# Interpretation of Bernoulli and Binomial random variables in a call simulation center

The question I am asking has its reference here. I will list it out simply below. Since I am learning probability properly for the first time, there are a lot of confusion that I want to clarify.

Imagine that we are data scientists tasked with improving the ROI (Return on Investment) of our company's call center, where employees attempt to cold call potential customers and get them to purchase our product. You look at some historical data and find the following:

• The typical call center employee completes on average 50 calls per day.
• The probability of a conversion (purchase) for each call is 4% .
• The average revenue to your company for each conversion is 90 .
• The call center you are analyzing has 100 employees.
• Each employee is paid 200 per day of work.

Before everything, I tried to define the random variables.

## Bernoulli

Let $$X$$ be the success conversion call status of a call made by 1 single randomly chosen employee. In other words, let $$X$$ be the outcome of a call made by 1 single randomly chosen employee where the outcome is either success or failure in conversion.

Then we can model the probability of success of a call made by a randomly chosen employee as a Bernoulli random variable $$X$$ with parameter $$p$$ where $$p$$ is the probability of success of a call made by that randomly chosen employee.

$$\color{red}{\text{Take note how I emphasized the word single, as everything done here is by a single employee.}}$$

## Binomial

Now the above details a single call made by a randomly chosen employee, which follows a Bernoulli distribution where $$X \sim \text{Bernoulli}(0.04)$$.

Now the transition from Bernoulli to Binomial should be simple, but I have a big confusion.

A single employee can make multiple calls. For example, in the example above, an employee can make 50 calls a day. We can define a sequence of calls $$X_1, X_2, \ldots, X_{50}$$. One confusion that arise here is what each $$X_i$$ mean. In this particular settings, would it be reasonable to assume that each $$X_i$$ is a copy of $$X$$ where $$X$$ is the randomly chosen person. This means once $$X$$ is determined (tagged to the person chosen, say person A), then $$X_1, X_2, \ldots, X_{50}$$ all refer to this person A's calls as it does not make sense that $$X_i$$ corresponds to a different person.

In any case, since $$n=50$$ and $$p=0.04$$, we can define our $$Y$$ as follows:

$$Y \sim \text{Binomial}(50, 0.04)$$

where $$Y = X_1 + X_2 + \ldots + X_{50}$$ since each call made by a random employee is assumed to be i.i.d.

Now the second confusion (assuming there is no conceptual error before this, but I doubt it) is the interpretation of $$Y_i$$.

In this context, can we be unambiguous and say that each $$Y_i$$ should be the number of successful conversion calls made by a single randomly selected employee.

To be more precise, each experiment for $$Y_i$$ is like a new randomly drawn 50 calls by a random person, so each time is a realisation of 50 calls from a randomly selected person. So if you have 100 people you can have 100 $$Y_i$$. Note the distinction, in $$X_i$$, this $$X_i$$ is bind to a single person (may be wrong here), here each $$Y_i$$ can be any member from the population.

I understand this is a long question, but would like to seek some validation on whether my understanding above is outright wrong, or somewhat close. I am taking the course alone and won't be able to ask others to clarify conceptual understanding.

Your calculations and doubts are correct.

In this context, can we be unambiguous and say that each $$Y_i$$ should be the number of successful conversion calls made by a single randomly selected employee.

Yes. If you want to be precise, you could add "assuming the average number of calls".

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

• Thanks for this, so in general my understanding is roughly correctly. Yes, I got very confused because I am unsure which one is the random one, the call, or the employee. As you pointed out, the call can be also random, which can make this problem much more complicated. I went 1 step further, but I assumed that the revenue per employee per day on average should be $Z = 100Y$, but did not manage to deduce it follows $\sim \mathcal{B}(100n, p)$. Instead I just calculated the expectation $\mathbb{E}(Z)$.
– nan
Oct 21, 2022 at 8:57
• @nan both. A single call is a Bernoulli random variable, all calls per employee per day are a binomial random variable (sum of the calls). The total per call center can also be a random variable made of summing all the employees, etc.
– Tim
Oct 21, 2022 at 9:00
• Got it, thanks for clearing the air. The working looks simple but if I digged deep, there's a lot of confusions that came out like this.
– nan
Oct 21, 2022 at 9:05