document.write( "Question 860856: Someone please help, it would be greatly appreciated!!! I have been stuck on this question for awhile now.\r
\n" ); document.write( "\n" ); document.write( "In a study of men at a college (ages 20-29), the mean height was 70.9 inches whith a standard deviation of 3.5 inches.\r
\n" ); document.write( "\n" ); document.write( "What height represents the 90th percentile?
\n" ); document.write( "What height represents the third quartile? \n" ); document.write( "
Algebra.Com's Answer #518690 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
In a study of men at a college (ages 20-29), the mean height was 70.9 inches whith a standard deviation of 3.5 inches.
\n" ); document.write( "What height represents the 90th percentile?
\n" ); document.write( "Find the z-value with a left tail of 90%:: invNorm(0.9) = 1.2816
\n" ); document.write( "Find the corresponding height: h = z*s+u = 1.2816*3.5+70.9 = 75.39 inches
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\n" ); document.write( "What height represents the third quartile?
\n" ); document.write( "Find the z-value with a left tail of 75%:: invNorm(0.75) = 0.6745
\n" ); document.write( "h = 0.6745*3.5+70.9 = 73.26 inches
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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