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. |