Question

Find a Doctor is a small startup that helps people find a physician that best meets their needs (location, insurance accepted, etc.). During a "slow" time for them, they have 9 staff members taking calls from customers. On average, one call arrives every 5 minutes. On average, each staff member spends 18 minutes with each customer.

what is the probability that one of their staff members is busy? round answer to 1 decimal place

Answer 1

The probability that one of the staff members is busy at Find a Doctor startup is approximately 3.37%.

Step-by-step explanation:

In order to find the probability that one of the staff members is busy, we need to consider the rate at which calls arrive and the time it takes for a staff member to handle a call. The rate at which calls arrive is 1 call every 5 minutes, and each staff member spends 18 minutes with each customer. This means that on average, a staff member can handle 18/5 = 3.6 calls in an hour. Since they have 9 staff members, the total number of calls they can handle in an hour is 9 * 3.6 = 32.4 calls. However, since they receive calls at a rate of 1 call every 5 minutes, in an hour they can receive a maximum of 60/5 = 12 calls. Therefore, if they receive more than 12 calls in an hour, one of the staff members will be busy. The probability of this happening can be calculated as 1 minus the probability of receiving 12 calls or less in an hour. Using the Poisson distribution, we can calculate this probability as:

P(X ≥ 12) = 1 - P(X ≤ 11) = 1 - e^(-λ) * ∑(k=0 to 11) (λ^k / k!),

where λ is the average rate of calls per hour, which is 12, and e is Euler's number (approximately 2.71828). Evaluating this expression, we find:

P(X ≥ 12) ≈ 0.0337.

Therefore, the probability that one of the staff members is busy is approximately 0.0337 or 3.37%.