Question

The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average there are 10 calls per hour. (a) What is the probability that there are exactly 5 calls in one hour? (b) What is the probability that there are 3 or fewer calls in one hour? (c) What is the probability that there are exactly 15 calls in two hours? (d) What is the probability that there are exactly 5 calls in 30 minutes?

Answer 1

Answer with Step-by-step explanation:

We know from Poisson distribution

The probability that a random process with an average arrival rate of λ occurs 'n' times in time interval of 't' is given by

`P(n,t)=((\lambda t)^ne^(-\lambda t))/(n!)`

Part a)

The probability for 5 calls in 1 hour is

`P(5,1)=((10* 1)^(5)e^(-10* 1))/(5!)\\\\P(5,1)=0.0378`

Part b)

Probability of 3 or fewer calls occurs in 1 hour is the sum of the following probabilities

1) Only 1 call occurs in 1 hour.

2) Only 2 calls occurs in 1 hour.

3) Only 3 calls occurs in 1 hour.

4) There is no cal in 1 hour.

Thus we can write

`P(n<3,1)=((10* 1)^0e^(-10))/(0!)+((10* 1)^1e^(-10))/(1!)+((10* 1)^2e^(-10))/(2!)+((10* 1)^3e^(-10))/(3!)\\\\P(n<3,1)=0.0103`

Part c)

The probability for 15 calls in 2 hours is

`P(15,2)=((10* 2)^(15)e^(-10* 2))/(15!)\\\\P(15,2)=0.051`

Part d)

The probability for 5 calls in 30 minutes or 0.5 hours is

`P(5,0.5)=((10* 0.5)^(5)e^(-10* 0.5))/(5!)\\\\P(5,0.5)=0.175`