# Help finding P($\overline{Y} \geq \overline{Y}^*$)

Let $Y_1, Y_2,...,Y_9$ be a random sample size of $9$ from a normal distribution where $\mu=2$ and $\sigma=2$. Let $Y_{1}^*, Y_{2}^*,...,Y_{9}^*$ be an independent random sample from a normal distribution having $\mu=1$ and $\sigma=1$. Find P($\overline{Y} \geq \overline{Y}^*$).

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hint:$$\overline{Y} - \overline{Y}^* is \quad normal\quad distribution$$ then compute $$P(\overline{Y} \geq \overline{Y}^*_ =P(\overline{Y} - \overline{Y}^*\geq0)$$
This helps a bit but I'm really unsure how to find $P(\overline{Y} - \overline{Y}^* \geq 0)$. – user59633 Jan 28 '13 at 3:55
Let $W$ be the difference we are worried about. For typing convenience we write $W=S-T$. As was pointed out above, $W$ has normal distribution. I am sure you can find the mean of $W$. Now we find the variance of $W$. The variance of $S$ is $\frac{4}{3}$. The variance of $T$ is $\frac{1}{3}$. Therefore the variance of $W$ is $\frac{4}{3}+\frac{1}{3}$. (Yes, I really mean plus.) So the standard deviation of $W$ is $\sqrt{5/3}$. Now I think you can handle the rest. – André Nicolas Jan 28 '13 at 4:32