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Let $Y_1 be order statistics from a random sample of size $2$ from a normal distribution, $\mathcal{N}(\mu,\sigma^2)$, where $\sigma^2$ is known. If $\bar{X}$ is the mean of this sample, how would I find the constant $c$ which solves this equation: $P(\bar{X} - c\sigma < \mu < \bar{X} + c\sigma)$. Thanks for the help.

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    Any progress on [this](http://math.stackexchange.com/q/33993/6179)?2011-11-16

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The expression $P(\bar{X} - c\sigma < \mu < \bar{X} + c\sigma)$ is not actually an equation until you put "$=\text{something}$" after it. Presumably you want it to be equal to some number between $0$ and $1$. Let's call that number $p$.

Notice that $\bar{X} = (Y_1+Y_2)/2$, and $\bar{X} \sim \mathcal{N}(\mu, \sigma^2/2)$. So $ \frac{\bar{X} - \mu}{\sigma/\sqrt{2}} \sim \mathcal{N}(0,1). $ Now find $c$ such that $P(-c if $Z\sim\mathcal{N}(0,1)$. You get this from a table or from software unless you want to go through a moderately elaborate discussion of how to do it numerically.

Then notice that $ -c < \frac{\bar{X}-\mu}{\sigma/\sqrt{2}} < c \text{ if and only if }\bar{X}-c\frac{\sigma}{\sqrt{2}} < \mu < \bar{X}+c\frac{\sigma}{\sqrt{2}}. $

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    The value of $c$ for which \Pr(-c is the same as the value for which \Pr(Z, and if I can trust the software package I'm using, that is 0.7733726.2011-11-15