document.write( "Question 398216: Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 degrees F and a standard deviation of 0.62 degrees F \r
\n" ); document.write( "\n" ); document.write( "If you have a body temperature of 99 degrees F what is your percentile score
\n" ); document.write( "Convert 99.00 degrees F to a standard score
\n" ); document.write( "What body temperature is the 95th percentile
\n" ); document.write( "What body percent is the 5th percentile
\n" ); document.write( "Is a body temperature of 99.00 degrees F unusual? Why or Why not
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Algebra.Com's Answer #282170 by stanbon(75887)\"\" \"About 
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Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 degrees F and a standard deviation of 0.62 degrees F
\n" ); document.write( "If you have a body temperature of 99 degrees F what is your percentile score
\n" ); document.write( "Convert 99.00 degrees F to a standard score
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\n" ); document.write( "z(99) = (99-98.2)/0.62 = 1.29
\n" ); document.write( "P(z < 1.29) = normalcdf(-100,1.29) = 90%ile
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\n" ); document.write( "What body temperature is the 95th percentile
\n" ); document.write( "invNorm(0.95) = 1.645
\n" ); document.write( "temp = zs+u
\n" ); document.write( "temp = 1.645*0.62+98.2 = 99.22 degree
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\n" ); document.write( "What body percent is the 5th percentile
\n" ); document.write( "invNorm(0.05) = -1.645
\n" ); document.write( "temp = -1.645*0.62+98.2 = 97.18 degrees
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\n" ); document.write( "Is a body temperature of 99.00 degrees F unusual?
\n" ); document.write( "Why or Why not
\n" ); document.write( "Not really. It's only 1.29 standard deviations above the mean.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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