Question

The time to failure, t, in hours, of a machine is often exponentially distributed with a probability density function f(t), 0 < t < [infinity], where k and a is the average amount of time that will pass before a failure occurs. Suppose that the average amount of time that will pass before a failure occurs is k. What is the probability that a failure will occur in a or less?

Answer 1

The probability that a failure will occur in time
`\(a\)` or less is given by the cumulative distribution function (CDF) of the exponential distribution and is expressed as
`\(P(T \leq a) = 1 - e^{-(a)/(k)}\).`

Step-by-step explanation:

In an exponential distribution, the probability density function (PDF) is given by
`\(f(t) = (1)/(k) e^{-(t)/(k)}\`), where
`\(k\)`is the average time to failure. To find the probability of failure within time
`\(a\)` or less, we integrate the PDF from 0 to \(a\) to obtain the cumulative distribution function (CDF).

`\[P(T \leq a) = \int_(0)^(a) f(t) dt = \int_(0)^(a) (1)/(k) e^{-(t)/(k)} dt\]`

Solving this integral yields
`\(P(T \leq a) = 1 - e^{-(a)/(k)}\)`. This formula gives the probability that the time to failure is less than or equal to
`\(a\)`. The exponential distribution is memoryless, meaning that the probability of failure in the next instant is the same regardless of how much time has already passed. This property is reflected in the simplicity of the CDF formula.

In summary, the probability that a failure will occur in time
`1−e − ka` providing a straightforward and concise expression for assessing the likelihood of failure within a specified time frame in an exponentially distributed system.