(a) To find the probability of exactly five calls in a five-minute period, we can use the Poisson distribution formula. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space. In this case, the mean is given as six calls per five minutes. The formula for the Poisson distribution is P(x; λ) = (e^(-λ) * λ^x) / x!, where x is the number of events, λ is the mean, e is the base of the natural logarithm, and x! represents the factorial of x. Substituting the given values, we have P(5; 6) = (e^(-6) * 6^5) / 5!. Calculating this expression will give us the probability of exactly five calls in a five-minute period. (b) Similarly, to find the probability of exactly four calls in a four-minute period, we can use the Poisson distribution formula with a mean of six calls per five minutes. However, since we are now considering a four-minute period, we need to adjust the mean accordingly. To do this, we can use the fact that the mean number of calls per minute is λ/5. Therefore, the mean number of calls in a four-minute period would be (4/5) * (6/5) = 24/25. Using this adjusted mean, we can calculate P(4; 24/25) using the Poisson distribution formula. (c) To find the probability of more than two calls in a five-minute period, we can subtract the probability of two or fewer calls from 1. This can be calculated using the cumulative distribution function (CDF) of the Poisson distribution. The CDF of the Poisson distribution is given by P(X ≤ k; λ) = ∑(i=0 to k) (e^(-λ) * λ^i) / i!, where k is the number of events. Therefore, P(X 2; 6) = 1 - P(X ≤ 2; 6). (d) To find the probability of fewer than five calls in a five-minute period, we can calculate P(X

Question

(a) To find the probability of exactly five calls in a five-minute period, we can use the Poisson distribution formula. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space. In this case, the mean is given as six calls per five minutes.

The formula for the Poisson distribution is P(x; λ) = (e^(-λ) * λ^x) / x!, where x is the number of events, λ is the mean, e is the base of the natural logarithm, and x! represents the factorial of x.

Substituting the given values, we have P(5; 6) = (e^(-6) * 6^5) / 5!.

Calculating this expression will give us the probability of exactly five calls in a five-minute period.

(b) Similarly, to find the probability of exactly four calls in a four-minute period, we can use the Poisson distribution formula with a mean of six calls per five minutes. However, since we are now considering a four-minute period, we need to adjust the mean accordingly.

To do this, we can use the fact that the mean number of calls per minute is λ/5. Therefore, the mean number of calls in a four-minute period would be (4/5) * (6/5) = 24/25.

Using this adjusted mean, we can calculate P(4; 24/25) using the Poisson distribution formula.

(c) To find the probability of more than two calls in a five-minute period, we can subtract the probability of two or fewer calls from 1. This can be calculated using the cumulative distribution function (CDF) of the Poisson distribution.

The CDF of the Poisson distribution is given by P(X ≤ k; λ) = ∑(i=0 to k) (e^(-λ) * λ^i) / i!, where k is the number of events.

Therefore, P(X > 2; 6) = 1 - P(X ≤ 2; 6).

(d) To find the probability of fewer than five calls in a five-minute period, we can calculate P(X < 5; 6) using the CDF of the Poisson distribution. This will give us the probability of two or fewer calls, which we can subtract from 1 to get the desired probability.

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