Skip to main content
Cross Validated

Return to Answer

added 183 characters in body
Source Link
Tim
Tim
  • 141.2k
  • 26
  • 270
  • 512

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 $Z + W \sim \mathcal{B}(n+m, p)$.

We 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.

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, 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.

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.

Source Link
Tim
Tim
  • 141.2k
  • 26
  • 270
  • 512

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, 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.