The average grade for an exam is 74, and the standard deviation is 7.
(A) If 12% of the students in the class are given A's, and the grades
are curved to follow a normal distribution, what is the lowest
possible score to receive an A? (B) What is the 60th percentile score?

Hello.

A way to solve problems like these is to use a standard normal
probability table, such as this one from Penn State:
STANDARD NORMAL PROBABILITY TABLE
http://www.stat.psu.edu/~antoniou/stat250.3/ZTable.html


(A)

The first step is to find the "z" value for the percentile that you need.

If 12% of the class are to be given A's, then you want the "z" value
for the 88th percentile.

Consulting the table, we see that 0.8810 corresponds to z = 1.18.

Now, we can determine the necessary score by setting the z value equal to:

(X - the average)
-------------------
standard deviation

(where X is the score to be determined)

So:

         X - 74
1.18  = ------- 
           7

Using algebra, we see that:


             X - 74
1.18 x 7  = --------- x 7
               7

so:

8.26 = X - 74

so

X = 82.26

Rounding down, 82 is the score needed to receive an A.


(B)

We can determine the 60th percentile score exactly the same way.

Again, we consult the table:
http://www.stat.psu.edu/~antoniou/stat250.3/ZTable.html

We see that for the value 0.6026, z=0.26

Thus, 

         X - 74
0.26  = ------- 
           7


Using algebra, we see that:

             X - 74
0.26 x 7  = --------- x 7
               7

so:

1.82 = X - 74

so

X = 75.82

Rounding up, we see that 76 is the 60th percentile score.

--------------------

search strategy:
"standard deviation"  percentile "what is the minimum"
This brought up a similar problem in this document from University of Rochester:
http://www.econ.rochester.edu/khan/lect7.pdf

I hope this helps.