Answer:
a) 0.9084
b) 0.2502
Step-by-step explanation:
We are given the following in the question:
The number of failures follows a Poisson distribution with mean
[tex]\mu = 0.012[/tex]
Formula:
[tex]P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}[/tex]
a) probability that the instrument does not fail in an 8-hour shift
[tex]\lambda = 0.012\times 8 = 0.096[/tex]
We have to evaluate:
[tex]P(x = 0) = \displaystyle\frac{(0.096)^0 e^{-0.096}}{0!} = 0.9084[/tex]
0.9084 is the probability that the instrument does not fail in an 8-hour shift
b) probability of at least 1 failure in a 24-hour day
[tex]\lambda = 0.012\times 24 = 0.288[/tex]
We have to evaluate:
[tex]P(x \geq 1) =1-P(x=0) =1-\displaystyle\frac{(0.288)^0 e^{-0.288}}{0!} = 0.2502[/tex]
0.2502 is the probability of at least 1 failure in a 24-hour day.
