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Arnold's Muffler Example

Arnold's Muffler Shop installs new mufflers on automobiles and small trucks. The mechanic can install new mufflers at a rate of about three per hour, and this service time is exponentially distributed. What is the probability that the time to install a new muffler would be an hour or less? Using Equation 2-19 we have

X - exponentially distributed service time

= average number that can be served per time period = 3 per hour

t = 1/2hour = 0.5 hour

P(X 0.5) = 1 e^-3(0.5)= 1 - e ^-1.5 = 1 - 0.2231 = 0.7769

Figure 2.18 shows the area under the curve from 0 to 0.5 to be 0.7769. Thus, there is about a 78% chance the time will be no more than 0.5 hour and about a 22% chance that the time will be longer than this. Similarly, we could find the probability that the service time is no more 1/3 hour or 2/3 hour, as follows:

P(X 1/3) = 1 - e^-3(1/3)= 1 - e^-1= 1 - 0.3679 = 0.6321

P(X 1/3) = 1 - e^-3(2/3)= 1 - e^-2= 1 - 0.1353 = 0.8647

While Equation 2-19 provides the probability that the time (X) is less than or equal to a particular value t, the probability that the time is greater than a particular value t is found by observing that these two events are complimentary. For example, to find the probability that the mechanic at Arnold's Muffler Shop would take longer than 0.5 hour, we have

P(X > 0.5) = 1 - P(X 0.5) = 1 - 0.7769 = 0.2231.

Exponential distribution - the random variable (X) is time
Average number per time period = =	3	per hour
t =	0.5000	hours
P(X) t) =	0.7769
P(X) t) =	0.2231

In the Arnold's Muffler example for the exponential distribution in this chapter, the average rate of service was given as 3 per hour, and the times were expressed in hours. Convert the average service rate to the number per minute and convert the times to minutes. Find the probabilities that the service times will be less than 1/2 hour, 1/3 hour and 2/3 hour. Compare these probabilities to the probabilities found in the example.