Question 932588
mean of 3.1 hours and a standard deviation of 0.5 . {{{z = blue(x - 3.1)/blue(.5)}}} 
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P(x < 4.1) = p (z < 1/.5) = normalcdf(-100, 2) = .9772  0r 97.72%
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  Area under the standard normal curve to the left of the z = 2 is the p-value = P(z < 2)
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))}}}