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?
Accepted Solution
A:
Answer:The probability is 0.4972Step-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]