document.write( "Question 872391: The average credit card debt of college seniors is $3962. If the debt is normally distributed with a standard deviation of $950, find these probabilities:
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\n" ); document.write( "\n" ); document.write( "b) The 10% of seniors that owe the most money owe more than what amount?\r
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\n" ); document.write( "\n" ); document.write( "c) 95% of all seniors owe how much debt (centered about the mean)?
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Algebra.Com's Answer #526118 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( "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 
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document.write( "and %50 to the right
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document.write( "Population: mean = 3962, SD = 950
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document.write( "a) P(x ≥ 1000) = -2962/950 = -3.118
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document.write( "   P(z > -3.118) = normalcdf( -3.118, 10) = .9991  0r 99.91%
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document.write( "b) invNorm(.90) = 1.28,   1.28*950 + $3962 = $5178 (10% owe more than this)
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document.write( "c) As a Rule: 2SD on either side of mean: is 95% of the Population
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document.write( "   3962 - 2*950 < x < 3962 + 2*950\r
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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% of the set 
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document.write( "two standard deviations from the mean account for about 95%
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document.write( "and three standard deviations from the mean account for about 99.7%.\r
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