1. Suppose that the three inspectors at a lm factory are supposed
to stamp the expiration data on each package of lm at the end of
assembly line. John , who stamps 25% of the packages, fails to stamp
the expiration date once in every 200 packages, Tom , w
3.5 Expected Values
3.5.9. Recall Question 3.4.4, where the length of time Y
(in years) that a malaria patient spends in remission has
pdf f Y (y) = 19 y 2 , 0 y 3. What is the average length of
time that such a patient spends in remission?
3.5.10. Let th
142 Chapter 3 Random Variables
Now, let j = k 1. It follows that
E(X ) = np
Finally, letting m = n 1 gives
#
n1 "
!
n1
j=0
E(X ) = np
j
m " #
!
m
j
j=0
p j (1 p)n j1
p j (1 p)m j
and, since the value of the sum is 1 (why?),
(3.5.3)
E(X ) = np
!
Comment Th
144 Chapter 3 Random Variables
negative for the disease, all fifty individuals must necessarily be free of the infection,
and no further testing would need to be done. If the pooled sample gave a positive
reading, of course, all fifty B samples would have
140 Chapter 3 Random Variables
and
p X (1) = P(X = 1) =
20 10
=
38 19
9
Then 19
of the time we will win $1 and 10
of the time we will lose $1. Intuitively,
19
then, if we persist in this foolishness, we stand to lose, on the average, a little more
than 5
3.5 Expected Values
143
r
Comment Let p represent the proportion of red balls in an urnthat is, p = r +w
.
The formula, then, for the expected value of a hypergeometric random variable
has the same structure as the formula for the expected value of a bino
146 Chapter 3 Random Variables
!
Let w = y/, so that dw = 1/ dy. Then E(Y ) = 0 wew dw. Setting u = w and
dv = ew dw and integrating by parts gives
"
(3.5.4)
E(Y ) = [wew ew ]" =
0
Equation 3.5.4 shows that is aptly namedit does, in fact, represent the a
148 Chapter 3 Random Variables
More specifically, the median lifetime for these bulbsaccording to Definition 3.5.2is the value m for which
! m
0.001e0.001y dy = 0.5
But
"m
0
0
0.001e
0.001y
dy = 1 e
0.001m
. Setting the latter equal to 0.5 implies that
m
Chapter
Special Distributions
4.1
4.2
4.3
4.4
4.5
4.6
Introduction
The Poisson Distribution
The Normal Distribution
The Geometric Distribution
The Negative Binomial Distribution
The Gamma Distribution
4
Taking a Second Look at Statistics
(Monte Carlo Simu
4.2 The Poisson Distribution
223
Proof We begin by rewriting the binomial probability in terms of :
lim
n
!n "
k
k
nk
p (1 p)
! n " # $k #
$
nk
1
= lim
n k
n
n
# $#
$ #
$
1
n!
k
n
k
= lim
1
1
n k!(n k)!
nk
n
n
#
$n
k
1
n!
=
1
lim
k
k! n (n k)! (n )
n
B
4.2 The Poisson Distribution
Calculate both a binomial answer and a Poisson answer. Is
the binomial model exact in this case? Explain.
4.2.6. A newly formed life insurance company has underwritten term policies on 120 women between the ages of
forty and f
228 Chapter 4 Special Distributions
Fitting the Poisson Distribution to Data
Poisson data invariably refer to the numbers of times a certain event occurs during
each of a series of units (often time or space). For example, X might be the weekly
number of
4.2 The Poisson Distribution
229
To see whether a probability function of the form p X (k) = e k /k! can
adequately model the observed proportions in the third column, we first need
to replace with the samples average value for X . Suppose the six observa
4.2 The Poisson Distribution
225
Case Study 4.2.1
Leukemia is a rare form of cancer whose cause and mode of transmission
remain largely unknown. While evidence abounds that excessive exposure to
radiation can increase a persons risk of contracting the dis
226 Chapter 4 Special Distributions
(Case Study 4.2.1 continued)
P(X 8) = 1 P(X 7)
=
1
7
!
e1.75 (1.75)k
k=0
k!
= 1 0.99953
= 0.00047
How close can we expect 0.00047 to be to the true binomial sum? Very
close. Considering the accuracy of the Poisson limi
222 Chapter 4 Special Distributions
by the same pdf). That said, it makes sense to single out these real-world pdfs and
investigate their properties in more detail. This, of course, is not an idea we are seeing for the first timerecall the attention given
Math 340
Piryatinska
minutes)
Name:
Fall 2013
Midterm (90
KEY
SHOW ALL WORK
1)An investor is monitoring stock from company A and company B, where each either increases
or decreases each day. On a given day, suppose that there is a probability of 0.38 that
3.5 Expected Values
Figure 3.5.4
147
Resultant
Wave 1
Wave 2
A Second Measure of Central Tendency: The Median
While the expected value is the most frequently used measure of a random variables central tendency, it does have a weakness that sometimes makes
3.5 Expected Values
145
Therefore,
E(X ) =
!
all k
2k p X (2k ) =
!
k=1
2k
1
=1+1+1+
2k
which is a divergent sum. That is, X does not have a finite expected value, so in order
for this game to be fair, our ante would have to be an infinite amount of mone
Math 340
Piryatinska
Spring 2011
Practice test, Midterm
1. Given that P (A) = P (B ) = 1 , and P (A B ) = 0.1, nd:
3
P (A B ), P (A B ), and P (A B ).
2. Players A, B, and C toss a fair coin in order. The rst to throw a head
wins. What is the probability
MATH 340
Syllabus
Spring 2011
Instructor: Dr. Alexandra Piryatinska
Office: TH 942
Office hours: T 1.00-2.00 pm, TH 8.30-9.30 or by appointment
Email: [email protected]
Class: TTH 9:35-10:50, TH 409
Prerequisites: MATH 228 with a grade of C or better (may
MATH 340
Syllabus
Spring 2012
Instructor: Dr. Alexandra Piryatinska
Office: TH 942
Office hours: T, TH 9.50-10.50 or by appointment
Email: [email protected]
Class: TTH 11:00-12:15, T-R 4
Prerequisites: MATH 228 with a grade of C or better (may be taken co