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**mirrormirror** Suppose that X1, X2,....Xm and Y1,Y2,....,Yn are independent random samples, with the variables Xi normally distributed with mean $\displaystyle \mu _1$ and variance $\displaystyle \sigma{_1}^{2}$ and the variables Yi normally distributed with mean $\displaystyle \mu_2$ and variance $\displaystyle \sigma{_2}^{2}$. The difference between the sample means, $\displaystyle \bar{X} - \bar{Y}$ is then a linear combination of m + n normally distributed random variables and is itself normally distributed.

Suppose that $\displaystyle \sigma{_1}^{2}$ = 2, $\displaystyle \sigma{_2}^{2}$ = 2.5, and m=n. Find the sample sizes so that ($\displaystyle \bar{X} - \bar{Y}$) will be within 1 unit of ($\displaystyle \mu _1 - \mu _2$) with probability .95.

I don't know how to solve this one. I've tried to set up P(($\displaystyle \mu _1 - \mu _2) - 1 $$\displaystyle \preceq $$\displaystyle (\bar{X} - \bar{Y} )$$\displaystyle \preceq $ $\displaystyle (\mu _1 - \mu _2) + 1) $ = .95

but I don't if that's right. And, if so, how do I go on?