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<v Instructor>Hello, students.</v>

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This is problem three, D.

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Now, this is not a problem in our text,

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but it's important for us to learn

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how to calculate percentile of a normal distribution.

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So we are extending problem number three

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and we want to know now, based on the values provided,

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what would be the total cholesterol of children

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in this population at the 95th percentile?

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In other words, what is the x value at the 95th percentile?

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So similar to the previous one, I am going to start

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with the values that were provided to us.

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So, our x

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at 95th percentile

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is what we are trying to figure out.

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That is what we are trying to figure out.

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The mu here is, again, 191.

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Sigma is 22.4. Okay.

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So next I am going to draw

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the normal distribution.

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And again, we need that z transformation,

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otherwise we cannot really do anything.

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And

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what we are looking for

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is this x, okay?

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So basically, this is the 5%

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and

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this is the 95%.

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And what is that x value?

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So basically, here we are kind of flipping the process

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because previously we looked at x less than some value

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transformed it to the z value or the z distribution

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or we looked at x above some value

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and, again, transformed it

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to the g value or the z distribution.

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And then we identified some value

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that was for z less than as per our table.

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And then we found out the probability of the z value

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being less than some value.

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Now, here what we are doing is,

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again, kind of flipping the process.

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Here we are saying that the probability of z

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less than some value is 0.95, okay?

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Or in the 95 percentile.

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So now, let's go to the back of our book

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and we are going to see what is the z value for 0.95.

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So basically, now we are going to go

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to the middle of our table,

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not on the columns or the rows, in the middle of our table,

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and we are going to look for the value 0.95.

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So what do we find here?

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We find that we don't have an exact 0.95 value.

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But what we do have is a value

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for 0.9495 and 0.9505.

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So in the back of our book, we have a value for zero.

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Oops, sorry, I gotta change this to black now.

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So let me go back to black.

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And we have the value 4.9495

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and per 0.9505.

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Okay? So what are these values?

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This value is 1.64.

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This value is 1.65.

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So if 0.95 were to be placed in our table,

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where would it be?

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It would be in between.

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So what do we do?

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We take these two numbers, add them up,

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divide them by two, and get the average.

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And the average, what would it that be?

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It would be 1.645.

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Now, if you do not want

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to go through all the complicated process,

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then you can really go to the next page

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and you can look at table 1A.

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And here we have the z values

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for some commonly used percentile.

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And for the 95th percentile, that is 1.645.

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Again, very important to use the z table

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at the back of our book.

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Now, we are going to the actual process.

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What we are going to do is here, again, use the same formula

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but the transformation is somewhat reverse.

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So here z, again, the same formula,

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x minus mu divided by sigma.

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The z value here is 1.645.

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Now, we have to do a little algebra here.

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And I'm going to show you the algebra

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before I actually plug in the numbers.

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So it's going to be z multiplied by sigma

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because that's going under the reverse side

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and then it's going to be plus mu equal to x.

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Because remember, we are looking at the x value here.

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So when we insert the numbers here,

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z here is 1.645.

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Sigma again is 22.4.

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Our mu again is 191,

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and that will give our x value.

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So what is our X value going to be?

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It's going to be 227.85,

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which is kind of 227.9, we can say.

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So what does this mean?

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So this means that given the values provided,

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the 95th percentile of cholesterol for children,

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10 to 15 years of age,

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in this population is 227.9.

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So basically, we can say that 5% of the children,

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and I'm gonna write this here,

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5% of the children, given the values,

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given the values

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and between the ages of

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10 to 15 years of age,

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will have a cholesterol above

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or basically total cholesterol will have a,

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and I'll abbreviate, total cholesterol above

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227.9, okay?

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All right.

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So again, please feel free to reach out

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if this is not clear, if you're having any confusion.

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Very important that you understand the process

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because it's important for you to be able to solve

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the homework, assignments, and also for a test.

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You will need to understand the process

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so you can perform well in that test

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and ultimately learn the material.

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So again, please feel free to reach out

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if you have any questions.