document.write( "Question 1128414:  The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 8.6 cm.\r
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document.write( "a. Find the probability that an individual distance is greater than 
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document.write( "210.00cm.
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document.write( "b. Find the probability that the mean for 20 randomly selected distances is greater than 196.00 cm.\r
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document.write( "A. 0.0735 I got this one.
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document.write( "B. Need help with this and please show work so I can understand. \n" );
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| Algebra.Com's Answer #744874 by Boreal(15235)     You can put this solution on YOUR website! z=(x-mean)/sd \n" ); document.write( "z>(210-197.5)/8.6 or z>1.45 \n" ); document.write( "That probability is 0.0735\r \n" ); document.write( "\n" ); document.write( "For 20, the denominator is sigma/sqrt(n) That is 8.6/sqrt(20) \n" ); document.write( "that would be 1.92, the z value is >-1.5/1.92 or >-0.78 \n" ); document.write( "The -1.5 is for 196-197.5 and that is divided by the 1.92 \n" ); document.write( "The probability is 0.7823 \n" ); document.write( " |