Question

1. At a 24-hour computer repair facility broken-down computers arrive at an average rate of 3 per day, Poisson distributed.
   1. What is the probability that on a given day no computers arrive for repair?
   2. What is the probability that on a given day at least 3 computers arrive for repair?
   3. What is the distribution of the time between arrivals of computers to this facility and what is the average time between arrivals?
   4. On one particular day no computer has arrived for repair until noon. What is your best estimate of when the next computer will arrive for repair? Hint: forgetfulness property of the exponential distribution.
   5. Suppose a computer has just arrived for repair. What is the probability that the next one will not arrive for at least another 5 hours?

Answer 1

a) probability that on a given day no computers arrive for repair =e^-3*3⁰/0! =0.0498

b) probability that on a given day at least 3 computers arrive for repair =P(X>=3) =1-P(X<=2)

=1-(P(X=0)+P(X=1)+P(X=2)) =1-(e^-3*3⁰/0!+e^-3*3¹/1!+e^-3*3²/2!) =0.8647

c)

distribution of the time between arrivals of computers to this facility is exponential with mean time between arrival =1/3 days

or (24/3) =8 hours

d)

best estimate for  next computer will arrive for repair =* hour after noon or at 8.00 pm

e)

probability that the next one will not arrive for at least another 5 hours =P(T>5)=e^-5/8 =0.5353