Quiz 3 Solution Sheet
Math 1680
Ms. Stonebraker
The GPA for college freshman approximately normally distributed with an average GPA
of 2.8 and a standard deviation of .4. Show all work. Round to two decimal places when
necessary. Use attached Z-chart. Draw pictures where applicable.
1. What percent of college freshmen have a GPA between 2.5 and 3.5?
a. First we need to find the z-scores associated with 2.5 and 3.5.
x  x 2.5  2.8
.3

   .75 (Worth 2pts)
SD
.4
.4
x  x 3.5  2.8 .7


  1.75 (Worth 2pts)
SD
.4
.4
z 2.5 
z 3.5
b. Draw a picture to better understand what we are looking at.
-.75
0
1.75
c. Now, we need to find the area associated with these z-scores.
z 2.5  .75  Area (Percent) = 54.67
z 3.5  1.75  Area (Percent) = 91.99
d. Remember that the area is representing the area between  z .(Worth 2pts)
Thus there is 91.99% of the data between  1.75, and 54.67% of the data
between  .75 . To find the area between -.75 and 1.75 it may be easiest to
look at the area between –0.75 and 0 and the area between 0 and 1.75.
Since the data is normally distributed, the data is symmetric, thus the area
between -.75 and 0 is the same as the area between 0 and .75. Therefore,
54.67
 27.34 . Likewise, the area between
the area between -.75 and 0 is
2
91.99
 46.00 .
0 and 1.75 is
2
Thus the area between -.75 and 1.75 is 27.34 + 46 = 73.34.
Thus 73.34% of the college freshmen have a GPA between 2.5 and 3.5.
2. At what percentile is a college freshman with a GPA of 3.0?
a. First we need to find the z-score associated with 3.0
z 3.0 
x  x 3.0  2.8
.2

   .5 (Worth 2pts)
SD
.4
.4
b. Draw a picture to represent the area we are dealing with.
0
.5
c. Find the area associated with the z-score.
z 3.0  .5  Area (Percent) = 38.29
d. (Worth 2pts) Remember that percentile is accumulative and this area
only give us the area between  .5. We still need the area below -.5.
Since there is 38.29% between  .5, there is 100 – 38.29 = 61.71% of
the data beyond  .5 (that is in the tails or above .5 and below -.5).
The graph is normally distributed and thus symmetric and both tail
hold the same area.
61.71
 30.86% of the data below -.5.
In particular, there is
2
Total area up to .5 is 30.86 + 38.29 = 69.15.
If you put the areas together in a different manner than mine, that is fine
as long as it is logical and valid.
Thus, a college freshman with a 3.0 GPA is of the 69.15th percentile.
3. What GPA is the 95th percentile for a college freshman?
a. Draw the area this is referring to.
95%
0
z
b. Notice, for some z-score, say z, the area below z is 95%, thus above
z there is 100 -95 = 5%. Since the data is normally distributed and
thus is symmetric, there is also 5% below –z. Since there is 5%
below –z and 5% above z, there is 100 - 5 – 5 = 90% between  z.
c. Now we can use the z-chart to find our z value because we have an
area that is representing  z.
Area (Percent) = 90  z = 1.65 (Worth 2pts)
d. Since we know that z = 1.65, we can “unwrap it” to find our GPA score
associate with that z value. (Worth 2pts)
z
xx
x  2.8
, thus 1.65 
, x  2.8  .4(1.65)  2.8  .66  3.46
SD
.4
Thus a college freshman at the 95th percentile would have a 3.46 GPA.
4. Sixty-five percent of college freshmen have a GPA between what two
values (centered on the mean)?
a. Draw a picture to describe the area we are referring to one another.
65%
-z
0
z
b. Now use the z-chart to find the z value that is associated with 65%. We
can go straight to the z-chart since we are looking for the z score such that
has 65% of the data between  z .
Area (Percent) = 65  z = .95 (Worth 2pts)
c. Thus 65% of the data falls between  .95. Now we need to “unwrap” our
z-scores to find the data pieces associated with them. (Worth 2pts)
x  x  z ( SD)  2.8  .95(.4)  2.8  .38
2.8  .38  3.18
2.8  .38  2.42
Thus, sixty-five percent of college freshman have a GPA between 3.18 and 2.42
(when centered on the mean).