View Full Version : Central Limit Theory?
11-28-2005, 06:44 PM
Here's my problem:
In a large population, 12.5% of all individuals have diabetes.
(a) Seven of these individuals are sampled at random. What is the probability
that two or more sampled individuals have diabetes?
(b) Seven hundred of these individuals are sampled at random. What is the
probability that eighty or fewer have diabetes? (Use a convenient approxi-
mation and justify it)
I solved a) using the binomial distribution and calculating 1-P(X=1)-P(X=0).
But I am not sure that is how you do it.
For b) I suppose I have to go with the central limit theory, but I have no idea how to apply it.
Can you help out, please?
I have been browsing for hours looking for some solved examples like this but I haven't found anything yet.
11-28-2005, 06:55 PM
(a) is correct
for (b), use the normal approximation to the binomial - we've definitely covered it on this site - if you can't find it here, your textbook definitely covers it....
11-28-2005, 07:23 PM
what I was trying to do was:
then I get stuck on how to calculate the z-value:
1st question: since the probability is for "eighty or fewer", how do I state it?
and if I want to calculate the probability without using the z-table, I just substitute the z value in the probability density function for the normal, right?
sorry if I am asking something too obvious, I am tired...
11-28-2005, 07:31 PM
z = (80-87.5)/8.75 = -0.857
Then find the proportion of the area under the normal curve below z. Usually the tables give you this directly.
11-28-2005, 07:54 PM
I just noticed that you do not use sqrt(n) in your z calculations. why is that?
My book has it like this:
z=sqrt(n) * (Xbar - mean)/s
but multiplying my previous answer by sqrt(n) gives me an impossible value (-22.67)
what is the difference between your calculation and my book?
thank you so much once again!
11-28-2005, 08:03 PM
z = sqrt(n) * (Xbar - mean)/s
also written as z = (Xbar - mean) / (s/sqrt(n))
is used when you're determining the probability of getting a particular sample mean
- in this particular case, we are trying to find the probability of a proportion, so we use:
mean = n*p
std dev = sqrt(n*p*q)
and it follows a normal distribution quite closely
11-28-2005, 09:36 PM
Thank you for your help, JohnM.
Your explanations also helped me to work out some of the other exercises I had to solve.
I still have some problems but on another subject: maximum likelihood estimation.
I am given a distribution function of a random variable X:
F(x) = 1 - x^-a
and I have to find the maximum likelihood estimator of a.
My approach would be to:
and then i'm not sure what to do next...
d/dx F(x) = a*x^(-a-1) (is this correct?)
L(a)=n*log a + SUM((-a-1)log xi)
that is as far as i am right now.
thank you for your help. hope you can understand my notations.
11-28-2005, 10:35 PM
Below are examples of MLE using normal and exponential distributions.
Maximum likelihood estimation (http://www.weibull.com/LifeDataWeb/maximum_likelihood_estimation_appendix.htm)
Your d/dx F(x) is correct.
I think after obtaining L(a), you calculate d/da L(a) then let it be zero and solve for a. The solution is the MLE.
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