document.write( "Question 1175807: The distribution of scores on a recent quiz in Mr. Waldron's statistics class is approximately normal. About 2.5% of the students scored below 25 on the quiz. 84% of the students scored below 40 on the quiz. Use the 68-95-99.7 rule to estimate the mean and standard deviation of the quiz scores in this class. Make sure to show your work and justify your answer. \n" ); document.write( "
Algebra.Com's Answer #801498 by Boreal(15235)\"\" \"About 
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2.5% is z=-2 SD by the empiric rule
\n" ); document.write( "84% is z=+1 sd by the empiric rule
\n" ); document.write( "z=(x-mean)/sd
\n" ); document.write( "so
\n" ); document.write( "-2=(25-mean)/sd
\n" ); document.write( "-2sd=25-mean
\n" ); document.write( "mean-2sd=25
\n" ); document.write( "-
\n" ); document.write( "and
\n" ); document.write( "+1=(40-mean)/sd
\n" ); document.write( "mean+1=40
\n" ); document.write( "-
\n" ); document.write( "two equations in two unknowns.
\n" ); document.write( "subtract
\n" ); document.write( "-3sd=-15
\n" ); document.write( "sd=5
\n" ); document.write( "mean=35
\n" ); document.write( "-
\n" ); document.write( "mean is 35 and sd is 1
\n" ); document.write( "68% are between 30 and 40
\n" ); document.write( "16% are above 40 or 84% are below 40
\n" ); document.write( "95% are between 25 and 45 so 97.5% are greater than 25 and 2.5% below.
\n" ); document.write( " \n" ); document.write( "
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