M/G/1 Queue Expected Waiting Time: W Calculation

Assume there is only one machine (server) at a factory that produces j types of products and $\lambda_j$ and $\mu_j$ are known individually.

For example, the 1st type of product has an arrival rate of $\lambda_1=0.10$ products/min and type 1 can be produced by the server with a service rate, $\mu_1=0.70$ products/min (deterministic). The 2nd type of product has an arrival rate of $\lambda_2=0.30$ products/min and type 2 can be produced by the server with a service rate, $\mu_2=0.40$ products/min (deterministic). By saying deterministic, I mean the service times for each j are constant and there is no variation within each type.

My questions:

In this case, would this queue be an M/G/1 queue?

If it is an M/G/1 queue, can you please explain how I can use the closed form formulation of $W$,

$$W=\frac{1}{\mu} +\frac{\lambda(\frac{1}{\mu^2}+CV^2)}{2(1-\frac{\lambda}{\mu)}}, where~~CV~is~Coefficient~of~Variation$$

to calculate the expected waiting time in the system?

How can I calculate $\lambda$ based on $\lambda_1$ and $\lambda_2$, similarly $\mu$ based on $\mu_1$ and $\mu_2$. Besides, how is CV calculated in such a situation?

EDIT1:

Well, it seems that from this answer: M/G/1 Queue or not?, this is an M/G/1 queue.

EDIT2:

On the other hand, from this answer: M/G/1 queuing system with two arrivals, it may not be an M/G/1 queue.

EDIT3:

Okay, I also knew that $\lambda=\lambda_1+\lambda_2$ and just learned how to calculate $\mu$ from the M/G/1 queuing system with two arrivals (though they had a little mistake), which is

$\frac{1} {\mu}=(\frac{λ_1}{λ_1+λ_2})\frac{1} {\mu_1}+(\frac{λ_2}{λ_1+λ_2})\frac{1}{\mu_2}$.

Assuming these calculations are true and this is an M/G/1 queue (which still needs a clarification), my question becomes:

How can I find $CV^2$?

EDIT4:

Actually, this post Queueing Delay(W) for M/D/1 queue with different value of service times kind of convinced me that my problem is an M/G/1 queue, furthermore, it has a specific name, called Multiclass M/G/1 queue. Also, I think I got the answer for $CV^2$ from the book "Analysis of Queues Methods and Applications" authored by Natarajan Gautam. Please correct me, if it is wrong.

$CV^2 = \frac1 \lambda \sum_{j=1}^J \lambda_j (\frac1 \mu)^2$

From the wiki page： The coefficient of variation (CV) is defined as the ratio of the standard deviation $\sigma$ to the mean $\mu$. $$CV=\frac {\sigma }{m}$$
CV I believe is for service time distribution, then let the service time be $t$, since $$\mu=\frac { \text{Unit Time} }{t}$$ then $$t = \frac { \text{Unit Time} }{\mu}$$ Unit Time in your example is 1 minute. \begin{align} CV & = \frac {\sigma }{m} \\ \sigma_{j} & = \sqrt{ \frac { \sum_{j=1}^{J}( t_{j}-{\overline {t} )^{2} } }{J-1} } \\ CV_{j} & = \frac { \sigma_{j} }{\overline { t_{j} }} \end{align}