document.write( "Question 928778: 16. The amount of time it takes to recover physiologically from a certain kind of sudden noise is found to be normally distributed with a mean of 80 seconds and a standard deviation of 10 seconds. Using the 50%–34%–14% figures, approximately what percentage of scores (on time to recover) will be:\r
\n" ); document.write( "\n" ); document.write( "Above 100?
\n" ); document.write( "Below 100?
\n" ); document.write( "Above 90?
\n" ); document.write( "Below 90?
\n" ); document.write( "Above 80?
\n" ); document.write( "Below 80?
\n" ); document.write( "Above 70?
\n" ); document.write( "Below 70?
\n" ); document.write( "Above 60?
\n" ); document.write( "Below 60?
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Algebra.Com's Answer #563888 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
For the normal distribution: Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.
\n" ); document.write( "Area under the standard normal curve to the left of the particular z is P(z)
\n" ); document.write( "Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50% to the right
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\n" ); document.write( "mean of 80 seconds and a standard deviation of 10 seconds
\n" ); document.write( "Using the 50%–34%(1sd)–14%(2sd) figures
\n" ); document.write( "Above 100: 50 - 48 = 2%
\n" ); document.write( " Below 100: 98%
\n" ); document.write( " Above 90: 50 -34 = 16%
\n" ); document.write( " Below 90: 84%
\n" ); document.write( " Above 80: 50%
\n" ); document.write( " Below 80: 50%
\n" ); document.write( " Above 70: 34 + 50 = 84%
\n" ); document.write( " Below 70: 16%
\n" ); document.write( " Above 60: 14 + 34 + 50 = 98%
\n" ); document.write( " Below 60: 2%
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