Normal distribution problem

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July 6th 2006, 04:50 PM
Jimmy23

Normal distribution problem

Hi, I'm trying to finish my homework and I'm stuck on a problem. Here it is: Suppose that the ages at a time of onset of a certain disease are normally distributed with a mean of 11.5 years and a standard deviation of 3 years.

In a population of 10,000 with the disease, how many would you expect to be between the ages of 7.5 and 13.5 years?

So here is what I did:

z=13.5-11.5 / 3 = 0.67

z=7.5-11.5 / 3 = -1.33

I look up those numbers in my book for the standard normal distribution and I get:
0.67 = .7486
-1.33 = .0918

So then I added .7486 and .0918 which gave me .8404 then I multiplied by 10,000 so I get (10,000)(.8404)=8,404 people between 7.5 and 13.5 have the disease. Did I do this right?
July 6th 2006, 04:53 PM
Quick

Quote:

Originally Posted by Jimmy23

Hi, I'm trying to finish my homework and I'm stuck on a problem. Here it is: Suppose that the ages at a time of onset of a certain disease are normally distributed with a mean of 11.5 years and a standard deviation of 3 years.

In a population of 10,000 with the disease, how many would you expect to be between the ages of 7.5 and 13.5 years?

Are you completely sure the mean is 11.5? It seems to me like they would be asking for 10.5 instead.
July 6th 2006, 04:58 PM
Jimmy23

Yeah I'm positive it is 11.5
July 6th 2006, 05:10 PM
Quick

Then I agree with your work.
July 6th 2006, 05:14 PM
Jimmy23

Awesome. Thanks, I appreciate it!
July 6th 2006, 05:24 PM
Quick

WAIT!!!

Quote:

I look up those numbers in my book for the standard normal distribution and I get:
0.67 = .7486
-1.33 = .0918

To solve this problem I think you need to find the normal distribution of the area between the different ages and the mean. I don't think the answer would yield 0.67=.7486

I think you're answers still right, but I don't have enough confidence in myself to guarantee it. (Maybe someone else can help you)
July 6th 2006, 05:28 PM
Jimmy23

I'm still here.
July 6th 2006, 05:52 PM
JakeD

Quote:

Originally Posted by Quick

WAIT!!!
To solve this problem you need to find the normal distribution of the area between the different ages and the mean. I don't think the answer would yield 0.67=.7486

I agree with this. The answer is the difference .7486 - .0918 which is $P(-1.33 \le Z \le .67) = P(Z \le .67) - P(Z \le -1.33)$ . For these kind of problems I like this normal distribution calculator. It gives you the numerical answer and a picture showing the area of the normal distribution it's using.
July 6th 2006, 06:26 PM
Jimmy23

Oh man now I'm confused! So Instead of adding my two values I mentioned, I should just take .7486-.0918 = .6568 and multiply .6568 x 10000 which is 6,568 So the answer is 6,568 people?
July 6th 2006, 08:15 PM
Quick

Quote:

Originally Posted by Jimmy23

Oh man now I'm confused! So Instead of adding my two values I mentioned, I should just take .7486-.0918 = .6568 and multiply .6568 x 10000 which is 6,568 So the answer is 6,568 people?

NO!!! [EDIT] Actually Yes (I was wrong :( )
Ahem, what I know about this problem I got from a book, and here is the exact quotes...

Quote:

Originally Posted by Statistics Hacks

If you want to know what proportion of scores falls between any two points under the curve, define those points by their z score and figure out the relevant proportion. Depending on whether both scores fall on the same side of the mean, one of two methods will give you the correct proportion between those points:

$\cdot$ If the z scores are on the same side of the curve, look up the proportion of scores in either the "larger area" or "smaller are" column for both z scores and subtract the lower value from the higher value.

$\cdot$ If the z scores fall on both sides of the mean with the mean between them, use the "Proportion of scores between the mean and z column. Look up the value for both scores and add them together.

The scores fall on different sides of the mean, therefore you add them together, my problem is that the book uses different wording than your text book :mad: . Otherwise I could verify your answer. [EDIT] This happens to be incorrect when using "normal distribution"
July 6th 2006, 08:19 PM
CaptainBlack

Quote:

Originally Posted by Jimmy23

Hi, I'm trying to finish my homework and I'm stuck on a problem. Here it is: Suppose that the ages at a time of onset of a certain disease are normally distributed with a mean of 11.5 years and a standard deviation of 3 years.

In a population of 10,000 with the disease, how many would you expect to be between the ages of 7.5 and 13.5 years?

So here is what I did:

z=13.5-11.5 / 3 = 0.67

z=7.5-11.5 / 3 = -1.33

I look up those numbers in my book for the standard normal distribution and I get:
0.67 = .7486
-1.33 = .0918

So then I added .7486 and .0918 which gave me .8404 then I multiplied by 10,000 so I get (10,000)(.8404)=8,404 people between 7.5 and 13.5 have the disease. Did I do this right?

0.7486 is the area under the normal density to the left of z=0.67,
0.0918 is the area under the normal density to the left of z=-1.33.

The area between the z-scores is the difference between these areas: 0.7486-0.0918=0.6568.

RonL