Finding Marginal Density functions with $Y\sim N_4(\mu,\Sigma)$ Suppose $Y$ is $N_4(\mu, \Sigma)$ where

$$\mu = ( 1,2,3,-2)'$$
  and
$$\Sigma =\begin{bmatrix} 
4& 2& -1& 2 \\
2& 6& 3& -2 \\
-1& 3& 5& -4 \\
2& -2& -4& 4
\end{bmatrix}$$

where $\Sigma$, the covariance matrix is a $4\times4$ matrix. 
(a) Find the marginal distribution of $Y_2$.
(b) Find the joint marginal distribution of $Y_1$ and $Y_3$.
(c) Find the distribution of $Z = Y_1 + 2Y_2 - Y_3 + 3Y_4$.
Some questions that I have:
1) What does $N_4$ mean?
Edits
2) For part a, is it right to say $Y_2$ is of $N(2, 6)$ from the above $\Sigma$ and $\mu$? I am unsure how the marginal distribution is gotten for $Y_2$ from a group of $Y_1,Y_2,Y_3,Y_4$, inside the vector $Y$.
 A: $N_4(\mu,\Sigma)$ is the pdf of the multivariate normal distribution. The marginal distributions are normal distributions whose parameters are given in the covariance matrix $\Sigma$ and in the expectation vector $\mu$.
If you recall the definition the entries of the covariance matrix, $\Sigma$, $$\sigma_{i,j}=E[(Y_i-\mu_i)(Y_j-\mu_j)].$$ and $$\mu_i=E[Y_i].$$ 
Specifically, the variance of $Y_2$, $\sigma_{Y_2}^2=\sigma_{2,2}=6$ and the expectation of $Y_2$, $m_{Y_2}=\mu_2=2$. So, the distribution of $Y_2$ is $N_1(2,6)$.
The common distribution of $Y_1$ and $Y_3$ is also normal ($N_2(\mu_{Y_1,Y_3},\Sigma_{Y_1,Y_3})$ with the covariance matrix
$$\Sigma_{Y_1,Y_3}=
\begin{bmatrix}
4&-1\\
-1&5\\
\end{bmatrix}$$
and expectation vector 
$$\mu_{Y_1,Y_3}=(1,3)'.$$
As far as the distribution of $Z = Y_1 + 2Y_2 - Y_3 + 3Y_4$; it is normal with expectation
$$m_Z=E[Z]=\mu_1+2\mu_2-\mu_3+3\mu_4=-4.$$
The variance of $Z$ can be easily calculated now considering that
$$\sigma_Z^2=E[Z^2]-E^2[Z].$$
($E[Z^2]=E[(Y_1 + 2Y_2 - Y_3 + 3Y_4)^2]$.)
Or
$$\sigma_Z^2=E[Z-m_Z)^2]=E[(Y_1-\mu_1)+2(Y_2-\mu_2)-(Y_3-\mu_3)+3(Y_4-\mu_4))^2].$$
After squaring we are going to have terms that can be calculated. For instance:
$$E[((Y_1-\mu_1)2(Y_2-\mu_2)]=2\sigma_{1,2}=2\cdot2=4.$$
