Chapter 7 Random Variables

INTERESTING NEWS:
http://news.yahoo.com/s/ap/20071213/ap_on_re_us/hiv_lawsuit


This test will be on Thursday, December 13, 2007.

You will need to create a chart to remind yourself about the formulas for this chapter until you have practiced enough to know them by heart. A sticky-note at pages 396 and 400 would also be helpful!

Mu = the POPULATION average. This is a parameter.
Sigma squared = the POPULATION variance. This is also a parameter.
Standard deviation is the square root of the variance.

X-bar is the sample average, the unbiased estimator of the population average. It is a statistic.
S-squared is the sample variance, the unbiased estimator of the population variance. It is also a statistic.

HW due Wednesday, December 12th: 7.13, 7.15, 7.17, 7.34, 7.42
HW due Tuesday, December 11th: 7.24, 7.28. Use the formulas and the examples in the text.

HW due Monday, December 10th : 7.2, 7.4, 7.7.

Chapter 6 Probability

The Chapter 6 test will be December 6th.
Previous tests will be returned to students as soon as they are graded.

Prepare for the test. Work problems from each section. Read the chapter and section summaries. Write down what you are doing. Draw the Venn diagram or the tree diagram for complicated situations. Ask questions on the blog. Take the practice test.

Here are some answers to even HW problems:
Problem 6.10 (a) S= {all numbers between 0 and 24}
(b) = {any whole number up to and including 11,000}
(c) S = {0, 1, . . . 12}
(d) S= {any dollar and cents amount up to [insert your maximum guess here]}
(e) S = {any positive or negative number}

Problem 6.12 Four outcomes for two coins: {HH, HT, TH, TT}, eight for three coins: {HHH. HHT, HTH, HTT, THH, THT, TTH, TTT}, and sixteen for four coins (do that one yourself!).

Problem 6.16 (a) YYY -0000 through YYY-9999 = 10,000 numbers. (b) YYY-ZXX-XXXX, each X having 10 possible numbers, except the number can't start with a 0 or a 1 meaning that Z has only 8 possible values, so this means 8 * 10^6 LESS the restricted numbers (911-xxxx, 411-xxxx, etc.)

Problem 6.20 P(moves to another class) = 1 - P(stays) = 1 - .46 = .54.

Problem 6.24 P(wins large battle) = .6, P(wins three small battles) = P(wins individual small battle)^3 = .8^3 = .512. Choose the strategy with the larger probability of occurring.

Problem 6.40
Venn diagram has two circles representing getting job A and getting job B.
Both jobs: intersection of the two circles, the overlapped part, the biscuit.
First but not second: the part of circle A that is not within circle B
Second but not first: the part of circle B that is not within circle A
Neither: the part that is in the background, in NEITHER circle.

Problem 6.44
P(W) = 856/1626
P(W given prof degree) = 30/74
These are not the same. so gender and professional degree are not independent

Problem 6.56
P(y <> x) = 1/8,
P(y > x) = 1/2,
P(y <> x) = P(y <> x) /P(y > x) = 1/4.

Problem 6.48: P(W) * P(Manager given W) = P(Woman AND Manager)

One pattern that shows up a lot is Marginal * Conditional = Joint

If you divide both sides by Marginal you get
Conditional = Joint / Marginal.

IFF means IF AND ONLY IF.

IFF P(A) * P(B) = P(both A&B), A and B are independent.

IFF P(A) = P(A given B), A and B are independent.

HW for Tuesday night: DO problems 6.33 and 6.48. Read 6.66 and be prepared to work the problem. Essential question: How do mathematical independence and our regular understanding of independence relate? The chapter 4 tests were returned today. HW for Monday night: 6.39, .40, .53, and .56. The problem we worked today in class was problem .65. You would be wise to work through this problem and problem .66.

Notes from Friday (11/30) are embedded in the purple sections below.

Conditional probability rules:

PLEASE NOTE THE CORRECTION! BLOGGER WON"T ACCEPT THE VERTICAL LINE SYMBOL!!!
P(A GIVEN B) = P(A and B)/P(B)
P(B GIVEN A) = P(A and B)/P(A)

so of course

P(B) P(A GIVEN B) = P(A and B), the joint probability of A and B. It may be helpful to think of it like cancelling factors in the numerator and denominator of a fraction EXCEPT that the result is the JOINT probability. Be careful.

P(A) P(B GIVEN A) = P(A and B), again, the joint probability of A and B.

These relationships can be represented in two-way tables, Venn diagrams, and tree diagrams. The count within a cell of a two-way table divided by the marginal total is a conditional probability. Likewise, the joint probability for that cell divided by the marginal probability is also the conditional probability.

Tree diagrams can be useful when you are trying to work the problems backwards.

I don't think that I made this clear in class today:
P(A) = P(A and B) + P(A and not B) = P(A)P(B given A) + P(not A) P(B given not A).

HW for the weekend is problems 6.44 and 45.

If A and B are independent, then P(A) = P(A GIVEN B) and P(B) = P(B GIVEN A).
Interpretation: If A and B are independent, then whether or not B happened has no relationship with whether A happened.
Likewise, if A and B are independent, then whether or not A happened has no relationship with whether B happened

Today we used a Venn Diagram, a two-way table, and a tree diagram to represent the outcomes and probabilities associated with throwing two strangely-marked dice. All of the methods yielded the same answer.

HW for Tuesday (11/27) night: Re-work the weird dice problem from the AP exam (the one with two dice, one has only 9s and 0s, the other has 11s and 3s.) This time, instead of using simulation, use formal probability rules and a tree diagram, table, or Venn diagram. Answer the question in complete sentences. For part B, reconcile the answer with the joint probabilities you found in part A. Figure out the guidelines in your own words that tell you whether a price/reward is fair.

Get the reading done! What are the big concepts?
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Sorry for the delay: just got home from KSU.
HW for Monday night, Nov 26: 6.19, 20, 21
plus. . .finish State of Fear.
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Add problems 6.24 and 6.25, due Monday, November 26.

Don't forget to read State of Fear.
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he complete list of HW problems due Tuesday: 6.9, 10, 12, 13, 16 (plus any others you feel like doing).

Don't forget to read State of Fear.

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Events that are mutually exclusive ARE NOT independent.

Addition principle: P(A or B) = P(A) + P(B) - P(A and B)

Multiplication principle: P(A) P(BA) = P(A and B), the joint probability.

When B and A are independent, P(BA) = P(B), A happening or not has not relationship to B happening, so P(A) P(B)=P(A and B). THAT IS ONLY WHEN THE EVENTS ARE INDEPENDENT.

Key vocabulary
parameter
sample space
event
probability
joint probability
independent
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This won't be so bad. The test will be Thursday, Dec. 6.

What are YOU doing to maximize your understanding of the material?

• Are you creating an outline of the chapter?
• Have you developed a glossary for the vocabulary and formulas?
• Have you worked all of the homework problems when assigned?
• Do you read the sections that relate to the homework?
• Are you part of a study group?
• Do you ask questions?
• Have you worked problems from a study guide?
• Have you worked the online quiz (see the link on the right panel of this blog)?

Do you try to see the big picture?

• Do you look for the similarities and differences in the ways data are processed?
• Do you work with problems long enough to understand why the formulas work the way they do?
• Have you made connections between current concepts and prior knowledge?
• Have you gone online to review concepts that you have forgotten?

Just "going through the motions" does not lead to the success that you desire in Advanced Placement courses. Take control of your learning.

Be safe.