document.write( "Question 634248: The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.\r
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Algebra.Com's Answer #399541 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi,
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document.write( "mean of 500 and a standard deviation of 50. z = \"%28x-mu%29%2Fsigma\"
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document.write( "P(x < 600)= P(z <  2)= .9773 or 97.73%  | using NORMSDIST Excel function to find P knowing z
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document.write( "P(x < 400)= P(z < -2)= .0228 or 2.28%
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document.write( "P(x > 350)= P(z > -3)= .9987 or 99.87%
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document.write( "P(x > 580)= P(z > 1.6) = .0548 or 5.48%\r
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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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