standard deviation is 9
to be in the top 20%, the z-score needs to be greater than 80%.
that would be the area to the left of the z-score.
if you look in the z-score table for an area of .8 to the left of the indicated z-score, you will find:
z-score is between .84 and .85.
interpolation yields .8416129032
raw score yields 71.57451613
that would be the score that he needs to be better than 80% of the other applicants.
you could save yourself a lot of grief by using a z-score calculator.
in that calculator, .....
provide the mean of 64.
provide the standard deviation of 9.
the calculator should give you the raw score indicated above, perhaps rounded a bit differently.
here's a picture from one of the easiest of them to use.
this calculator can be found at http://davidmlane.com/hyperstat/z_table.html
if you do not have access to a calculator and must use the z-score tables, then the procedure would be as shown below:
look in the z-score for an area to the left of the z-score closest to .8.
one value above.
one value below.
i found:
z-score area to the left
.84 .79955
.85 .80234
difference .01 .00279
the area to the left of the z-score you want is .8.
.8 - .79955 = .00045
your ratio is:
x / (.01) = .00045 / .00279
solve for x to get:
x = .00045 /.00279 * .01 = .0016129032.
add this to the z-score of .84 to get:
your z-score is .84 + .0016129032 = .8416129032.
use the z-score formula to find the raw score.
z = (x - m) / s becomes .8416129032 = x - 64) / 9
solve for x to get:
x = 9 * .8416129032 + 64 = 71.57451612
if you round this to 3 decimal places, it will be 71.575, same as the calculator provided.
the x that i used for interpolation is not the same x that i used to find the raw score.
same variable name used in two different procedures is acceptable.
they are not referring to the same thing.
manual interpolation can be tricky, but it does work reasonable well.
with so many calculators that do z-scores and raw scores around, it's usually not necessary.