sollectexamp7 - Solutions to Chapter 1 Lecture Examples 1....

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to Chapter 1 Lecture Examples 1. 0.0344; 0.9826; 0.5359, respectively. 2. 0.7939; 0.1170; 0.1788, respectively. 3. Round z to 1.42, which gives 0.0778. 4. From the table in the text, the best we can do is: ≤ . 0002 ; ≥ . 9998 . 5. From the table in the text, the best we can do is: ≤ . 0002 ; ≥ . 9998 . 6. Class discussion example, so there is no ‘solution.’ 7. A CM has a sample space that consists of four elements, denoted: a, b, c and d. As- suming the ELC, find the probabilities of each of the following events. (a) P ( A ) = 0 . 25 (b) P ( B ) = 0 . 50 (c) P ( C ) = 0 . 75 8. Refer to the previous problem. Now, in- stead of the ELC, assume that the probabil- ities of a, b, c and d follow the ratio 9:3:3:1. (a) Call the probabilities 9 r , 3 r , 3 r and r . These must sum to one: 16 r = 1 , hence r = 1 / 16 . Thus, the probabil- ities of a, b, c and d are 9 / 16 , 3 / 16 , 3 / 16 and 1 / 16 . (b) P ( A ) = 9 / 16 ; P ( B ) = 12 / 16 = . 75 ; and P ( C ) = 7 / 16 . 9. You are given the following information: the events A and B are disjoint; P ( A ) = . 40 ; and P ( B ) = 0 . 25 . Calculate the fol- lowing probabilities. (a) P ( A or B ) = 0 . 40 + 0 . 25 = 0 . 65 . (b) P ( A c ) = 1- . 40 = 0 . 60 . (c) P ( B c ) = 1- . 25 = 0 . 75 . 10. P ( A or B ) = 0 . 25 + 0 . 45- . 20 = 0 . 50 . 11. What is wrong with each of the following? (a) P ( A ) = 0 . 20 ; P ( B ) = 0 . 55 ; and P ( AB ) = 0 . 25 : This violates Rule 5; AB is a subset of A , so its probability cannot be larger. (b) P ( A ) = 0 . 60 ; P ( B ) = 0 . 55 ; and A and B are disjoint: This violates Rule 2. By Rule 3, P ( A or B ) = 0 . 60 + . 55 = 1 . 15 , which is too large. 1 Solutions to Chapter 2 Lecture Examples 1. In the table below, in each cell I have placed the value of X = X 1 + X 2 . For example, in the center cell, corresponding to X 1 = 2 and X 2 = 2 , I have placed 2 + 2 = 4 . X 2 X 1 1 2 3 1 2 3 4 2 3 4 5 3 4 5 6 The nine cells are equally likely, so prob- abilities are obtained by simply counting, yielding the following table. x : 2 3 4 5 6 P ( X = x ) : 1 / 9 2 / 9 3 / 9 2 / 9 1 / 9 2. In the table below, in each cell I have placed the value of X = X 1 X 2 . For example, in the cell corresponding to X 1 = 2 and X 2 = 3 , I have placed 2 × 3 = 6 . X 2 X 1 1 2 3 4 1 1 2 3 4 2 2 4 6 8 3 3 6 9 12 4 4 8 12 16 The 25 cells are equally likely, so probabili- ties are obtained by simply counting, yield- ing the following table. x P ( X = x ) x P ( X = x ) 0.36 6 0.08 1 0.04 8 0.08 2 0.08 9 0.04 3 0.08 12 0.08 4 0.12 16 0.04 3. B/c the outcomes of each trial are not equally likely, the cells in the table are not equally likely. Instead, we obtain the prob- ability of each cell by using the multiplica- tion rule. For example, P ( X 1 = 2 ,X 2 = 3) = 0 . 3(0 . 5) = 0 . 15 . These probabilities are presented below....
View Full Document

This note was uploaded on 10/23/2009 for the course STAT STATS 371 taught by Professor Professorwardrop during the Fall '09 term at Wisconsin.

Page1 / 13

sollectexamp7 - Solutions to Chapter 1 Lecture Examples 1....

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online