A large service depot has a clientele of large compaies with extensive computer

Question

A large service depot has a clientele of large compaies with extensive computer equipment. The average number of service calls they receive every day over an 8-hour period is 3. They employ three highly trained technicians and other supporting staff. Each service call requires one technician for the entirety of the day on which they are called.

a) What is the probability that on a given day, there will be exactly 4 service calls?

b) What is the probability that there will be more than three calls in one day so that the three technicians will not be sufficient?

c) One day, there were no service calls in the morning and one technician was allowed to take the afternoon off. In the remaining four hours of the afternoon, what is the probability that there will be more than two service calls in the afternoon so that the service depot will not be able to meet the full service needs with only two technicians?

Explanation / Answer

a)

Assuming that the service calls they receive every day follows an exponential distribution, the pdf of the number of service calls per day (8 hr period) is given as,

P(X = x) = (1/3) exp(-x/3) (Mean of the distribution is 3)

where X denotes the number of  service calls they receive every day.

P(X = 4) = (1/3) exp(-4/3) = 0.0879

The probability that on a given day, there will be exactly 4 service calls is 0.0879

b)

The CDF of the distribution of number of  service calls receive every day is given as,

P(X > x) = exp(-x/3)

The probability that there will be more than three calls in one day = P(X > 3)

= exp(-3/3) = exp(-1) = 0.3679

(c)

As, the exponential distribution is memoryless distribution, the probability that there will be more than two service calls in the afternoon is same as the probability that there will be more than two service calls in 8 hr period

= P(X > 2) = exp(-2/3) = 0.5134

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