a)Incoming phone calls generally are thought to be Poisson distributed. If an op

Question

a)Incoming phone calls generally are thought to be Poisson distributed. If an operator averages 2.2 phone calls every 30 seconds, what is the expected (average) amount of time between calls (in seconds)?

b)Incoming phone calls generally are thought to be Poisson distributed. If an operator averages 2.2 phone calls every 30 seconds, what is the probability that a minute or more would elapse between incoming calls?

c)incoming phone calls generally are thought to be Poisson distributed. If an operator averages 2.2 phone calls every 30 seconds, what is the probability that at least two minutes would elapse between incoming calls?

Explanation / Answer

Poisson Distribution

Solution:

Here, we have to find the expected value for the amount of time between calls.

We are given a average phone calls = 2.2

We know that the average value of the Poisson distribution is equal to the expected value of the Poisson random variable X.

That is, E(X) = µ

E(X) = 2.2

Required answer = 2.2

Solution:

Here, we have to find the probability that a minute or more would elapse between incoming calls.

P (X1) = 1 – P(X=0)

The average for 30 seconds is given as 2.2 phone calls

So, average for 1 minute = 2.2*2 = 4.4

P(X=0) =[ (e^(-4.4)*4.4^0] / 0!

P(X=0) = 0.012277*1/1 = 0.012277

P (X1) = 1 – P(X=0) = 1 – 0.012277 = 0.987723

Required Answer = 0.987723

Solution:

Here, we have to find the probability that at least two minutes would elapse between incoming calls.

P (X2) = 1 – [ P(X=0) + P(X=1) ]

P (X2) = 1 – P(X=0) - P(X=1)

The average for 30 seconds is given as 2.2 phone calls

So, average for 1 minute = 2.2*2 = 4.4

P(X=0) =[ (e^(-4.4)*4.4^0] / 0!

P(X=0) = 0.012277*1/1 = 0.012277

P(X=1) =[ (e^(-4.4)*4.4^1] / 1! = 0.05402

P (X1) = 1 – P(X=0) - P(X=1) = 1 – 0.012277 - 0.05402 = 0.933702

Required Answer = 0.933702

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