document.write( "Question 929439: According to the data published by the U.S. Dept. of Agriculture, the average annual amount of chicken eaten by American adults (excluding vegetarians) is normally distributed with a mean of 55 pounds and a standard deviation of 9.2 pounds.\r
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document.write( "If a randomly selected American adult (excluding vegetarians) is selected, what is the probability that this adult eats at least 52 pounds of chicken per year.
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document.write( " A .13
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document.write( " B .26
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document.write( " C .37
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document.write( " D .63
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document.write( " E .73 \n" );
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Algebra.Com's Answer #564291 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! mean= 55lb, SD = 9.2 pounds. \n" ); document.write( "P(x ≥ 52) = P(z > -3/9.2) = normalcdf(-3/9.2, 100) = .6278 0r 62.78% \n" ); document.write( "D \n" ); document.write( " \n" ); document.write( " |