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2.9 The Exponential Distribution

The exponential distribution, also called the negative exponential distribution, is used in dealing with queuing problems. The exponential distribution often describes the time required to service a customer. The exponential distribution is a continuous distribution. Its probability function is given by

f (X) = μ e^{− μ x}

$f (X) = μ e^{− μ x}$ (2-14)

where

$X = random variable (service times)$ $X = random variable (service times)$
$μ = average number of units the service facility can handle in a specific period of time$ $μ = average number of units the service facility can handle in a specific period of time$
$e = 2.718 (the base of the natural logarithm)$ $e = 2.718 (the base of the natural logarithm)$

The general shape of the exponential distribution is shown in Figure 2.16. Its expected value and variance can be shown to be

Expected value = \frac{1}{μ} = Average service time

$Expected value = \frac{1}{μ} = Average service time$ (2-15)

Variance = \frac{1}{μ^{2}}

$Variance = \frac{1}{μ^{2}}$ (2-16)

As with any other continuous distribution, probabilities are found by determining the area under the curve. For the normal distribution, we found the area by using a table of probabilities. For the exponential distribution, the probabilities can be found using the exponent key on a calculator with the formula below. The probability that an exponentially distributed time, X, required to serve a customer is less than or equal to time t is given by the formula

P (X \leq t) = 1 - e^{− μ t}

$P (X \leq t) = 1 - e^{− μ t}$ (2-17)

The time period used in describing $μ$ $μ$ determines the units for the time t. For example, if $μ$ $μ$ is the average number served per hour, the time t must be given in hours. If $μ$ $μ$ is the average number served per minute, the time t must be given in minutes.

The y axis of this graph is labelled f open parens x closed parens. The X axis is labelled, simply, X.

Figure 2.16 Full Alternative Text

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 will be $^{1} /_{2}$ $^{1} /_{2}$ hour or less? Using Equation 2-17, we have

$X = exponentially distributed service time$ $X = exponentially distributed service time$
$μ = average number that can be served per time period = 3 per hour$ $μ = average number that can be served per time period = 3 per hour$
$t =^{1} /_{2} hour = 0.5 hour$ $t =^{1} /_{2} hour = 0.5 hour$
$P (X \leq 0.5) = 1 - e^{− 3 (0.5)} = 1 - e^{− 1.5} = 1 - 0.2231 = 0.7769$ $P (X \leq 0.5) = 1 - e^{− 3 (0.5)} = 1 - e^{− 1.5} = 1 - 0.2231 = 0.7769$

Figure 2.17 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 $^{1} /_{3}$ $^{1} /_{3}$ hour or $^{2} /_{3}$ $^{2} /_{3}$ hour, as follows:

$P (X \leq \frac{1}{3}) = 1 - e^{− 3 (\frac{1}{3})} = 1 - e^{− 1} = 1 - 0.3679 = 0.6321$ $P (X \leq \frac{1}{3}) = 1 - e^{− 3 (\frac{1}{3})} = 1 - e^{− 1} = 1 - 0.3679 = 0.6321$
$P (X \leq \frac{2}{3}) = 1 - e^{− 3 (\frac{2}{3})} = 1 - e^{− 2} = 1 - 0.1353 = 0.8647$ $P (X \leq \frac{2}{3}) = 1 - e^{− 3 (\frac{2}{3})} = 1 - e^{− 2} = 1 - 0.1353 = 0.8647$

A line graph showing a sloped line moving from the top of the y axis down to the x axis. The first section of the graph is shaded and labelled zero point seven seven six nine. — Figure 2.17 Probability That the Mechanic Will Install a Muffler in 0.5 Hour

Figure 2.17 Full Alternative Text

A screenshot showing the Excel output for exponential distribution when the random variable is time. — Program 2.5A Excel 2016 Output for the Exponential Distribution

Figure 2.5A Full Alternative Text

A screenshot showing the Excel formulas used to calculate exponential distribution in the prior figure. — Program 2.5B Function in an Excel 2016 Spreadsheet for the Exponential Distribution

Figure 2.5B Full Alternative Text

While Equation 2-17 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 complementary. 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 \leq 0.5) = 1 - 0.7769 = 0.2231

$P (X > 0.5) = 1 - P (X \leq 0.5) = 1 - 0.7769 = 0.2231$

Programs 2.5A and 2.5B illustrate how a function in Excel can find exponential probabilities.

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Table of Contents for
2.9 The Exponential Distribution

2.9 The Exponential Distribution

Figure 2.16 Exponential Distribution

Arnold’s Muffler Example

Figure 2.17 Probability That the Mechanic Will Install a Muffler in 0.5 Hour

Program 2.5A Excel 2016 Output for the Exponential Distribution

Program 2.5B Function in an Excel 2016 Spreadsheet for the Exponential Distribution

Table of Contents for 2.9 The Exponential Distribution

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Table of Contents for
2.9 The Exponential Distribution