Question

The number of wrongly dialed phone calls you receive can be modeled as a Poisson process with the rate of one per month. For simplicity, assume there are four weeks in one month.

a. Find the probability that it will take between two and three weeks to get the first wrongly dialed phone call.
b. Suppose that you have not received a wrongly dialed phone call for two weeks. Find the expected value and variance of the additional time until the next wrongly dialed phone call.

Answer 1

Part a: P (14/30<x<21/30) = 0.1304

Partb: Expected value= Variance= 0.866

Explanation:

The poisson distribution is given by

P(x)= μˣ . e^ -u/ x!

In this question

x= 1

n= 30

μ= 1/30

P(x)= μˣ . e^ -u/ x!

= 0.333. e⁻¹/³⁰/1!

= 0.33*0.967/1

= 0.32208

For 2 weeks

x= 14

n= 30

μ= 14/30

P(x)= μˣ . e^ -u/ x!

= 0.467. e⁻¹⁴/³⁰/14!

= 0.467*0.627/14!

= 0.29285/14!

=3.359*e⁻¹²

= 0.0139

For 3 weeks

x= 21

n= 30

μ= 21/30

P(x)= μˣ . e^ -u/ x!

= 0.7* e⁻²¹/³⁰/21!

= 0.7*0.4965/21!

= 0.3476/21!

=6.8037*e⁻²¹

= 0.11648

Part a:

P (14/30<x<21/30) = P (x= 14/30) + P (x=21/30)

(0.0139 +0.11648) =0.1304

Part b:

Probability of Not receiving the call for two weeks = 1- P (x= 14/30)=0.9861

The mean and the variance of the Poisson distribution are equal to μ

For x= 14

Expected value not getting wrongly dialed phone call in 2 week = μ= 1-14/30 = 1-0.467= 0.533

Variance of not getting wrongly dialed phone call in 2 week= μ=1- 14/30= 1-0.467= 0.533

Expected value of the additional time until the next wrongly dialed phone call

Expected value not getting wrongly dialed phone call in 2 weeks + Expected value of next wrongly dialed phone call in a month

= 0.533+0.33=0.866

Variance of the additional time until the next wrongly dialed phone call

= Expected value of the additional time until the next wrongly dialed phone call =0.866