Question

Suppose that calls arrive at a switchboard according to a Poisson process at an average rate of 30 per hour.

a)What is the probability that no calls arrive in a three-minute interval?

b) What is the probability that more than five calls arrive in a six-minute interval?

Answer 1

The probability of no calls arriving in a three-minute interval, given the average rate of 30 calls per hour, is e^-1.5. To find the probability of more than five calls in a six-minute interval, one must calculate the sum of probabilities for 0 to 5 calls and subtract from 1.

Step-by-step explanation:

The subject of this question involves understanding the Poisson distribution, which is used for modeling the number of times an event occurs in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

Part a) Probability of No Calls in a Three-Minute Interval

To calculate the probability of no calls occurring within a three-minute interval, we need to convert the average rate to the current interval of time. Since the average rate is 30 calls per hour, this is equivalent to 0.5 calls per minute (30 calls/hour divided by 60 minutes/hour). For a three-minute period, the average rate (λ) would be 1.5 (0.5 calls/minute multiplied by 3 minutes).

The formula for the Poisson probability is P(X = k) = (λk * e-λ) / k!, where k is the number of occurrences and λ is the average rate. In this case, we're looking for the probability of no calls, so k = 0.

Substituting the values into the formula, we get: P(X=0)= (1.50 * e-1.5) / 0! which simplifies to e-1.5. Therefore, the probability that no calls arrive in a three-minute interval is e-1.5.

Part b) Probability of More than Five Calls in a Six-Minute Interval

For a six-minute interval at the given rate, the average number of calls λ is 3 (0.5 calls/minute multiplied by 6 minutes).

To find the probability of more than five calls, we sum the probabilities of 0 through 5 calls occurring and subtract that sum from 1:

P(X > 5) = 1 - P(X ≤ 5) = 1 - ⢬( e-3 * λk / k! ) for k = 0, 1, 2, ... 5.

This means we need to calculate the probability for 0, 1, 2, 3, 4, and 5 calls using the Poisson formula, sum them up, and then subtract from 1 to get the probability of getting more than five calls in a six-minute interval.