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Let $Y_1<Y_2$ 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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  • $\begingroup$ The order statistics don't even matter in this case because you are considering their mean. $\endgroup$ – Emre Nov 14 '11 at 22:02
  • $\begingroup$ Any progress on this? $\endgroup$ – Did Nov 16 '11 at 15:19
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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 $P(-c<Z<c)= p$ 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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  • $\begingroup$ Let's say p = 0.5, what would you get for c? I get c to be 0.67 to satisfy this: P(−c<Z<c)= 0.5; however, my solutions for some reason get 0.65. Am I incorrectly doing this? Thanks. $\endgroup$ – icobes Nov 14 '11 at 23:38
  • $\begingroup$ The value of $c$ for which $\Pr(-c<Z<c)=0.5$ is the same as the value for which $\Pr(Z<c)=0.75$, and if I can trust the software package I'm using, that is 0.7733726. $\endgroup$ – Michael Hardy Nov 15 '11 at 17:58

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