Given a sample of size n from a normal distribution, estimate the probability of picking a value larger than X from this distribution

Question

Say I picked 10 random samples from a normal distribution with unknown parameters: 1.7, 2.6, 3.0, 4.4, 1.6, 2.1, 2.4, 2.7, 5.2, 3.3

What is the probability that I will pick a value larger than 100.0 from this distribution?

Here is the process that I have in mind, but I don’t know how to do the second step:

Step 1. Find out the unbiased estimators of the population mean and the standard deviation by calculating the mean and the appropriate standard deviation of the sample.

Step 2. Evaluate the following integral: $$ \int_{\sigma=0}^{+\infty} \int_{\mu=-\infty}^{+\infty} P(X > 100 | \mu, \sigma, X\tilde~ N(\mu, \sigma) )f(\mu)g(\sigma) \,d\mu d\sigma $$

where $f$ and $g$ are the probability distribution functions of $\mu$ and $\sigma$ respectively.

$\frac{\mu - \bar X}{\sqrt{s^2/n}}$ follows Student's t-distribution with n-1 degrees of freedom: $t_{n-1}$ (In my example n=10, $\bar X$ is the sample mean, and $s$ is the unbiased standard deviation of the sample.)

So my questions are:
1. What distribution does $\sigma$ follow?
2. How to evaluate the integral shown above? Is there a closed form solution?

Thanks!

Answer 1

What I was looking for is Prediction Interval (wikipedia article: section 2.2.3: unknown mean, unknown variance). Thanks @whuber for guiding me in the right direction.

With that in mind, the specific question:

Say I picked 10 random samples from a normal distribution with unknown parameters: 1.7, 2.6, 3.0, 4.4, 1.6, 2.1, 2.4, 2.7, 5.2, 3.3

What is the probability that I will pick a value larger than 100.0 from this distribution?

could be answered as follows. Given that

$\frac{X_{n+1}-\overline{X}_n}{s_n\sqrt{1+1/n}} \sim t_{n-1}\,,$

where the sample mean and standard deviation are 2.9 and 1.148, and given the accompanying assumptions, the probability that $X_n$ exceeds 100 is $P(T>\frac{100-2.9}{1.148(1.1)})$ where $T\sim t_9$.