a) $\int_{2}^{-1}\Phi (x)dx$, $\Phi$ - distribution $N(0,1)$
b) $\int_{2}^{-1}\frac{1}{\sqrt{2\pi}}e^{^{\frac{-x^{2}}{2}}}dx$
c) $\frac{1}{\sqrt{2\pi}}e^{^{\frac{-1^{2}}{2}}}+\frac{1}{\sqrt{2\pi}}e^{^{\frac{2^{2}}{2}}}$
d) $\Phi(2)+\Phi(1)$, $\Phi$ - distribution $N(0,1)$
e) $\Phi(2)+\Phi(1)-1$, $\Phi$ - distribution $N(0,1)$
So its Gaussian distribution with parameter $(-1,2)$
Formula for depth: $f(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{^{\frac{-(x-\mu )^{2}}{2\sigma^{2}}}}$ , $\mu \in R, \sigma>0$
I don't really understand this, I would bet on c) since it has pow $-1$ and $2$,
$e^{^{\frac{-1^{2}}{2}}}$ + $e^{^{\frac{2^{2}}{2}}}$ which are parameters of this distribution...
But there might be more then 1 correct answer so I am not sure.
It isn't c), because that is just the normal density evaluated at two points. The graph is clearly showing area under the curve, i.e., the integral of the density, which is represented by the CDF. Also, the parameters of the distribution are not $-1$ and $2$, since the peak of the curve is clearly at zero (indeed your first sentence says that it is a standard normal distribution).
Since $\Phi\left(x\right) = P\left(X \leq x\right)$ is the integral of the density from $-\infty$ to $x$, the area shown here can be written as
$$\Phi\left(2\right) - \Phi\left(-1\right).$$
However, that is not one of your options. But, by the symmetry of the normal distribution, $P\left(X < -x\right) = P\left(X > x\right)$. You can use this fact to convert the above expression into one of the given options.