document.write( "Question 1047922: Average daily prices are expected to be $400 with a standard deviation of $6.50.\r
\n" ); document.write( "\n" ); document.write( "What can a person expect in what range will the price be 68% of the time? 95% of the time? What does he have to assume?
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Algebra.Com's Answer #663486 by ewatrrr(24785)\"\" \"About 
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one standard deviation from the mean accounts for about 68% of the set
\n" ); document.write( "two standard deviations from the mean account for about 95%
\n" ); document.write( "and three standard deviations from the mean account for about 99.7%.
\n" ); document.write( "What can a person expect in what range will the price be 68% of the time
\n" ); document.write( "400 - 6.50 to 400 + 6.50
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\n" ); document.write( "What can a person expect in what range will the price be 95% of the time
\n" ); document.write( "400 - 13.00 to 400 +13.00
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