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exam2_solution - Exam II STAT 211 Fall 2010 NAME: STUDENT...

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Unformatted text preview: Exam II STAT 211 Fall 2010 NAME: STUDENT ID: You have 75 minutes to complete this test. You may use two one-page cheat sheets and a calculator. If I provide partial resultsassume they are correct and use them even if they are not. If there is no correct answer or if multiple answers are correct, select the best answer. Mark your answer on the scantron, and turn in both the scantron and this test. This test consists of 20 questions. Make sure you have all questions and Good luck! Useful information: The probabilities for a normal random variable to fall within 1, 2, 3 standard deviations are respectively 68 % , 95 . 45 % , 99 . 7 %. Additionally, we have Z . 005 = 2 . 5758 Z . 01 = 2 . 3263 Z . 025 = 1 . 96 Z . 05 = 1 . 645 Z . 1 = 1 . 282 . 1 1. In a family day care center, 3 girls and 4 boys are registered. The joint probability model for the number of girls ( X ) and boys ( Y ) to show up on any given day is given below. X 1 2 3 0.01 0.03 0.02 0.02 1 0.02 0.02 0.04 0.05 Y 2 0.03 0.02 0.05 0.42 3 0.01 0.01 0.02 0.05 4 0.01 0.02 0.01 K The missing value K in the table is: A. K = 0 . 14 B. K = 0 . 46 C. K = 0 . 96 D. K = 0 . 01 E. none of the above. 2. In the above problem, is the girls attendence and boys attendence independent of each other and why? A. Yes, because the boys and girls do not have any agreement to attend. B. No, because the probability P (no boy and no girl at the center) is not the same as the product of the probability P (no boy at the center) and the probability P (no girl at the center)....
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This note was uploaded on 09/24/2011 for the course STAT 211 taught by Professor Parzen during the Spring '07 term at Texas A&M.

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exam2_solution - Exam II STAT 211 Fall 2010 NAME: STUDENT...

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