t 0 be the process which counts the marked events.
We can use Deniti
198
23. PROPERTIES OF THE POISSON PROCESS
Interarrival Times of the Poisson Process:
We can think of the Poisson process as a counting process with a given
interarrival distribution That is, N (t) is the number of events that have
occurred up to time t, w
21
The Exponential Distribution
From Discrete-Time to Continuous-Time:
In Chapter 6 of the text we will be considering Markov processes in continuous time. In a sense, we already have a very good understanding of
continuous-time Markov chains based on our
194
22. THE POISSON PROCESS: INTRODUCTION
Denition 2 of a Poisson Process:
A continuous-time stochastic process cfw_N (t) : t 0 is a Poisson
process with rate > 0 if
(i) N (0) = 0.
(ii) It has stationary and independent increments.
(iii) P (N (h) = 1) = h
188
22. THE POISSON PROCESS: INTRODUCTION
Stationary and Independent Increments:
We rst dene the notions of stationary increments and independent
increments. For a continuous-time stochastic process cfw_X(t) : t 0,
an increment is the dierence in the proc
191
For a given t of the form nh, we know the exact distribution of B(t).
Up to time t there are n independent trials, each with probability h
of success, so B(t) has a Binomial distribution with parameters n and
h. Therefore, the mean number of successes
184
21. THE EXPONENTIAL DISTRIBUTION
2
1 + 2
1
+ P (B in system|you nish before A)
1 + 2
P (B in system) = P (B in system|A nishes before you)
Now P (B in system|you nish before A) = 1 since B will still be
waiting in line when you move to server 2. On th
180
21. THE EXPONENTIAL DISTRIBUTION
To see how this works, imagine that at time 0 we start an alarm clock
which will ring after a time X that is exponentially distributed with
rate . Let us call X the lifetime of the clock. For any t > 0, we
have that
P