document.write( "Question 1120867: 9. A normal distribution of utility bills shows the mean to be $100 and the standard deviation of $12.\r
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document.write( "a) What percent of the utility bills are more than $125.
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document.write( "b) If 300 utility bills are randomly selected, about how many of them would you expect to be less than $90?
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Algebra.Com's Answer #736613 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z=(x-mean)/sd or 25/12 or +2.167 \n" ); document.write( "probability z>2.17 is 0.0150 \n" ); document.write( "for 300 bills, this would be 300*0.015 or 4.5. Four or five bills. \n" ); document.write( " |