Question 929413
mean =140sec, SD = 20 seconds, {{{z = blue(x - 140)/blue(20)}}}    
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P(x < 120) = p( z < -20/20) = normalcdf(-100,-1) = .1587
P(x > 120) = P(z > -1) = normalcdf(-1,100) = .8413
P(x > 160) = P(z > 20/20)= normalcdf(1,100) = .1587 0r 15.87% ***
.5000 = .5000
D.
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For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
Area under the standard normal curve to the left of the particular z is P(z)
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))}}}