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
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document.write( "Above 100?
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document.write( "Below 100?
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document.write( "Above 90?
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document.write( "Below 90?
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document.write( "Above 80?
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document.write( "Below 80?
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document.write( "Above 70?
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document.write( "Below 70?
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document.write( "Above 60?
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document.write( "Below 60?
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Algebra.Com's Answer #563888 by ewatrrr(24785) 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 \n" ); document.write( " \n" ); document.write( "......... \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% \n" ); document.write( " \n" ); document.write( " |