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

Answer:

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

Step-by-step 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.

$P(X=x)=\frac{e^{-\lambda t}(\lambda t)^x}{x!}$

Given that,

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

The average rate of call is $\frac{1.2}{5}=0.24$ per minute.

Here,

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

$\therefore P(X=0)=\frac{e^{-0.24\times 10}(0.24\times 10)^0}{0!}$

                  =0.0907

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