Probability over normal distribution Given a normal distribution with mean $\mu$ and unknown $\sigma$, what's the probability that, a sample of size $N$ will have mean in $[k-\epsilon, k+\epsilon]$  (as a function of $\sigma$)?
I can see that this is related to the problem of checking whether or not a coin is fair, but I can't see how to use that approach here.
 A: The distribution of the mean $\overline X$ of an i.i.d. sample from $N(\mu,\sigma^2)$ is $N(\mu, \sigma^2/n)$.  So we have
\begin{align}
& \Pr(k-\varepsilon<\overline X < k+\varepsilon) = \Pr\left( \frac{k-\varepsilon-\mu}{\sigma/\sqrt{N}} < \frac{\overline X-\mu}{\sigma/\sqrt{N}}< \frac{k+\varepsilon-\mu}{\sigma/\sqrt{N}} \right) \\[10pt]
= {} & \Pr\left( \frac{k-\varepsilon-\mu}{\sigma/\sqrt{N}} < Z < \frac{k+\varepsilon-\mu}{\sigma/\sqrt{N}} \right) = \Phi\left(\frac{k+\varepsilon-\mu}{\sigma/\sqrt{N}}\right) - \Phi\left(\frac{k-\varepsilon-\mu}{\sigma/\sqrt{N}}\right)
\end{align}
where $Z\sim N(0,1)$ and $\Phi$ is the c.d.f. of $N(0,1)$.
A: If $X$ is ${\cal N}(\mu_X, \sigma_X^2)$, $Y$ is ${\cal N}(\mu_Y, \sigma_Y^2)$
and $X,Y$ are independent then
$X+Y$ is ${\cal N}(\mu_X+\mu_Y, \sigma_X^2+\sigma_Y^2)$.
If $X_1,...,X_N$ are iid. with distribution ${\cal N} (\mu, \sigma^2)$ then it
is straightforward to obtain the distribution of $\sum_k X_k$. It is also
straightforward to scale the distribution to get the distribution of
${1 \over N} \sum_k X_k$.
