Question

An automated phone system can answer three calls in a five-minute period. Assume that calls occur at an average rate of 1.2 every five minutes and follow a Poisson probability distribution. Calculate the probability that no calls will occur during the next ten minutes.

Answer 1

The probability that no calls will occur during the next ten minutes is 0.0907.

Explanation:

Poisson distribution:

Poisson distribution is a statistical distribution that helps to find out the number of events is likely occur in a specific time period.

`P(X=x)=(e^(-\lambda t)(\lambda t)^x)/(x!)`

Given that,

Calls occur at an average rate of 1.2 every 5 minutes.

The average rate of call is
`(1.2)/(5)=0.24` per minute.

Here,

`\lambda =0.24`, t=10 and x=0

`\therefore P(X=0)=(e^(-0.24* 10)(0.24* 10)^0)/(0!)`

=0.0907

The probability that no calls will occur during the next ten minutes is 0.0907.