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
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document.write( "A. what percentage of adult men can fit through the doors without bending?\r
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document.write( "B. what doorway height would allow 99% of adult men to fit without bending? \n" );
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Algebra.Com's Answer #705795 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! 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. \n" ); document.write( " \n" ); document.write( " |