HW 1 Solutions

HW 1 Solutions - HW 1 Solutions 1.11 a. b. c. c = c + c + +...

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HW 1 Solutions 1.11 a. = n i c 1 = c + c + … + c = nc . b. i n i y c = 1 = c ( y 1 + … + y n ) = = n i i y c 1 c. ( 29 = + n i i i y x 1 = x 1 + y 1 + x 2 + y 2 + … + x n + y n = ( x 1 + x 2 + … + x n ) + ( y 1 + y 2 + … + y n ) Using the above, the numerator of s 2 is = - n i i y y 1 2 ) ( = = + - n i i i y y y y 1 2 2 ) 2 ( = = - n i i y 1 2 = + n i i y n y y 1 2 2 Since = = n i i y y n 1 , we have = - n i i y y 1 2 ) ( = = - n i i y n y 1 2 2 . Let = = n i i y n y 1 1 to get the result. 2.16 a. 1/3 b. 1/3 + 1/15 = 6/15 c. 1/3 + 1/16 = 19/48 d. 49/240 2.76 Define the events: U : job is unsatisfactory A : plumber A does the job a. P ( U | A ) = P ( A U )/ P ( A ) = P ( A | U ) P ( U )/ P ( A ) = .5*.1/.4 = 0.125 b. From part a. above, 1 – P ( U | A ) = 0.875. 2.90 a. (1/50)(1/50) = 0.0004. b. P (at least one injury) = 1 – P (no injuries in 50 jumps) = 1 = (49/50) 50 = 0.636. Your friend is not correct. 2.133
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This note was uploaded on 04/21/2009 for the course STAT 120B taught by Professor Bennett during the Spring '09 term at UCSB.

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HW 1 Solutions - HW 1 Solutions 1.11 a. b. c. c = c + c + +...

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