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p139_exercises_4-8

p139_exercises_4-8 - -‘ e 4—24 Lack of;a> property of...

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Unformatted text preview: -‘ e 4—24 Lack of ;a> property of 1' nential 1', ution. if n Lack of Memory ., Property 4—8 EXPONENTIAL DISTRIBUTION 139 Example 4-22 illustrates the lack of memory property of an exponential random vari- able and a general statement of the property follows. In fact, the exponential distribution is the only continuous distribution with this property. For an exponential random Variable X, P(X< t1 + tle> t1) = P(X< t2) Figure 4-24 graphically illustrates the lack of memory property. The area of region A divided by the total area under the probability density function (A + B + C + D = 1) equals P(X < t2). The area of region C divided by the area C + D equals P(X < t1 + t2 I X > t1). The lack of memory property implies that the proportion of the total area that is in A equals the proportion of the area in C and D that is in C. The mathematical verification of the lack of memory property is left as a mind-expanding exercise. The lack of memory property is not that surprising when you consider the development of , a Poisson process. In that development, we assumed that an interval could be partitioned into small intervals that were independent. These subintervals are similar to independent Bernoulli trials that comprise a binomial process; knowledge of previous results does not affect the prob- abilities of events in future subintervals. An exponential random variable is the continuous ana- log of a geometric random variable, and they share a similar lack of memory property. The exponential distribution is often used in reliability studies as the model for the time un- til failure of a device. For example, the lifetime of a semiconductor chip might be modeled as an exponential random variable with a mean of 40,000 hours. The lack of memory property of the exponential distribution implies that the device does not wear out. That is, regardless of how long the device has been operating, the probability of a failure in the next 1000 hours is the same as the probability of a failure in the first 1000 hours of operation. The lifetime L of a de- vice with failures caused by random shocks might be appropriately modeled as an exponential random variable. However, the lifetime L of a device that suffers slow mechanical wear, such as bearing wear, is better modeled by a distribution such that P(L < t + At] L > t) increases with . t. Distributions such as the Weibull distribution are ofien used, in practice, to model the failure time of this type of device. The Weibull distribution is presented in a later section. ’t‘VRCISES FOR SECTION 4—8 ,. Suppose X has an exponential distribution with A = 2. 4—77. Suppose X has an exponential distribution with mean u ' e the following: equal to 10. Determine the following: tors o) (b) P(X a 2) (a) P(X > 10) (b) P(X> 20) V- XS 1) (d)P(1< X< 2) (c) P(X< 30) 1' ind the value of x such that P(X < x) = 0.05. (d) Find the value of x such that P(X < x) = 0.95. ...
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