Download Presentation Tuesday January 15

96.37% of statistics are made up.
Professor: David Sumner
Office: LeConte 400H
Phone: 777-7529
Email: [email protected]
Office Hours: Tuesday - Wednesday - 2:00 - 3:30PM
And by Appointment. I also respond promptly to email.
Textbook: Mathematical Statistics, Wackerly, et al 7th Edition
Room: LeConte 310 – Tu - Th 11:00AM - 12:15PM
WEB Site
A great deal of material will be available on the class web
site at:
http://www.math.sc.edu/~sumner/math511spring2013
This will include problem sets, problem set solutions,
review materials, practice exams, quiz solutions,
handouts, software and much more.
Problem Sets and other assignments will be posted on the
WEB site by 6:00PM each Tuesday and Thursday.
Grading
Three exams 100 points each
300 points
Twelve quizzes 10 points each
100 points
Final Exam
150 points
90 - 100 A
88 - 89
B+
80 - 87
B
78 - 79
C+
70 - 77
C
68 - 69
D+
60 - 67
D
Total: 550 points
Homework will be assigned but
not collected and not graded.
Exam Schedule (tentative)
Exam 1: Thursday, February 7
Exam 2: Thursday, March 7
Exam 3: Thursday, April 11
Final Exam: Friday, May 3 9:00AM
These dates (other than the final exam) are tentative. But
you will have a week’s notice if they should change.
What is
probability?
When the weather anchor says there is a 30%
chance of rain, what does that mean?
If we roll a pair of dice what is the
probability of getting a sum of 7?
Possible Sums are:
2 3 4 5 6 7 8 9 10 11 12
That’s 11 possible sums.
So, is the probability of getting 7 equal to
?
There are actually 36 equally likely
things that can happen when tossing
a pair of dice.
Which of these yield
a sum of seven?
There are 6 ways to get a sum of 7.
The probability of a sum of 7 is
So suppose that S is the set of all outcomes of some
experiment. We refer to S as the sample space of
outcomes. If S is finite or countable, then it is said to be
a discrete sample space.
Suppose that S is finite and that all the outcomes in S are
equally likely to occur.
An event in this context simply refers to a subset A of S
i.e., to a set of possible outcomes.
We usually describe an event in ordinary English.
So suppose that S is the set of all outcomes of rolling a
pair of fair dice. Then there are 36 possible equally
likely outcomes that we would typically consider.
S = {(1, 1), (1, 2), ..., (3, 4), ..., (5, 5),..., (6,6) }
Subsets of the sample space are called events.
Event: Roll doubles –
{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
1
6
The probability of this event is
=
36
6
Event: Roll a sum of 10 –
{(4, 6), (5, 5), (6, 4)}
1
3
The probability of this event is
=
36 12
Event: Sum is 8 but no doubles –
{(2, 6), (3, 5), (5, 3), (6, 2)}
1
4
The probability of this event is
=
36
9
So, although there is still much to say about sample spaces
of experiments in which not all the sample points are
equally likely, it is evident that many problems dealing
with discrete probability can be settled by basic counting
(enumeration).
If S is a sample space of an experiment and all the
elements of S are equally likely to occur, then for any
event A in the sample space, the probability P(A) of the
event is
Number of ways A
can occur
A
P(A) =
Number of equally
S
likely outcomes
Suppose that we toss three coins. What is the probability
that exactly two heads?
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Each of these 8 outcomes is as likely as any of the others.
Three of these outcomes consist of two heads and one tail.
The probability is 3/8.
More generally, we may have sample spaces S where the
outcomes are not equally likely. In such a case we will
endeavor to find a way to assign probabilities properly to
the events of the sample space.
So if S is a finite sample space, a function P defined on the
events of S is a probability function on S if
(i). 0 ≤ P(A) ≤ 1 for each event A.
(ii). P(S) = 1
(iii). If A and B are disjoint events, then
P ( A ∪ B ) = P(A) + P(B)
(iii). If S is countably infinite and A1 , A2 , A3 ,… An ,…
⎛ ∞ ⎞ ∞
are disjoint events in S, then P ⎜  Ai ⎟ = ∑ P ( Ai )
⎝ i=1 ⎠ i=1
Disjoint events are said to be mutually exclusive.
So if A and B are mutually exclusive then
P ( A ∩ B ) = 0, and P ( A ∪ B ) = P ( A ) + P ( B )
For now we will concentrate on discrete probability
where the sample space S is finite (or occasionally
countable).
In order to do this effectively, we need to learn how to
count well.
Set Notation
Suppose that A and B are both subsets of the
Sample Space set S. We express this by,
A ⊆ S and B ⊆ S
A ∩ B = { x : x ∈A and x ∈B}
The intersection of A and B.
A ∪ B = { x : x ∈A or x ∈B}
The union of A and B.
Key conjunctions!
Set Notation
Suppose that A and B are both subsets of the
Sample Space set S. We express this by,
A ⊆ S and B ⊆ S
A − B = { x ∈A and x ∉B} = A ∩ B
The difference A minus B.
A = {x ∈S : x ∉A} = S − A
The complement of A.
“Not A”
Also, A = A′ and A − B = A − A ∩ B
S = {a, b, c, d, e, f , g}
A = {a, c, f } , B = {a, c, d}
A ∩ B = {a, c}
A ∪ B = {a, c, d, f }
A− B ={f}
A = {b, d, e, g}
A =3
A∪ B = 4
A Fundamental Law of Counting
One of the more effective ways to
determine the number of objects of some
type is to devise a procedure that will
produce each object exactly once and
then count the number of ways to carry
out that procedure.
A × B = {(a,b) : a ∈A, b ∈B}
The Cartesian Product of A and B
{a, 1, 2} × {a, b} =
{(a, a), (a, b), (1, a), (1, b), (2, a), (2, b)}
A× B = A ⋅ B
Rule of Product
If one action can be done in n ways, and
then after that a second action can be done in
m ways, then the total number of ways to
perform the first followed by the second is
the product nm.
This example will be useful someday when you’re
in an institution different from this one.
How many license plates are there that consist of
3 letters followed by 3 digits?
26 x 26 x 26 x 10 x 10 x 10
This example will be useful someday when you’re
in an institution different from this one.
What if the letters and digits must
all be different?
How many license plates are there that consist of
3 letters followed by 3 digits?
26 x 25 x 24 x 10 x 9 x 8