Question

1. 7 During the period of time that a local university takes phone-in registrations, calls come in at the rate of one every two minutes.
   1. What is the probability of receiving NO calls in a 10-minute period?
   2. What is the probability of receiving more than five calls in a 10-minute period?
   3. What is the probability of receiving less than seven calls in 15-minutes?
   4. What is the probability of receiving at least three but no more than 10 calls in 12 minutes?

USING EXCEL

Answer 1

If you have a Poisson process with parameter λ (λ is the average number of events occurring in an interval of given unit length), and if X is the number of events occurring in an interval of length t, then X has Poisson distribution with parameter λt. That is

Here λ = 0.5

a) t = 10, therefore mean() = λ * t = 5

In Excel use the function

POISSON.DIST(x,mean,cumulative)

where x .-> No. of events, mean = and cumulative which takes two options True or False.

True for probability inclusive of x and False for exactly x.

POISSON.DIST(0,5,FALSE) = 0.006738

b) We will find Prob for x between 0 & 5.

POISSON.DIST(5,5,TRUE) = 0.615

Required probability of more than 5 calls would be = 1 - 0.615 = 0.385

c) For 15 minutes,

Mean = = 15 * 0.5 = 7.5

so function would be

POISSON.DIST(7,7.5,TRUE) = 0.524

d) For 12 minutes,

= 12 * 0.5 = 6

First we need to find probability for calls between 0 to 10 using

POISSON.DIST(10,6,TRUE) = 0.957

Now we need to find probabilities of exactly 0, 1 & 2 calls in the same period. The sum of probability of all these three would need to be subtracted from prob of 0 to 10 calls to find probability of 3 to 10 calls.

0 call --> POISSON.DIST(0,6,FALSE) = 0.00247

1 call --> POISSON.DIST(1,6,FALSE) = 0.0148

2 calls --> POISSON.DIST(2,6,FALSE) = 0.044618

P( 3 to 10 calls) = 0.957 - 0.00247 - 0.0148 - 0.044618 = 0.895112