Final answer:
The probability of receiving exactly 3 calls in each of two separate one-minute periods when calls follow a Poisson distribution with an average rate of 2.6 calls per minute is 0.050.
Step-by-step explanation:
The student's question involves calculating the probability of receiving exactly 3 calls in each of two separate one-minute periods, given that calls to a call center follow a Poisson distribution with an average of 2.6 calls per minute.
We will denote X as the number of calls received in one minute.
Given that X follows a Poisson distribution with a mean (λ) of 2.6, the probability of receiving exactly 3 calls in one minute is calculated using the formula P(X = k) = (λ^k * e^{-λ}) / k!, where k is the number of calls.
Therefore, we compute P(X = 3) for one-minute period:
P(X = 3) = (2.6^3 * e^{-2.6}) / 3! = 0.224.
To find the probability of this event occurring in both of the two one-minute periods independently, we multiply the probabilities for each period:
P(3 calls in each of two periods) = P(X = 3) * P(X = 3)
= 0.224 * 0.224
= 0.050176
Rounded to three decimal places, the probability is 0.050.