Question

Telephone calls are received at an emergency 911 number as a non-homogeneous Poisson process such that, λ(t)-0.5 calls/hr for 0<ts? hr, λ (t)-0.9 calls/hr for 7<ts17 hr, and λ(t)-1.3 calls/hr for 17<ts24. a. Find the probability that there are no calls between 6 am and 8 am. b. Find the probability that there are at most 2 calls before noon. c. What is the probability that there is exactly one call between 4:50 pm and 5:10 pm? d. What is the PMF of the number of calls received in a day?

Answer 1

a )expected number of calls between 6am and 8 am =expected calls from 6am to 7 am +expected calls from 7am to 8am =0.5+0.9 =1.4

therefore P(no calls between 6 am and 8am) =P(X=0)=e^-1.4*1.4⁰/0! =0.2466

b)

expected number of calls before noon =expected calls from 0 am to 7 am +expected calls from 7am to 12 am =0.5*7+0.9*5 =8

P(at most 2 calls before noon) =P(X<=2) =P(X=0)+P(X=1)+P(X=2)

=e^-88⁰/0!+e^-88¹/1!+e^-88²/2!=0.0138

c)

expected number of calls between 4:50 pm and 5:10 pm (20 min) =1.3*20/60=1.3/3

P(one call between 4:50 pm and 5:10 pm) =P(X=1)=e^-1.3/3*(1.3/3)¹/1! =0.2809

d)

expected number of calls in a day (24 hours )=7*0.5+10*0.9+7*1.3 =21.6

therefore number of calls in a day follows Poisson distribution with parameter $\lambda$ =21.6

and pmf =P(X=x)=e^-21.6*21.6^x/x!