midterm1+fall+2011_solutions

# midterm1+fall+2011_solutions - ECI 114 Fall 2011 Name...

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Fall 2011 Name Solutions Midterm 1 (100 points) Wednesday, October 19, 2011 Show your work. It’s not necessary to give a final numerical answer unless specifically asked, but you need to provide all the information necessary to compute the final answer. Please circle or otherwise highlight your answer. Turn in your “crib sheet” with your exam. GOOD LUCK! 1. For each of the following statements, circle the letter “T” if it is true, and “F” if it is false. (3 * 9 =27 pts) The correct answer has been bolded. T F a. S = {… -2.0, -1.6, -1.2, -0.8, -0.4, 0.0, 0.4, 0.8, 1.2, 1.6, 2.0, …} is an example of a continuous sample space. There are still only a countably infinite number of points in the sample space. T F b. For any event A, (A S) (A C S) = S, where S represents the entire sample space. A S=A, A C S= A C , S=A A C by definition of complement. T F c. Pr[A|B c ] = 1 – Pr[A|B] . For it to be true, the complement should be on the A: Pr[A c |B] = 1 – Pr[A|B]. T F d. If two events form a partition, then they are always independent. Because they are M.E., and M.E. events are dependent – knowing one occurs affects the probability of the other occurring (you know it cannot occur, since they are M.E.). T F e. The mean and median are identical for symmetric distributions. T F f. If Pr[A] + Pr[B] + Pr[C] = 1, then the set of events {A, B, C} forms a partition. The converse of this is true. But suppose, e.g., B=C, and Pr[A] = Pr[B] = Pr[C] = 1/3. Then the sum = 1, but {A, B, C} is not a partition since B and C are not M.E. (nor are {A, B, C} C.E.). You could also devise an example in which B and C overlap but are not equal. T F g. 1 7 7 6 n n n -   × = ×  ÷ ÷   LHS: 7 ! ! 7!( 7)! 6!( 7)! n n n n × = - - , RHS: ( 1)! ! 6!(

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## This note was uploaded on 01/19/2012 for the course ECI 114 taught by Professor Mocktarian during the Fall '08 term at UC Davis.

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midterm1+fall+2011_solutions - ECI 114 Fall 2011 Name...

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