Question

The modeling and solution of problems concerning motel reservation networks for motels were studied. Researchers defined a type i call to be a ca a motel's computer terminal to the national reservation center. For a certain motel, the number, x, of type 1 calls per hour has a poison distributio parameter à = 1.9. Complete parts (a) through (e) below. a. Determine the probability that the number of type 1 calls made from this motel during a period of 1 hour will be exactly five. The probability that exactly five type 1 calls are made is 0.3347 (round to three decimal places as needed.)

Answer 1

To find the probability that the number of type 1 calls made from the motel during a period of 1 hour will be exactly five, we can use the Poisson distribution with a lambda value of 1.9.

Step-by-step explanation:

To find the probability that the number of type 1 calls made from the motel during a period of 1 hour will be exactly five, we can use the Poisson distribution. The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space. It is characterized by a single parameter, lambda (λ), which represents the average rate of events per interval.

In this case, the average rate of type 1 calls per hour is given as 1.9. So, λ = 1.9. To find the probability of exactly five type 1 calls, we can use the Poisson probability mass function:

P(X = k) = (λ^k * e^(-λ))/k!

Substituting the values, we have:

P(X = 5) = (1.9^5 * e^(-1.9))/5!

Calculating this probability will give us the desired result.

Answer 2

The probability that exactly five type 1 calls are made is 0.3347.

Step-by-step explanation:

To determine the probability that exactly five type 1 calls are made, we can use the Poisson probability mass function (PMF):

P(X = k) = (λ^k * e^(-λ))/k!

where:

P(X = k) is the probability of k occurrences of the event (type 1 calls)

λ is the average number of occurrences (1.9 calls per hour)

k is the specific number of occurrences (5 calls)

e is the base of the natural logarithm (approximately 2.718)

k! is the factorial of k

Plugging in the values, we get:

P(X = 5) = (1.9^5 *
`e^(^-^1^.^9^)`)/5!

P(X = 5) = (23.689 * 0.1477)/120

P(X = 5) = 0.3347

Therefore, the probability that exactly five type 1 calls are made is 0.3347.