Let $x_1, \ldots, x_n$ be a sample from a normal population having unknown mean and variance. Let $\bar{x}$ be the average of the first n of them.
- What is the distribution of $x_{n+1} - \bar{x}$?
- If $\bar{x}=4$, give an interval that, with 90% confidence, will contain the value of $x_{n+1}$.
The distribution of $\bar{x}$ would be normal with mean equal to the population mean and standard deviation equal to the population standard deviation divided by $\sqrt{n}$. But how do we find the distribution of $x_{n+1}$?
We could say that the distribution of $x_1+x_2+\ldots+ x_{n+1}$ would be normal with mean $\bar{x}_{n+1}$ and variance $S^2_{n+1}$ (sample variance). Assuming that all of these variables are identically distributed and independent, we could say that the distribution of $x_{n+1}$ could be normal with $\bar{x}_{n+1}/(n+1)$ and $S^2_{n+1}/(n+1)$ using the central limit theorem.If this approach is correct how do I proceed from here?