The distribution of wages (in monetary units) of male workers in a large factory can be considered as N (μ, 4), and that of female workers as N (μ, 5). Two independent samples are drawn, one with 16 men and the other with 16 women. Determine $P(| \overline{X} - \overline{Y}| > 1)$
$\overline{X}={\sum_{i=0}^N{x_i}\over N}$ is a random variable, so we are able to calculate the expected value:
$E[\overline{X}] = E[{\sum_{i=0}^N{x_i}\over N}] = {1 \over N} E[\sum_{i=0}^N{x_i}] = {1 \over N} \sum_{i=0}^N{E[x_i]} $ Since E is a linear function.
Considering that $x_i$ is normally distributed $E[x_i]=\mu$. Then,
${1 \over N} \sum_{i=0}^N{E[x_i]} = {1 \over N} N*\mu = \mu$
In this way, we get $E[\overline{X}] = \mu$ so $\overline{X}$ is also a normally distributed variable.
However, I am not sure on what I am supposed to do next to solve the determining question of this problem, I was wondering if I could get some help.
You can use a similar argument to compute the variance of $\overline{X}$ since the sample is made of independent variables in the following way $$\text{Var}(\overline{X})=\text{Var}\left( \frac{\sum_{j=1}^{16}x_i}{16} \right)=\frac{16*4}{16^2}=\frac{4}{16}.$$ Moreover, you can compute the mean and variance for $\overline{Y}$ in the same way getting $$\text{E}(\overline{Y})=\mu\hspace{2mm}\text{ and }\hspace{2mm}\text{V}(\overline{Y})=\frac{16*5}{16^2}=\frac{5}{16}.$$ Now observe that since both $\overline{X}$ and $\overline{Y}$ are made of linear combinations of independant normal variables, they are also normal variables. Repeating the argument once more with $\overline{X}-\overline{Y}$ you can easily see that $$\overline{X}-\overline{Y}\sim N\left(0-0,\frac{4}{16}+\frac{5}{16}\right)=N\left(0,\frac{9}{16}\right).$$ Finally, since you have the distribution of $\overline{X}-\overline{Y}$ you can easily see that
$$P(|\overline{X}-\overline{Y}|>1)=2\left(1-P\left(\frac{(\overline{X}-\overline{Y})-0}{\sqrt{9/16}} < \frac{1-0}{\sqrt{9/16}} \right)\right)=2(1-P(Z<1))\approx 0.182422$$