The time to​ failure, t, in​ hours, of a machine is often exponentially distributed with a probability density function ​f(t) = k e^(-kt)​, 0 to infinity​, where k= 1/a 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 80 hr. What is the probability that a failure will occur in 55 hr or​ less?

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Answer:

The probability is 0.4972

Step-by-step explanation:

The probability density function for the time of failure is:

f(t) = k e^(-kt)​

Where k=1/a and a = 80hr so f(t) is equal to:

[tex]f(t)=\frac{1}{a}e^{\frac{-t}{a} }[/tex]

[tex]f(t)=\frac{1}{80}e^{\frac{-t}{80} }[/tex]

Then, if the probability density function follow a exponential distribution, the probability distribution function is:

[tex]F(t)=1-e^{-t/a}[/tex]

[tex]F(t)=1-e^{-t/80}[/tex]

The probability distribution function give as the probability that a failure will occur in t hours or less, so the probability that a failure will occur in 55 hr or​ less is calculated as:

[tex]F(55)=1-e^{-55/80}=0.4972[/tex]