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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\n" ); document.write( "\n" ); 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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\n" ); document.write( "\n" ); document.write( " A .13
\n" ); document.write( " B .26
\n" ); document.write( " C .37
\n" ); document.write( " D .63
\n" ); document.write( " E .73
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Algebra.Com's Answer #564291 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
mean= 55lb, SD = 9.2 pounds. \"z+=+blue%28x+-+55%29%2Fblue%289.2%29\"
\n" ); document.write( "P(x ≥ 52) = P(z > -3/9.2) = normalcdf(-3/9.2, 100) = .6278 0r 62.78%
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