Question

PLEASE DONT COPY AND PASTE AN ANSWER EXPLAIN

The number of requests for assistance received by a towing service is a Poisson process with a mean rate of 5 calls per hour.

a. What is the probability that 3 or more requests are received during a one hour period?

b. If the operator of the towing service takes a 30-minute break for lunch, what is the probability that they do not miss any requests for assistance?

Answer 1

a)

expected number of calls per hour =5 =

from Poisson distribution:

probability that 3 or more requests are received during a one hour period =P(X>=3)=1-P(X<=2)

=1-(P(X=0)+P(X=1)+P(X=2))

=1-(e^-55⁰/0!+e^-55¹/1!+e^-55²/2!)

=1-(0.0067+0.0337+0.0842)

=0.8753

b)

expected number of calls per 30 minute=5/2 =2.5 =

P( they do not miss any requests for assistance in 30 minute break) =P(0 calls in 30 minute) =(e^-2.5*2.5⁰/0!)

=0.0821