Question 431046
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Hi
*Note: {{{z = blue(x - mu)/blue(sigma)}}}
mean of 50 and a std. deviation of 10
 z = -20/10 = -2  and z = 15/10 = 1.5
P(between 30 and 65) = P( -2 > z < 1.5) = .9332 - .0228 = .9104  
91.04% of them are between 30 and 65
Area under the graph between z = -2 and z = 1.5
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(-2,0,-2,exp(-2^2/2)),line(1.5,0,1.5,exp(-1.5^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}