Question

The number of requests for assistance are received by a towing service at an average steady rate of 4 per hour. If, during a normal day the requests are received independently of one another, compute the probability that exactly ten requests are received during a particular 2-hour period of a normal day. Suppose, that the dispatcher ignored at least one call between 2:25 and 2:45pm. What is the probability that no more than one call was received during this time period? What is the probability that no more than one call is received between 2:45 pm and 3:00pm?

Answer 1

1)

expected number of calls in 2 hour period=λ =2*4 =8

probability that exactly ten requests =P(X=10) =e-8810/10! =0.0993

2)between 2:25 and 2:45pm

expected number of calls in 20 minutes between 2:25 and 2:45pm=4*20/60=4/3

probability that no more than one call was received during this time period

=P(X<=1) =P(X=0)+P(X=1)=e-4/3*(4/3)0/0!+e-4/3*(4/3)1/1! =0.6151

3)

expected number of calls in 15 minute  between 2:45 pm and 3:00pm =15*4/60=1

probability that no more than one call is received between 2:45 pm and 3:00pm:

=P(X<=1) =P(X=0)+P(X=1)=e-1*(1)0/0!+e-1*(1)1/1! =0.7358