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

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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: 0 . 0002 ; 0 . 9998 . 5. From the table in the text, the best we can do is: 0 . 0002 ; 0 . 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 = 0 . 75 ; and P ( C ) = 7 / 16 . 9. You are given the following information: the events A and B are disjoint; P ( A ) = 0 . 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 - 0 . 40 = 0 . 60 . (c) P ( B c ) = 1 - 0 . 25 = 0 . 75 . 10. P ( A or B ) = 0 . 25 + 0 . 45 - 0 . 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 + 0 . 55 = 1 . 15 , which is too large. 1
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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 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 6 8 3 0 3 6 9 12 4 0 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 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. X 2 X 1 1 2 3 1 0.04 0.06 0.10 2 0.06 0.09 0.15 3 0.10 0.15 0.25 The values of X = X 1 + X 2 for these nine cells were given above in the solution to ex- ample 1. Using these two tables and adding probabilities, one gets the following distri- bution for X . x : 2 3 4 5 6 P ( X = x ) : 0.04 0.12 0.29 0.30 0.25 4. We (well, I) discussed this briefly in lec- ture. I remarked that in Chapter 5 we will learn that based on these data, I should dis- card the assumption of the ELC for my blue round-cornered die.
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