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

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!

• First, the limits on $\sigma$ should be from 0, not from -\infty$... This problem is natural to formulate in a Bayesian way, then you can just do a numerical integration inR ... Feb 13, 2015 at 18:43 •$\sigma$does not have a distribution unless you assume it's a random variable, in which case you need to assume a prior distribution for it. Regardless, search our site for normal prediction interval. Incidentally, if it is this particular dataset you are interested in, the answer is zero (to an accuracy of around$10^{-1500}\$).
– whuber
Feb 13, 2015 at 18:57

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 $$T\sim t_9$$.