document.write( "Question 898677: according to j.d. power and associates 2006 initial quality study , consumers reported on average 1.7 problems per vehicle with new 2006 volkswagens. in a randomly new volkswagen, find the probability of:
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document.write( "a) at least one problem;
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document.write( "b) no problem;
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document.write( "c) more than three problems \n" );
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Algebra.Com's Answer #544954 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! Poison Distribution: avg = 1.7 (stattrek.com Calculator) \n" ); document.write( "a) at least one problem: .8173 0r 81.73%% 0r 1 - P(0) = 1-.1827 = .8173 \n" ); document.write( " b) no problem: .1827 0r 18.27% \n" ); document.write( " c) more than three problems: .0932 0r 9.32% \n" ); document.write( "OR \n" ); document.write( " P(x>3) = Ti: 1- poissioncdf(1.7, 3) \n" ); document.write( " |