Question 1050738
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, 0,exp(-x^2/2)), blue(line( 1.54,0, 1.54,exp(-1.54^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)
)}}}
Shade to right of z = 1.54
p(z > 1.54) = 1- p(z <= 1.54) = 1 - normalcdf(-10,1.54) = .0617

{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, 0,exp(-x^2/2)), blue(line( .37,0, .37,exp(-.37^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)
)}}}
Shade to Left of z = .37
p( z < .37) = normalcdf(-10,.37) = .6443
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, 0,exp(-x^2/2)), blue(line( -1.31,0, -1.31,exp(-1.31^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)
)}}}
Shade to Left of z = -1.31
p(z < -1.31) = normalcdf(-10,-1.91) = .0951
Since this is a continuous function, we have P(z < value) = P(z &#8804; value).
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2 Sketch the area under the standard normal curve <u>between</u> the z-values given.

P(-1.50 &#8804; z &#8804; 1.90) = normalcdf(-1.50, 1.90)
P(-1.86 &#8804; z &#8804; 0) = normalcdf(-1.86, 0)