document.write( "Question 1129792: The mean of a certain random variable, X, is 41 and the standard deviation is 1.3.\r
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document.write( "Find the boundaries that separate the middle 72% of the data. \n" );
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Algebra.Com's Answer #746502 by rothauserc(4718)![]() ![]() You can put this solution on YOUR website! Assume X is a normally distributed random variable \n" ); document.write( ": \n" ); document.write( "100 - 0.72 = 0.28 \n" ); document.write( ": \n" ); document.write( "we have 0.14 lower boundary percent and 0.72 + 0.14 = 0.86 upper boundary percent \n" ); document.write( ": \n" ); document.write( "assume your problem is looking for the associated X values \n" ); document.write( ": \n" ); document.write( "z-score = (X - mean)/standard deviation \n" ); document.write( ": \n" ); document.write( "look at table of z-scores for associated z-scores for the lower and upper percents \n" ); document.write( ": \n" ); document.write( "z-score(lower) = -2.99 \n" ); document.write( ": \n" ); document.write( "z-score(upper) = 1.08 \n" ); document.write( ": \n" ); document.write( "-2.99 = (X - 41)/1.3 \n" ); document.write( ": \n" ); document.write( "X - 41 = -3.887 \n" ); document.write( ": \n" ); document.write( "X = 37.113 is approximately 37 \n" ); document.write( ": \n" ); document.write( "1.08 = (X - 41)/1.3 \n" ); document.write( ": \n" ); document.write( "X - 41 = 2.34 \n" ); document.write( ": \n" ); document.write( "X = 43.34 is approximately 43 \n" ); document.write( ": \n" ); document.write( "********************* \n" ); document.write( "lower boundary is 37 \n" ); document.write( ": \n" ); document.write( "upper boundary is 43 \n" ); document.write( "********************* \n" ); document.write( ": \n" ); document.write( " |