Question 856112
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Hi,
P(z < 1.05) = NORMSDIST(1.05)= .8531  
(85.31% of the area under the normal curve is to the left of z = 1.05)
P(0< z &#8804; .24)  = NORMSDIST(.24) - .5 = .5948 - .5 = .0948
Normal distribution: 
Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
Note: z = 0 (represents 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))}}}