Answer:
a) 0.9084
b) 0.2502
Explanation:
We are given the following in the question:
The number of failures follows a Poisson distribution with mean
![\mu = 0.012](https://img.qammunity.org/2021/formulas/mathematics/high-school/pzt5mzhi41r5h3dh5pafb32ebl6tcefmdq.png)
Formula:
a) probability that the instrument does not fail in an 8-hour shift
![\lambda = 0.012* 8 = 0.096](https://img.qammunity.org/2021/formulas/mathematics/high-school/7eagnxz94dy650ken0e5g0b2zo6x3yyk01.png)
We have to evaluate:
![P(x = 0) = \displaystyle((0.096)^0 e^(-0.096))/(0!) = 0.9084](https://img.qammunity.org/2021/formulas/mathematics/high-school/f8pvv5z6w8x5lbz5sz3ytzzkwgli2keur9.png)
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
![\lambda = 0.012* 24 = 0.288](https://img.qammunity.org/2021/formulas/mathematics/high-school/g5le86seggltzdjulnw2nozrtz0vq4ykam.png)
We have to evaluate:
![P(x \geq 1) =1-P(x=0) =1-\displaystyle((0.288)^0 e^(-0.288))/(0!) = 0.2502](https://img.qammunity.org/2021/formulas/mathematics/high-school/h0bokze1ljfqnn266pxletrnn98ivjakr2.png)
0.2502 is the probability of at least 1 failure in a 24-hour day.