Answer:
a) There is an 81.87% probability that the instrument does not fail in an 8-hour shift.
b) There is a 45.12% probability of at least one failure in a 24-hour day.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
a. What is the probability that the instrument does not fail in an 8-hour shift?
The mean for an hour is 0.025 failures.
For 8 hours, we have ![\mu = 8*0.025 = 0.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%208%2A0.025%20%3D%200.2)
This probability is P(X = 0).
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-0.2}*(0.2)^{0}}{(0)!} = 0.8187](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-0.2%7D%2A%280.2%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.8187)
There is an 81.87% probability that the instrument does not fail in an 8-hour shift.
b. What is the probability of at least one failure in a 24-hour day?
The mean for an hour is 0.025 failures.
For 24 hours, we have ![\mu = 24*0.025 = 0.6](https://tex.z-dn.net/?f=%5Cmu%20%3D%2024%2A0.025%20%3D%200.6)
Either we have at least one failure, or we have no failures. The sum of the probabilities of these events is decimal 1. So
![P(X = 0) + P(X \geq 1) = 1](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%2B%20P%28X%20%5Cgeq%201%29%20%3D%201)
![P(X \geq 1) = 1 - P(X = 0)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%201%29%20%3D%201%20-%20P%28X%20%3D%200%29)
In which
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-0.6}*(0.2)^{0}}{(0)!} = 0.5488](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-0.6%7D%2A%280.2%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.5488)
So
![P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5488 = 0.4512](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%201%29%20%3D%201%20-%20P%28X%20%3D%200%29%20%3D%201%20-%200.5488%20%3D%200.4512)
There is a 45.12% probability of at least one failure in a 24-hour day.