Question 855242
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Hi,
Find the z-scored for which 70% of the distributions area lies between -z and z.
NORMSINV(.30/2) = -1.0364
The z-scores are -1.0364, 1.0364
Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
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))}}}