SOLUTION: Help me find the z-score for which 70% of the distribution's area lies to its right.

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.



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