document.write( "Question 1135449: The number of adults sleep per night is normally distributed with a mean of 6.8 hours. assume that the standard deviation is unknown.\r
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document.write( "If 65% of adults sleep more than 6.2 hr/night. what is the variance? \n" );
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Algebra.Com's Answer #753601 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z>-0.385 for 65% being more than that value \n" ); document.write( "z=(x-mean)/sd \n" ); document.write( "-0.385=(6.2-6.8)/sd \n" ); document.write( "-0.385 s=-0.6 \n" ); document.write( "s=+1.56 \n" ); document.write( "variance is s^2=2.43 hr^2 (rounding at end) \n" ); document.write( " \n" ); document.write( " |