Instructor: Hello, students.
This is problem three, D.
Now, this is not a problem in our text,
but it's important for us to learn
how to calculate percentile of a normal distribution.
So we are extending problem number three
and we want to know now, based on the values provided,
what would be the total cholesterol of children
in this population at the 95th percentile?
In other words, what is the x value at the 95th percentile?
So similar to the previous one, I am going to start
with the values that were provided to us.
So, our x
at 95th percentile
is what we are trying to figure out.
That is what we are trying to figure out.
The mu here is, again, 191.
Sigma is 22.4. Okay.
So next I am going to draw
the normal distribution.
And again, we need that z transformation,
otherwise we cannot really do anything.
And
what we are looking for
is this x, okay?
So basically, this is the 5%
and
this is the 95%.
And what is that x value?
So basically, here we are kind of flipping the process
because previously we looked at x less than some value
transformed it to the z value or the z distribution
or we looked at x above some value
and, again, transformed it
to the g value or the z distribution.
And then we identified some value
that was for z less than as per our table.
And then we found out the probability of the z value
being less than some value.
Now, here what we are doing is,
again, kind of flipping the process.
Here we are saying that the probability of z
less than some value is 0.95, okay?
Or in the 95 percentile.
So now, let's go to the back of our book
and we are going to see what is the z value for 0.95.
So basically, now we are going to go
to the middle of our table,
not on the columns or the rows, in the middle of our table,
and we are going to look for the value 0.95.
So what do we find here?
We find that we don't have an exact 0.95 value.
But what we do have is a value
for 0.9495 and 0.9505.
So in the back of our book, we have a value for zero.
Oops, sorry, I gotta change this to black now.
So let me go back to black.
And we have the value 4.9495
and per 0.9505.
Okay? So what are these values?
This value is 1.64.
This value is 1.65.
So if 0.95 were to be placed in our table,
where would it be?
It would be in between.
So what do we do?
We take these two numbers, add them up,
divide them by two, and get the average.
And the average, what would it that be?
It would be 1.645.
Now, if you do not want
to go through all the complicated process,
then you can really go to the next page
and you can look at table 1A.
And here we have the z values
for some commonly used percentile.
And for the 95th percentile, that is 1.645.
Again, very important to use the z table
at the back of our book.
Now, we are going to the actual process.
What we are going to do is here, again, use the same formula
but the transformation is somewhat reverse.
So here z, again, the same formula,
x minus mu divided by sigma.
The z value here is 1.645.
Now, we have to do a little algebra here.
And I'm going to show you the algebra
before I actually plug in the numbers.
So it's going to be z multiplied by sigma
because that's going under the reverse side
and then it's going to be plus mu equal to x.
Because remember, we are looking at the x value here.
So when we insert the numbers here,
z here is 1.645.
Sigma again is 22.4.
Our mu again is 191,
and that will give our x value.
So what is our X value going to be?
It's going to be 227.85,
which is kind of 227.9, we can say.
So what does this mean?
So this means that given the values provided,
the 95th percentile of cholesterol for children,
10 to 15 years of age,
in this population is 227.9.
So basically, we can say that 5% of the children,
and I'm gonna write this here,
5% of the children, given the values,
given the values
and between the ages of
10 to 15 years of age,
will have a cholesterol above
or basically total cholesterol will have a,
and I'll abbreviate, total cholesterol above
227.9, okay?
All right.
So again, please feel free to reach out
if this is not clear, if you're having any confusion.
Very important that you understand the process
because it's important for you to be able to solve
the homework, assignments, and also for a test.
You will need to understand the process
so you can perform well in that test
and ultimately learn the material.
So again, please feel free to reach out
if you have any questions.