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Asked By :  Tripti
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Suppose the useful lifetime in years of a personal computer pc

Suppose the useful lifetime, in years, of a personal computer (PC) is exponentially distributed with parameter A-0.25. A student entering a four-year undergraduate program inherits a two-year-old PC from his sister who just graduated. Find the probability the useful life time of the PC the student inherited will last at least until the student graduates.






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To solve this problem, we need to understand the concept of the exponential distribution and how it applies to the lifetime of the PC.

Given that the lifetime XX of a personal computer is exponentially distributed with a rate parameter λ=0.25\lambda = 0.25, we have:

f(x)=λeλxf(x) = \lambda e^{-\lambda x}

The probability that the PC lasts at least until the student graduates (4 more years) is found by calculating the survival function. Since the PC is already 2 years old, we are interested in the probability that it lasts an additional 4 years (totaling a useful life of 6 years).

The survival function for an exponential distribution is given by:

S(x)=P(X>x)=eλxS(x) = P(X > x) = e^{-\lambda x}

We want P(X>6X>2)P(X > 6 \mid X > 2), which is the probability that it lasts more than 6 years given that it is already 2 years old. By the memoryless property of the exponential distribution, this conditional probability is:

P(X>6X>2)=P(X>4)=eλ4P(X > 6 \mid X > 2) = P(X > 4) = e^{-\lambda \cdot 4}

Plugging in the given rate λ=0.25\lambda = 0.25:

P(X>4)=e0.25×4=e10.3679P(X > 4) = e^{-0.25 \times 4} = e^{-1} \approx 0.3679

Therefore, the probability that the PC will last at least until the student graduates is approximately 0.3679, or 36.79%.


Answered By

Joe Evans

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