Confidence/Tolerance interval for a percentage of a population

Question

I have a problem I'm not sure how to solve. It goes something like:

Suppose we are manufacturing motherboards and we take a sample measuring their
areas: [400, 400.1, 400.2, 399.99, 399.1, 400.3, 399.2, 399.99]. Give a 99% 
tolerance interval that will contain 95% of all manufactured cards (suppose it 
is a normal distribution).

From the information Provided:

n = 8
X (Sample mean) = 399.86 
S (Sample STDEV) = 0.4525

Now, I know how to get 99% confidence interval for the mean using a small sample with the T distribution but I think that includes the whole population.

Question: How can I solve this problem?

I've found examples here and there but they are for Bernoulli distributions (success vs failure proportions).
The only relevant example I found is this, but I'm not sure how to use the given formulas to solve my problem.

Answer 1

This article/book has all the information required to solve the problem.
The tolerance interval can be found with:

Lower: X-KS
Upper: X+KS

Using the K table provided in the article:

K(n=8, alpha=0.05, gamma=0.99) = 4.287
Lower side: 399.86 - 4.287*0.4525 = 397.9201
Upper side = 399.86 + 4.287*0.4525 = 401.7999

Where alpha is obtained from:

100(1-alpha) = 95% of the motherboards
1-alpha = 0.95
alpha = 0.05

And gamma is the confidence level.