# Find CDF of the difference between two sample mean?

So here are 2 random samples drawn from the same population which has normal distribution with mean 80 and standard deviation 6. One with sample size 100 and the other 150. Denote the sample mean of the first sample as $$\overline{X}$$ and the second $$\overline{Y}$$. Want to know $$Prob(|\overline{X}-\overline{Y}| >2)$$?

I have calculated the $$\mathbb{E}(\overline{X}-\overline{Y})=\mathbb{E}(X)-\mathbb{E}(Y)=0$$ and $$Var(\overline{X}-\overline{Y})=\frac{Var(X)}{100}+\frac{Var(Y)}{150}=0.6$$. But I have no idea how to find the probability, without knowing the CDF? Or is there a way to calculate the CDF? Thanks.

Edit: $$\overline{X}-\overline{Y} \text{~} N(0, \sqrt{0.6})$$

Note that $$\bar{X}-\bar{Y}$$ follows a normal distribution of which you already know that mean ($$0$$) and standard deviation ($$\sigma$$).
\begin{align}Pr(|\bar{X}-\bar{Y}| > 2) &= Pr(\bar{X}-\bar{Y}>2) + Pr(\bar{X}-\bar{Y}<-2) \\ &=Pr\left(Z>\frac{2}{\sigma}\right) + Pr\left(Z<-\frac{2}{\sigma}\right)\\ &= 2Pr\left(Z<-\frac{2}{\sigma}\right)\end{align}
where $$Z$$ follows the standard normal distrbution. Computing the CDF should be doable at least numerically on many software or you can leave it in terms of CDF of the standard normal distribution.
• Hi, thanks for the answer! Can you briefly explain why $\overline{X}-\overline{Y}$ follows normal distribution? Except from that I got it. Thanks again. Commented Sep 27, 2019 at 6:07