Your calculations and doubts are correct. If you want to go further from your last step, notice that sum of independent binomials with the same probability $p$ [is binomial][1], so if we have $Z \sim \mathcal{B}(n, p)$ and $W \sim \mathcal{B}(n, p)$, then $Z + W \sim \mathcal{B}(n+m, p)$.

We are told that the probability $p$ is fixed:

> - The probability of a conversion (purchase) for each call is 4% .

But it is also said that

> - The typical call center employee completes on average 50 calls per day.

So it doesn't say that the number of calls is always 50. We are not given the distribution of the number of calls per employee, so you can only assume the average per employee. In such a case, the total is $\sum_i Y_i$ as you noticed.

However, if you were given the distribution, then the counts of calls would be also a random variable. In such a case, the model would be more complicated and the most practical solution would be probably to run a simulation where for each employee you first simulate the number of calls, then use it as a parameter for the binomial, so that you can simulate the number of purchases, and sum it for all employees. Repeating it many times would give you a distribution of interest.


  [1]: https://en.wikipedia.org/wiki/Binomial_distribution#Sums_of_binomials