Are you completely sure the mean is 11.5? It seems to me like they would be asking for 10.5 instead.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?
WAIT!!!
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=.7486I look up those numbers in my book for the standard normal distribution and I get:
0.67 = .7486
-1.33 = .0918
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)
I agree with this. The answer is the difference .7486 - .0918 which is . 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.Originally Posted by Quick
NO!!! [EDIT] Actually Yes (I was wrong )Originally Posted by Jimmy23
Ahem, what I know about this problem I got from a book, and here is the exact quotes...
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 . Otherwise I could verify your answer. [EDIT] This happens to be incorrect when using "normal distribution"Originally Posted by Statistics Hacks
0.7486 is the area under the normal density to the left of z=0.67,Originally Posted by Jimmy23
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