Let’s solve each part of the problem:
A. The average rate of calls is 6 calls per ten minutes. To find the expected number of calls in 5 minutes, we first need to convert the rate to calls per minute:
6 calls / 10 minutes = 0.6 calls per minute.
Then, for 5 minutes:
Expected number of calls = (0.6 calls/minute) * 5 minutes = 3 calls.
So, the expected number of calls in 5 minutes is 3.
B. To find the probability that exactly 2 calls come and go unanswered, we’ll use the Poisson probability formula:
P(X = k) = (λ^k * e^(-λ)) / k!
Where:
• λ (lambda) is the average rate of calls (in this case, 3 calls in 5 minutes).
• k is the number of occurrences (in this case, 2 calls).
So, substituting the values:
P(X = 2) = (3^2 * e^(-3)) / 2!
= (9 * e^(-3)) / 2
≈ (9 * 0.0498) / 2
≈ 0.4482
Therefore, the probability that exactly 2 calls come and go unanswered is approximately 0.4482.
C. To find the probability that more than 3 calls go unanswered, we need to find the probability of 0, 1, 2, and 3 calls and subtract it from 1.
Let’s calculate the probability of 0, 1, 2, and 3 calls:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= (3^0 * e^(-3)) / 0! + (3^1 * e^(-3)) / 1! + (3^2 * e^(-3)) / 2! + (3^3 * e^(-3)) / 3!
= (1 * e^(-3)) + (3 * e^(-3)) + (9 * e^(-3)) / 2 + (27 * e^(-3)) / 6
= (e^(-3)) + (3e^(-3)) + (4.5e^(-3)) + (4.5e^(-3))
≈ 0.0498 + 0.1494 + 0.2241 + 0.2241
≈ 0.6474
So, the probability that more than 3 calls go unanswered is:
P(X > 3) = 1 - P(X ≤ 3) = 1 - 0.6474 = 0.3526.
Therefore, the probability that more than 3 calls go unanswered is approximately 0.3526.
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