document.write( "Question 1091258: the mark vi monorail used at disney world has doors with a height of 72 inches. heights of men normally distributed with a mean of 69.4 inches and a standard deviation of 2.2 inches\r
\n" ); document.write( "\n" ); document.write( "A. what percentage of adult men can fit through the doors without bending?\r
\n" ); document.write( "\n" ); document.write( "B. what doorway height would allow 99% of adult men to fit without bending?
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Algebra.Com's Answer #705795 by Boreal(15235)\"\" \"About 
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z=(x-mean)/sd
\n" ); document.write( "want z for a mean of 69.4
\n" ); document.write( "percent of men over 72 inches
\n" ); document.write( "z=(72-69.4)/2.2=1.182
\n" ); document.write( "z <+1.182 is the probability desired,of 0.8814 or 88.1%
\n" ); document.write( "for 99%, need the value of z (0.99), which is +2.328
\n" ); document.write( "multiply that by the sd and get 5.1216. That is the difference between the mean and the height of the door.
\n" ); document.write( "The door height needs to be 69.4+5.1 or 74.5 inches.
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