Question 872391
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Hi
Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
Note: z = 0 (x value the mean) 50% of the area under the curve is to the left 
and %50 to the right
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(1,0,1,exp(-1^2/2)),line(-1,0,-1,exp(-1^2/2))),green(line(2,0,2,exp(-2^2/2)),line(-2,0,-2,exp(-2^2/2))),green(line(3,0,3,exp(-3^2/2)),line(-3,0,-3,exp(-3^2/2))),green(line( 0,0, 0,exp(0^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}
Population: mean = 3962, SD = 950
a) P(x &#8805; 1000) = -2962/950 = -3.118
   P(z > -3.118) = normalcdf( -3.118, 10) = .9991  0r 99.91%
b) invNorm(.90) = 1.28,   1.28*950 + $3962 = $5178 (10% owe more than this)
c) As a Rule: 2SD on either side of mean: is 95% of the Population
   3962 - 2*950 < x < 3962 + 2*950

For the normal distribution: 
one  standard deviation from the mean accounts for about 68% of the set 
two standard deviations from the mean account for about 95%
and three standard deviations from the mean account for about 99.7%.