Question: Suppose that a random sample of 16 observations is drawn from a normal distribution with mean μ and standard deviation 12; and that independently another random sample of 25 observations is drawn from a normal distribution with the same mean μ and standard deviation 20. Let $\bar{X}$ and $\bar{Y}$ denote the sample means of the two samples (i.e $\bar{X} = \frac{\sum_{i=1}^{16}X_i)}{16}$, $\bar{Y} = \frac{\sum_{i=1}^{25}y_i)}{25}$ ). Evaluate $Pr(|\bar{X}-\bar{Y}|\leq5))$ .
My attempt:
$E(\bar{X})=E(\bar{Y})= μ$
$var(\bar{X}) = 12^2/16$
$var(\bar{Y}) = 20^2/25$
Converting to standard normal distribution $z_x= (x-μ)/\sqrt{var(\bar{X})}$ and $z_y= (y-μ)/\sqrt{var(\bar{Y})}$. But, i'm not sure what to do next? I would really appreciate some help.
My book claims $\bar{X}-\bar{Y}$ has a mean of 0 and variance of 25, but i'm not sure how my book got those values.
If $U$ and $V$ are normally distributed and independent.
Then $Z:=U-V$ is normally distributed.
This with $\mu_Z=\mu_U-\mu_V$ and $\sigma^2_Z=\sigma^2_U+\sigma^2_V$.
Apply this on $U=\overline X$ and $V=\overline Y$.