Question 933739
 µ = 100 and σ = 20
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one score is randomly selected:   {{{z = blue(x - 100)/blue(20)}}}  
P( 80 < x <120) = P( -20/20 < z < 20/20) = normalcdf(-1,1) = .6826
0r
P( 80 < x <120) = P(z < 1) - P(z < -1) = .8413 - .1587 = .6826
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  For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
68.26% of the Area under the standard normal curve is between z = -1  and z = 1
Note: 34.13% of the Area under the standard normal curve is between z = 0  and z = 1
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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))}}}