00:01
All right, in your question, you're given information about a university and how they receive their grades.
00:06
They're receiving their grades as z scores, and you're asked to actually convert them back to gpas, given that the z scores are based on gpas that have a mean of 2 .7 and a standard deviation of 0 .4.
00:24
So i have a simple formula here for you.
00:26
You would just take, we'll do this formula each time.
00:30
You're just going to take your mean plus the z score times the standard deviation for each of these questions.
00:40
So the first one would be 2 .7 plus 2 times 0 .4, which that should work out to 3 .5 for the gpa.
00:53
For the second one, it's 2 .7 plus negative 1 times 0 .4, which would work out to be.
01:03
A 2 .3 gpa.
01:07
For the next one it's 2 .7 plus 0 .5 times 0 .4, which would equal to a 2 .9 gpa.
01:24
Finally, we have 2 .7 plus negative 2 .5 times 0 .4, which works out to be 1 .7 gpa.
01:51
And that answers your question for part a now part b we're supposed to figure out says students with c scores below i actually have my work mislabeled here sorry that's part c all right part b is saying students with z scores below negative 1 .6 are put on probation what is the corresponding gpa so it's a similar question to part a was, we just have a different z score.
02:27
So i'm going to use the same formula, 2 .7 plus negative 1 .6 times standard deviation.
02:39
And typing that in my calculator here gives me a 2 .06.
02:51
So the boundary, the corresponding probation gpa, would be a 2 .06.
02:57
All right, now on to part c, now part c, what i want to do is i want to do is i want to to draw a normal model.
03:08
Okay, because we're working with our z scores, we're going to assume we have a standardized model here.
03:14
And a standardized model, normal model, if you go out one standard deviation from the mean in either direction, we would expect 68 % of the data to fall within those two boundaries.
03:32
But the question would be how much falls outside the boundaries.
03:36
So if i did 100 minus 68, i get 32.
03:40
If i divide that by 2, i get 16.
03:44
So there's 16 % above and 16 % below.
03:50
So the z score boundary that blocks off the top 16 % would be one standard deviation or a z score of 1.
04:00
Now we also have to figure out the top 2 .5%...