Application of MAP criteria
The MAP criterion minimizes the average probability of error, which in your case reads
\begin{equation}
\begin{aligned}
P_e &amp;= P \left( -1 \right) P \left( Y&gt;\theta \mid -1 \right) + 
P \left( +1 \right) P \left( Y&lt;\theta \mid +1 \right) \\
&amp;= 0.2 \int_{\theta}^{+1} f_Y \left( y \mid -1 \right) \mathit{dy} + 
0.8 \int_{-1}^{\theta} f_Y \left( y \mid +1 \right) \mathit{dy} \,.
\end{aligned}
\end{equation}
Therefore, the optimum threshold, call it $\hat{\theta}$, is optimum in terms of minimizing $P_e$.
Exploiting the previous equation and the definitions of the random variables $X$ and $N$, you get
\begin{equation}
\begin{split}
P_e &amp;= 0.2 \times 0.25 \times \left(1-\theta\right) + 
0.8 \times 0.25 \times \left(1+\theta\right) \\
&amp;= \frac{1}{20} \times \left(1-\theta\right) + 
\frac{4}{20} \times\left(1+\theta\right) .
\end{split}
\end{equation}
Minimizing $P_e$ w.r.t. $\theta$, you obtain $\hat{\theta} = -1$.
An easier way to solve your problem would be: look at the plot, and select $\hat{\theta}$ such that it minimizes the sum between the right tail of the left function (i.e., the grey area) and the left tail of the right function. In this way you easily get the solution, but the analytical way is, in my opinion, preferable to understand the concepts.