Week 15 Mon Dec 6

# Week 15 Mon Dec 6 - 1 Posted after class WEEK 15 Monday Dec...

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Unformatted text preview: 1 Posted after class. WEEK 15, Monday, Dec 6 Today I hope to be able to review some of the material in the course that derives from Chapters 2 and 3. On Wednesday, we will review some of the material in the course that derives from Chapters 4 (normal distributions), 5 (sampling distributions), 6 (confidence intervals), and 7 (hypothesis testing). Chapter 1 (descriptive statistics) is very basic; please ask questions on it if you have them. CHAPTER 2 . Sections 2.1 – 2.4 . Example 1. Figure showing sample space with 11 outcomes. If the model specifies them as equally likely, then we know each has probability 1/11. The names of outcomes not shown in figure below; they are simply depicted by dots. . . . . . . . . . . . S Event . Given a probability model with sample space S , any subset of the sample space is called an event . We commonly use upper case letters A , B , C , … to denote events. The probability of an event is the sum of the probabilities of the outcomes that are in the event. Using mathematical notation for a sample space denoted by S = { o 1 , o 2 ,…, o N }, we can write Since the total probability of all outcomes is 1, we can write P( S ) = 1. 2 Example 2. Figure showing sample space with 11 outcomes and an event A . Suppose that the model specifies them as equally likely. Then P( A ) = 4/11. . . . . . . A . . . . . S For model with equally likely outcomes (2-1) Example 3. Model for independent tosses of three fair coins (say a nickel, a dime and a quarter). In the table below, the first letter denotes the outcome of the nickel, the second letter denotes the outcome of the dime, and the third letter denotes the outcome of the quarter. There are 2 3 = 8 possible outcomes for a toss of the three coins. If the coins are fair, these are equally likely outcomes. N D Q S Outcome TTT TTH THT HTT THH HTH HHT HHH Prob 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 Let A be the event “outcome has at least two H’s in it.” Then P(A) = P(THH) + P(HTH) + P(HHT) + P(HHH) = 1/8 + 1/8 + 1/8 + 1/8 = 4/8. Let B be the event “outcome has exactly two H’s in it.” Then P(B) = P(THH) + P(HTH) + P(HHT) = 1/8 + 1/8 + 1/8 = 3/8. Complement. If A is an event, then so is its complement which we denote by . consists of all outcomes not in A . It follows that (2-3) Note that and that . Intersection. If A and B are events then their intersection is an event. It consists of all outcomes that are in both A and B . For modeling we take as a basic rule 3 (2-8) Spoken as “probability of B given A ” Spoken as “probability of A given B ” and are called conditional probabilities . Reasoning and the situation being modeled often lead one to appropriate values for conditional probabilities. We will use the rules (2-8) many times. We call these Multiplication Rules for the Probability of an Intersection ....
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Week 15 Mon Dec 6 - 1 Posted after class WEEK 15 Monday Dec...

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