document.write( "Question 453135: assume body temp of healthy adults are normally distributed with a mean of 98.20F and a standard deviation of 0.62F. what body temperature is the 95th and 5th percentile?? \n" ); document.write( "
Algebra.Com's Answer #311490 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! assume body temp of healthy adults are normally distributed with a mean of 98.20F and a standard deviation of 0.62F. what body temperature is the 95th and 5th percentile?? \n" ); document.write( "---------------------- \n" ); document.write( "Find the z-value with a 95% left-tail: Ans: z = 1.645 \n" ); document.write( "---- \n" ); document.write( "Find the z-value with a 5% left-tail: Ans: z = -1.645 \n" ); document.write( "----- \n" ); document.write( "95%ile = 1.645*0.62+98.2 = 99.22 degrees \n" ); document.write( "--- \n" ); document.write( "5%ile = -1.645*0.62+98.2 = 97.18 degrees \n" ); document.write( "===================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |