Answer:
a) 75.34%
b) 13.53%
Step-by-step explanation:
First, let us consider the cumulative exponential distribution
[tex]F(t) = P[T\leq t] = 1 - e^{-rt}[/tex]
Where
[tex]r=\frac{1}{m}[/tex] and m denotes the mean
Therefore, our cumulative distribution for this exercise is as it follows:
[tex]P[T\leq t] = 1-e^{-\frac{1}{15}t }[/tex]
a) According to the distribution, the first question can be calculated as it follows:
[tex]P[T\leq 21]=1-e^{-\frac{1}{15}*21 } =1-e^{-\frac{7}{5} }[/tex]
[tex]P[T\leq 21]=0.7534[/tex]
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:
[tex]P[T\geq 30]=1-P[T<30][/tex]
[tex]P[T\geq 30]=1-(1-e^{-\frac{1}{15}*30 })=e^{-2}[/tex]
[tex]P[T\geq 30]=0.1353[/tex]
Therefore, we have that the probability that the generator will operate for 30 days without a breakdown is 13.53%