Answer:
a) 75.34%
b) 13.53%
Step-by-step explanation:
First, let us consider the cumulative exponential distribution
![F(t) = P[T\leq t] = 1 - e^{-rt}](https://tex.z-dn.net/?f=F%28t%29%20%3D%20P%5BT%5Cleq%20t%5D%20%3D%201%20-%20e%5E%7B-rt%7D)
Where
and <em>m</em> denotes the mean
Therefore, our cumulative distribution for this exercise is as it follows:
![P[T\leq t] = 1-e^{-\frac{1}{15}t }](https://tex.z-dn.net/?f=P%5BT%5Cleq%20t%5D%20%3D%201-e%5E%7B-%5Cfrac%7B1%7D%7B15%7Dt%20%7D)
a) According to the distribution, the first question can be calculated as it follows:
![P[T\leq 21]=1-e^{-\frac{1}{15}*21 } =1-e^{-\frac{7}{5} }](https://tex.z-dn.net/?f=P%5BT%5Cleq%2021%5D%3D1-e%5E%7B-%5Cfrac%7B1%7D%7B15%7D%2A21%20%7D%20%3D1-e%5E%7B-%5Cfrac%7B7%7D%7B5%7D%20%7D)
![P[T\leq 21]=0.7534](https://tex.z-dn.net/?f=P%5BT%5Cleq%2021%5D%3D0.7534)
Hence, the probability that the piece will break down in the next 21 days is 75.34%
b) Now, take into account that we now need the probability complement for applying the cumulative distribution like this:
![P[T\geq 30]=1-P[T](https://tex.z-dn.net/?f=P%5BT%5Cgeq%2030%5D%3D1-P%5BT%3C30%5D)
![P[T\geq 30]=1-(1-e^{-\frac{1}{15}*30 })=e^{-2}](https://tex.z-dn.net/?f=P%5BT%5Cgeq%2030%5D%3D1-%281-e%5E%7B-%5Cfrac%7B1%7D%7B15%7D%2A30%20%7D%29%3De%5E%7B-2%7D)
![P[T\geq 30]=0.1353](https://tex.z-dn.net/?f=P%5BT%5Cgeq%2030%5D%3D0.1353)
Therefore, we have that the probability that the generator will operate for 30 days without a breakdown is 13.53%