60% is to the right of z = -0.25
90% is to the left of z = +1.28
"z" is (x-x(mean))/{standard deviation)
You need a table of the error function or normal distribution to do this kind of problem.
For a normal distribution, identify the z-score location that would separate the distribution into sections so that there is:
a)60% in the body on the right-hand side
b)90% in the body on the left-hand side
1 answer