document.write( "Question 847359: Youth volleyball practice times are normally distributed with a mean of 75 minutes and standard deviation of 10 minutes.
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\n" ); document.write( " 65 minutes
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\n" ); document.write( " all of the above
\n" ); document.write( " none of the above
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Algebra.Com's Answer #510388 by ewatrrr(24785)\"\" \"About 
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document.write( "Hi,
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document.write( " mean of 75 minutes and standard deviation of 10 minutes.
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document.write( "  50 minutes   z = -2.5 an unusual practice
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document.write( " 65 minutes    z = -1
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document.write( " 90 minutes    z = 1.5
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document.write( "For the normal distribution: 
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document.write( "one  standard deviation from the mean accounts for about 68.2% of the set 
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document.write( "two standard deviations from the mean account for about 95.4%
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document.write( "and three standard deviations from the mean account for about 99.7%.
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document.write( "Important to Understand z -values as they relate to the Standard Normal curve:
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document.write( "Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
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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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