Answer by Augustin for random variable follows

Let $Y$ a Bernoulli distributed random variable with parameter $p=0.6$, that is $P(Y=1)=1-P(Y=0)=0.6$.

Then you have the conditional distribution of $X_1$ given $Y=1$:

$$X_1\mid Y=1\sim\mathcal{N}(0,1)$$Considering the pdf:$$f_{X_1\mid Y=1}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$You also have:$$X_1\mid Y=0\sim\mathcal{N}(0,2)$$and$$f_{X_1\mid Y=0}(x)=\frac{1}{2\sqrt{\pi}}e^{-\frac{x^2}{4}}$$

Finally, the pdf of $X_1$ is given by:$$f_{X_1}(x)=P(Y=1)f_{X_1\mid Y=1}(x)+P(Y=0)f_{X_1\mid Y=0}(x)=0.6\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}+0.4\frac{1}{2\sqrt{\pi}}e^{-\frac{x^2}{4}}$$