Question 622062
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
Am not quite sure what it is You are not understanding. z = {{{(x- mu)/sigma}}}
Here is the standard normal curve. z=values of 0,±1, ±2,±3 shown
for ex: 50% of the area under the normal curve is to the left and right of z=0
one  standard deviation from the mean(z= ±1)  accounts for about 68.2% of the set 
two standard deviations from the mean(z= ±2) account for about 95.4%
and three standard deviations from the mean(z= ±2) account for about 99.7%.

{{{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))}}}
(A) P(z > 0.90) = 1 -P(z &#8804; .90) = 1- NORMSDIST(.90) = 1-.8159 = .1841 or 18.41% 
 18.41% is the % of the Area to the right of the blue line
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, 0,exp(-x^2/2)), blue(line( .8159,0, .8159,exp(-.8159^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}
(B) P(-1.17 < z < 0) = .50 - NORMSDIST(-1.17)= .50-.121
(C) P(-0.58 < z < 1.57)  = NORMSDIST(1.57) - NORMSDIST(-.58) = .9418-.2801
 Here, the Excel function NORMSDIST is being used for the z-values.