-: Hello students.
Now we are going to continue working
on problem three and we are going to go to part B.
So again, I will read the problem.
Total classroom in children age 10 to 15 years
of age is assumed to follow a normal distribution
with Amina 191 and a standard deviation of 22.4.
Part B is asking us what proportion
of children 10 to 15 years of age would be classified
as hyperlipidemia.
Assume that hyperlipidemia is defined
as total cholesterol level over 200.
So again, we have the same format and the same values.
So as a first step, I'm going to write down the information
we have and draw the normal curve.
So our mu is equal to 191, our sigma is equal to 22.4.
Okay? And this is our normal curve.
Here is 191, here is X.
So the question is asking us what is the area
under the curve past 200?
So what I'm going to do is I'm going to use another color to
highlight that area.
So let me use
this color.
It's kind of like purple.
So here it is, 200 and we are looking for this area, okay?
The area past 200.
So as before, we have to now
transform this
to a Z distribution because otherwise we will not
be able to perform the calculation.
So for a Z distribution, we will have a mean of zero.
So to find the respective values right now we have to
go back to our formula, which is Z, equal
to X minus mu divided by sigma.
And here what we are trying to do is find the probability
that X is greater than 200.
So as we perform the transformation, we will be
using the formula again, and that is 200 minus 191 divided
by 22.4.
Okay, oops, went 2, 2, 4.
So again, this is the formula.
So after we do the calculation here
basically the algebra, we find out Z is greater than 0.40.
So again, what do we do?
We go to the back of our book.
Now we will go to the positive side of the table.
So first we look at the column
and stop at the 0.4 value and look to the right.
And there it is.
And again, from the way our text is set up, the table here,
the way the C table is set up for our text,
what we have found is the probability
of Z being less than 0.40.
So here what we have is
and again there is a little space issue here.
So we have found Z is less than 0.40
and that is equal to 0.6554.
So basically now to highlight this in our table
and I'm going to use a different color
I'm going to use blue.
So this is the value our table has given us.
Okay? This part, okay,
this part and let me go and color it much nicely.
This is the part our table has given us, but we are looking
for the part that is highlighted in purple.
So how are we going to find it?
So as I mentioned previously, we can do this very easily.
Why? Because we know that the area under the curve,
the entire area under the curve is one.
So all we have to do right now is a little subtraction.
So to find out P, the probability that C
is greater than 0.40, we have to deduct
from one that the probability Z is less than 0.40.
Okay? So when we do that, what do we get?
One minus 0.6554.
And that answer will be 0.3446, or 34.46%.
That is our answer.
Okay, so what does this mean?
This means that the given,
if someone asks us what is the probability
that the child is hyperlipidemia, we would say it is 34.46%.
And again, please keep in mind that it's based
on the values that are provided in this problem.