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Unformatted text preview: Genetics 320, Problem Set 8 Solutions 1 (4 points) : Consider the following data obtained by measuring two values ( x, y ) in a number of individuals: x y Frequency 0.10 1 0.30 1 0.45 1 1 0.15 Using this data, compute (a) The mean of x , E [ x ] : E [ x ] = 1 · (0 . 30 + 0 . 15) = 0 . 45 (b) The mean of y , E [ y ] : E [ y ] = 1 · (0 . 45 + 0 . 15) = 0 . 60 (c) E [ x 2 ] and the variance of x , V ar ( x ) : E [ x 2 ] = 1 2 · (0 . 30 + 0 . 15) = 0 . 45 V ar ( x ) = E [ x 2 ] ( E [ x ]) 2 = 0 . 45 . 45 2 = 0 . 2475 (d) E [ y 2 ] and the variance of y , V ar ( y ) : E [ y 2 ] = 1 2 · (0 . 45 + 0 . 15) = 0 . 60 V ar ( y ) = E [ y 2 ] ( E [ y ]) 2 = 0 . 6 . 6 2 = 0 . 24 (e) E [ x · y ] and the covariance of x and y , Cov ( x, y ) : E [ x · y ] = 1 · 1 · (0 . 15) = 0 . 15 Cov ( x, y ) = E [ x · y ] ( E [ x ])( E [ y ]) = 0 . 15 . 45 · . 60 = . 12 (f) The correlation of x and y , p ( x, y ) p ( x, y ) = Cov ( x, y ) p V ar ( x ) · V ar ( y ) = . 12 √ . 2475 · . 24 = . 492...
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This note was uploaded on 12/09/2009 for the course ECOL UNIVERSITY taught by Professor Walsh during the Spring '09 term at University of Arizona Tucson.
 Spring '09
 walsh

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