document.write( "Question 947860: If the mean height of college males is 70 inches with a standard deviation of 3 inches, what percentage of college males would be between 6' and 6'4\"? Assume a normal distribution. \r
\n" ); document.write( "\n" ); document.write( "Using the information in problem 1, what height would someone have to be in order to be in the 99th percentile? \n" ); document.write( "
Algebra.Com's Answer #578523 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
If the mean height of college males is 70 inches with a standard deviation of 3 inches, what percentage of college males would be between 6' and 6'4\"? Assume a normal distribution.
\n" ); document.write( "z(72) = (72-70)/3 = 2/3
\n" ); document.write( "z(76) = (76-70)/3 = 2
\n" ); document.write( "----
\n" ); document.write( "P(72 < x < 76) = P(2/3 < z < 2) = 0.2297
\n" ); document.write( "---------------------------------------------\r
\n" ); document.write( "\n" ); document.write( " Using the information in problem 1, what height would someone have to be in order to be in the 99th percentile?
\n" ); document.write( "Find the z-value with a left tail of 0.99
\n" ); document.write( "invNorm(0.99) = 2.3264
\n" ); document.write( "---
\n" ); document.write( "Find the corresponding height value
\n" ); document.write( "x = 2.3264*3+70 = 76.98 inches
\n" ); document.write( "------------------
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
\n" ); document.write( "---------------- \n" ); document.write( "
\n" );