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Unformatted text preview: Statistics 403 Midterm Exam October 26th, 2010 1. Answer the questions given below for the following probability distribution (your an swers may depend on a and p ): x a a P ( X = x ) p/ 2 1 p p/ 2 (a) What is the expected value EX of this distribution? Solution: The expected value is a p/ 2 + 0 (1 p ) + a p/ 2 = 0 . (b) What is the variance var( X ) of this distribution? Solution: var( X ) = E ( X EX ) 2 = a 2 p/ 2 + 0 2 (1 p ) + a 2 p/ 2 = a 2 p (c) What value of p maximizes the variance of this distribution? Solution: p = 1 (d) What is P ( X 0) for this distribution? Solution: P ( X 0) = P ( X = a ) + P ( X = 0) = p/ 2 + 1 p = 1 p/ 2 (e) What value of p maximizes P ( X 0) for this distribution? Solution: p = 0 1 (f) What is E [1 / (1 + X 2 )] for this distribution? Solution: ( p/ 2) / (1 + a 2 ) + 1 p + ( p/ 2) / (1 + a 2 ) = 1 p + p/ (1 + a 2 ) (g) Suppose the value of a is restricted so a 1. What value of a maximizes the value of E [1 / (1 + X 2 )] for this distribution? Solution: a = 1 2 2. Suppose we are comparing two samples of data, X 1 ,...,X n , and Y 1 ,...,Y 2 n (note that one sample is twice as large as the other). Our goal is to learn about the relationship between their population means EX and EY . The two populations have the same variance, which we denote 2 ....
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This note was uploaded on 02/06/2012 for the course STAT 403 taught by Professor Kerbyshedden during the Winter '12 term at University of MichiganDearborn.
 Winter '12
 KerbyShedden
 Statistics, Probability

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