Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Working with Sets - Lecture Notes | STA 100, Study notes of Data Analysis & Statistical Methods

Material Type: Notes; Professor: Thistleton; Class: Statistical Methods; Subject: Statistics; University: SUNY Institute of Technology at Utica-Rome; Term: Unknown 2008;

Typology: Study notes

Pre 2010

Uploaded on 08/09/2009

koofers-user-ics
koofers-user-ics ๐Ÿ‡บ๐Ÿ‡ธ

10 documents

1 / 8

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Prof. Thistleton
STA100 Statistical Methods
Lecture 6
Text Sections: Chapter 4 Sections 2
Working With Sets
Recall the table from Lecture 5.
Gender
Male
Female
Owns a Pet
Yes
15
27
No
6
3
We can think about this in a couple of ways. There are 51 students in this sample drawn from
the population of SUNYIT students. With each of these students we have associated two
variables: Gender and Owns a Pet. Both of these variables are categorical, as we discussed in the
first lecture. You could also say that each is dichotomous. All we mean by that is that each one
can occur in two ways: Yes or No. Later weโ€™ll call that Success or Failure.
We could also think a little differently with each individual having membership in Male or
Female and in Owns a Pet or Does Not Own A Pet. You might recall from earlier math courses
that we can begin to talk about membership in a set. A picture, called a Venn Diagram, often
helps.
Before proceeding further we need to develop a few ideas about collections of objects, or sets.
Male Owns
a Pet
pf3
pf4
pf5
pf8

Partial preview of the text

Download Working with Sets - Lecture Notes | STA 100 and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity!

Prof. Thistleton STA100 Statistical Methods Lecture 6

Text Sections: Chapter 4 Sections 2

Working With Sets

Recall the table from Lecture 5.

Gender Male Female

Owns a Pet

Yes 15 27 No 6 3

We can think about this in a couple of ways. There are 51 students in this sample drawn from the population of SUNYIT students. With each of these students we have associated two variables: Gender and Owns a Pet. Both of these variables are categorical, as we discussed in the first lecture. You could also say that each is dichotomous. All we mean by that is that each one can occur in two ways: Yes or No. Later weโ€™ll call that Success or Failure.

We could also think a little differently with each individual having membership in Male or Female and in Owns a Pet or Does Not Own A Pet. You might recall from earlier math courses that we can begin to talk about membership in a set. A picture, called a Venn Diagram, often helps.

Before proceeding further we need to develop a few ideas about collections of objects, or sets.

Male

Owns

a Pet

  1. Definition 1 A set is a collection of objects. These objects are called elements of the set. We typically will use an upper case letter, such as A, for a set and a lower case letter, such as a, for an element of a set. We write a โˆˆ A with the Greek letter โ€œlittle epsilonโ€.

Example: If Mike owns a pet we would say that ๐‘€๐‘–๐‘˜๐‘’ โˆˆ ๐‘‚๐‘Š๐‘๐‘† ๐ด ๐‘ƒ๐ธ๐‘‡. Note that we would also say that ๐‘€๐‘–๐‘˜๐‘’ โˆˆ ๐‘€๐ด๐ฟ๐ธ.

  1. Definition 2 A subset, say B, of a set A, is itself a set, every element of which also belongs to the set A. We write ๐ต โІ ๐ด. That is, ๐ต is just a collection of some (and maybe all) of the elements of A.

Example: the New York Yankees 2008 40 man roster http://newyork.yankees.mlb.com/team/roster_40man.jsp?c_id=nyy includes subsets corresponding to pitchers, catchers, infielders, outfielders, and designated hitters. Note that these subsets donโ€™t have any overlap- no one is both a pitcher and an infielder.

3. Definition 3 Two sets, A and B, are said to be equal if each element of A is also in B, and if each element of B is also in A. We write B = A. 4. Definition 4 We will call the set C the union of the sets A and B if every element of C is also in at least one of A or B (though maybe in both). We write ๐ถ = ๐ด ๐‘œ๐‘Ÿ ๐ต. Note that other books will write ๐ถ = ๐ด โˆช ๐ต.

Example: Suppose we toss a fair coin 3 times and observe the faces. What does our sample space look like? We can list these outcomes out in a straightforward way by noting that on one toss the sample space looks like ๐ป, ๐‘‡. (Note: exactly 2 outcomes). Build up from here by noting that if we toss twice then the sample space looks like ๐ป๐ป, ๐ป๐‘‡, ๐‘‡๐ป, ๐‘‡๐‘‡. (Note: exactly 4 outcomes. To help see this, draw a โ€œtree diagramโ€ as follows

outcomes. For the first subset, call it ๐ธ 1 we can list the outcomes with โ€œHeads on Firstโ€. Can you see that ๐ธ 1 = ๐ป๐ป๐ป, ๐ป๐ป๐‘‡, ๐ป๐‘‡๐ป, ๐ป๐‘‡๐‘‡? Now let the second event, call it ๐ธ 2 be those outcomes with โ€œHeads on Secondโ€. We can write this out as ๐ธ 2 = ๐ป๐ป๐ป, ๐ป๐ป๐‘‡, ๐‘‡๐ป๐ป, ๐‘‡๐ป๐‘‡. Hereโ€™s something to think about: The event ๐ธ 1 has 4 outcomes. The event ๐ธ 2 has 4 outcomes. How many does ๐ธ 1 ๐‘œ๐‘Ÿ ๐ธ 2 have? Itโ€™s not 8- just write them out, being careful not to write any outcomes out twice: ๐ธ 1 ๐‘œ๐‘Ÿ ๐ธ 2 = ๐ป๐ป๐ป, ๐ป๐ป๐‘‡, ๐ป๐‘‡๐ป, ๐ป๐‘‡๐‘‡, ๐‘‡๐ป๐ป, ๐‘‡๐ป๐‘‡ This is a terrifically important result for us. To state it best we also need the following.

5. Definition 5 We will call the set C the intersection of the sets A and B if every element of C is also in both of ๐ด ๐‘Ž๐‘›๐‘‘ ๐ต. We write ๐ถ = ๐ด ๐‘Ž๐‘›๐‘‘ ๐ต. Note that other books will write ๐ถ = ๐ด โˆฉ ๐ต.

Example: still working with the above, we have ๐ธ 1 ๐‘Ž๐‘›๐‘‘ ๐ธ 2 = ๐ป๐ป๐ป, ๐ป๐ป๐‘‡. That is, we include all those tosses with Heads on First and also Heads on Second. This allows us to say that โ€œthe number of elements in A or B is equal to the number of elements in A plus the number of elements in B minus the number of elements in both. Using the symbol ๐‘› ๐ด for โ€œthe number of elements in Aโ€ we can write this succinctly as

๐‘› ๐ด ๐‘œ๐‘Ÿ ๐ต = ๐‘› ๐ด + ๐‘› ๐ต โˆ’ ๐‘› ๐ด ๐‘Ž๐‘›๐‘‘ ๐ต

In our coin toss example this gives us

๐‘› ๐ป ๐‘œ๐‘› 1 ๐‘ ๐‘ก ๐‘œ๐‘Ÿ ๐ป ๐‘œ๐‘› 2 ๐‘›๐‘‘ = ๐‘› ๐ป ๐‘œ๐‘› 1 ๐‘ ๐‘ก + ๐‘› ๐ป ๐‘œ๐‘› 2 ๐‘›๐‘‘ โˆ’ ๐‘› ๐ป ๐‘œ๐‘› 1 ๐‘ ๐‘ก ๐‘Ž๐‘›๐‘‘ ๐ป ๐‘œ๐‘› 2 ๐‘›๐‘‘

In our Pets and Gender example this gives us:

Gender Male Female

Owns a Pet

Yes 15 27 42 No 6 3 9 21 30 51

And so we see that ๐‘› ๐‘‚๐‘ค๐‘›๐‘  ๐‘Ž ๐‘ƒ๐‘’๐‘ก = 42 and ๐‘› ๐‘€๐‘Ž๐‘™๐‘’ = 21_. We also have that_ ๐‘› ๐‘€๐‘Ž๐‘™๐‘’ ๐‘Ž๐‘›๐‘‘ ๐‘‚๐‘ค๐‘›๐‘  ๐‘Ž ๐‘ƒ๐‘’๐‘ก = 15 so that, finally,

๐‘› ๐‘‚๐‘ค๐‘›๐‘  ๐‘Ž ๐‘ƒ๐‘’๐‘ก ๐‘œ๐‘Ÿ ๐‘€๐‘Ž๐‘™๐‘’ = 42 + 21 โˆ’ 15 = 48

Bringing this back home to probability (remember Kolmogorovโ€™s third axiom!!)

๐‘ƒ ๐‘‚๐‘ค๐‘›๐‘  ๐‘Ž ๐‘ƒ๐‘’๐‘ก ๐‘œ๐‘Ÿ ๐‘€๐‘Ž๐‘™๐‘’ =^4251 +^2151 โˆ’ 1551 = (^4851)

6. Definition 6 Suppose we have ๐ต โІ ๐‘†. Then we denote the complement of the set B in ๐‘† as the set of elements which are in ๐‘† but not in ๐ต. That is, just those elements not in ๐ต. We write ๐‘›๐‘œ๐‘ก ๐ต.

In particular, consider the set of all outcomes of an experiment. We will call this set the sample space of the experiment and denote it as S. An event is then just a subset of a sample space and we may speak of an event ๐ธ and its complement ๐‘›๐‘œ๐‘ก ๐ธ and their probabilities.

We have a few relationships to help us do our work:

  1. ๐ด ๐‘‚๐‘… ๐‘›๐‘œ๐‘ก ๐ด = ๐‘†. That is, an event and its complement together make up the sample space.
  2. ๐‘›๐‘œ๐‘ก ๐ด ๐ด๐‘๐ท ๐ต = ๐‘›๐‘œ๐‘ก ๐ด ๐‘‚๐‘… ๐‘๐‘œ๐‘ก ๐ต. That is, the complement of the intersection is the union of the complements.
  3. ๐‘›๐‘œ๐‘ก ๐ด ๐‘‚๐‘… ๐ต = ๐‘›๐‘œ๐‘ก ๐ด ๐ด๐‘๐ท ๐‘›๐‘œ๐‘ก ๐ต. That is, the complement of the union is the intersection of the complements.

๐‘๐‘Ÿ๐‘œ๐‘ ๐ธ ๐‘‚๐‘… ๐น = ๐‘๐‘Ÿ๐‘œ๐‘ ๐ธ + ๐‘๐‘Ÿ๐‘œ๐‘ ๐น โˆ’ ๐‘๐‘Ÿ๐‘œ๐‘ ๐ธ ๐ด๐‘๐ท ๐น and so ๐‘๐‘Ÿ๐‘œ๐‘ ๐ธ ๐‘‚๐‘… ๐น = 0.5 + 0.2 โˆ’ 0.15 = 0.

  1. Calculate the probability that you do not stop. This is just the complement of the above and so has probability 1 โˆ’ 0.55 = 0.
  2. Calculate the probability that you stop exactly once. To answer this we can revisit our diagram and embellish it a little. Since ๐ธ has two parts which together give us a probability of 0.5, and since the part taken up by the intersection is of size 0.15, that leaves us 0.35 left over for the event โ€œstop at first but not at secondโ€.

Similarly, the probability that we stop at the second but not the first is 0.

- 0.15 = 0.05. So, the probability we stop just once is 0.35 + 0.05 = 0.4.

0.35 0.1^5

0.

0.4 5

Example: You will roll a fair die three times.

  1. How many possible sequences (outcomes) are there?
  2. What is the probability of that you will obtain 0 twos?
  3. What is the probability of that you will obtain 1 two?
  4. What is the probability of that you will obtain 2 twos?
  5. What is the probability of that you will obtain 3 twos?
  6. What is the probability of that, if you add the face values, you will obtain a sum of 17 or more?