Question 1128738

<pre>
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), 
 
locate(4.8,-.01,z),locate(4.8,.2,z)
  
)}}}

The area under the entire normal curve is 1.  So the area under each side
of the normal curve is .5. We want to find the value of z such that 70% or .70
of the area is under the curve to the right of that
z-value. 

So .50 of the .70 is to the right of z=0 and that leaves
.70 - .50 = .20 for the area on the left of z=0. 

So we look in the body of the normal table and find the nearest value to
.20 for the area to the left of z=0 that we need to make up the rest of the
70%.  We find that .1985 is the closest area value in the table.  Then we
observe that .1985 occurs where z = -0.52. We know to take it negative since
only negative values of z have more than 60% of the area to the right of them.
So that is the value we are looking for.  I'll indicate it with a green line
at z=-0.52, just a little past halfway between z=0 and z=-1.

{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), 
 
green(line(-0.52,0,-.52,exp(-.52^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)
  
)}}}

The area under the curve to the right of the green line at z=-0.52 is 70%
of the entire area
 
solved by:

Edwin</pre>