Q:

Why does the p-value for z=x correspond to the negative x in a z-score table?

I was given the following problem on a practice final exam and I think the professor posted the wrong answer, but with the final tomorrow I don't have time to ask him via email so I figured I'd look for the solution here:

Q. 7
  A  sample  of
  n
  =  36  adults  participated  in  a  study  measuring  the  effectiveness  of  ginkgo  biloba  and
  ginseng as tools for improving memory.  After 90 days of taking the supplements, these adults took a
  standardized memory test.  Their average score on the test was  ̄
  x
  = 84.
  a.
  Compared to the general population, whose mean score on this test was
  μ
  = 80 with
  σ
  = 18, use
  the critical-value approach to determine if there evidence that taking these supplements improves
  memory when
  α
  = 0
  .
  05.
  b.
  What is the
  p
  -value for the test conducted in
  a.
  ?

answer to a:
z = 1.33 (which I got as well), Z0.05= 1.645,
answer to b:
p-value: p = 0.0918. What I don't get is the p-value of 0.0918 corresponds to -1.33 from the Z score table. How did he get that p-value? Is he correct? 

A:

He did not really use the negative value although it looks like it. He used $$P(z \geq a) = 1-P(z < a)$$ which happens to be equal for continuous distributions that are symmetric around zero, $$1-P(y < a)=P(y \leq -a)$$ The value $P(z \leq a)$, the cumulative distribution function, is what you most commonly find in the tables (check which is the case). 

BTW note the round--of-error.
$$1-P(z < 4/3)=0.0912$$  is slightly different than $$1-P(z < 1.33)=0.0918$$   (if you work with this precision, which I find a bit superfluous/exaggerated in this case, then also do it correctly)

