Question

Between 2 and 4 pm the average number of calls per minute getting into the switch board of a company is 2.35. Find the probability that during one particular minute there will be at most 2 phones calls​

Answer 1

To solve this problem, we need to use the Poisson distribution, which describes the probability of a certain number of events occurring in a fixed interval of time or space, given the average rate of occurrence.

Let λ be the average number of calls per minute. From the problem statement, we have λ = 2.35.

Now we need to find the probability of having at most 2 phone calls in one minute. Let X be the random variable representing the number of phone calls in one minute. Then we have:

P(X ≤ 2) = e^(-λ) * (λ^0/0!) + e^(-λ) * (λ^1/1!) + e^(-λ) * (λ^2/2!)

Substituting λ = 2.35, we get:

P(X ≤ 2) = e^(-2.35) * (2.35^0/0!) + e^(-2.35) * (2.35^1/1!) + e^(-2.35) * (2.35^2/2!)

≈ 0.422

Therefore, the probability that during one particular minute there will be at most 2 phone calls is about 0.422, or 42.2%.