document.write( "Question 1180807: Suppose the wing lengths of houseflies are normally distributed with a mean of 45.5 millimeters and a
\n" ); document.write( "standard deviation of 3.92 millimeters. Use the standard normal distribution and a calculator or table to
\n" ); document.write( "estimate the following.
\n" ); document.write( "a. The percent of houseflies with wing lengths over 35 millimeters
\n" ); document.write( "b. The percent of houseflies with wing lengths over 50 millimeters
\n" ); document.write( "c. The percent of houseflies with wing lengths over 55 millimeters
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Algebra.Com's Answer #810598 by Boreal(15235)\"\" \"About 
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z(35)>(35-45.5)/3.92=-2.68. Probability is 0.9963 or 99.63%
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\n" ); document.write( "over 50 is z>(50-45.5)/3.92 or 1.148. Probability is 0.1255 or 12.55%
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\n" ); document.write( "over 55 is z>(55-45.5)/3.92 or 2.423. Probability is 0.0077 or 0.77%
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