document.write( "Question 971287: 9. Suppose heights of U.S. adults are normally distributed with a mean of 66.3 and a standard deviation of 4.4 inches.
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document.write( "(a) Find the height such that 95% of the population is less than this height. (the 9 percentile)
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document.write( "(b) What percent of US adults are taller than 64 inches?
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document.write( "(c) What is the probability that a sample of 18 adults will have a mean height greater than 64 inches?
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Algebra.Com's Answer #593822 by Boreal(15235) You can put this solution on YOUR website! z=( x-66.3)/4.4 \n" ); document.write( "1.644 is the z value for the 95th percentile.\r \n" ); document.write( "\n" ); document.write( "Multiply by 4.4 inches \n" ); document.write( "7.24 inches from the mean. That is x\r \n" ); document.write( "\n" ); document.write( "66.3+7.24=73.54 inches.\r \n" ); document.write( "\n" ); document.write( "95% of the population is <= 73.5inches in height.\r \n" ); document.write( "\n" ); document.write( "Taller than 64 inches\r \n" ); document.write( "\n" ); document.write( "z=(64-66.3)/4.4\r \n" ); document.write( "\n" ); document.write( "= -(2.3)/4.4\r \n" ); document.write( "\n" ); document.write( "= -.5227 \r \n" ); document.write( "\n" ); document.write( "I want the percentage of the distribution greater than a z-value of -0.5227\r \n" ); document.write( "\n" ); document.write( "This is 0.699 or 70% of the adults are greater than 64 inches tall.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |