document.write( "Question 524473: An accerlerated life test on a large number type-D alkaline batteries revealed that the mean life for a particular use before they failed is 19.0 hours. The distribution of the lives approximated a normal distriburtion. The standard deviation of the distribtuion was 1.2 hours About 95% of the batteries failed between what two values? \n" ); document.write( "
Algebra.Com's Answer #347749 by stanbon(75887)\"\" \"About 
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An accerlerated life test on a large number type-D alkaline batteries revealed that the mean life for a particular use before they failed is 19.0 hours. The distribution of the lives approximated a normal distriburtion. The standard deviation of the distribtuion was 1.2 hours About 95% of the batteries failed between what two values?
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\n" ); document.write( "Find the z-value with a left-tail of 2.5%:
\n" ); document.write( "invNorm(0.025) = 1.96
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\n" ); document.write( "Note: 95% of the data lies between -1.96 and +1.96
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\n" ); document.write( "Find the corresponding raw-score values using x = zs+u
\n" ); document.write( "x = -1.96*1.2+19 = 16.65 hrs.
\n" ); document.write( "x = +1.95*1.2+19 = 21.35 hrs.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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