Question

The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a mean of 0.025 failure per hour.
a. What is the 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?

Answer 1

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ 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 $\mu = 8*0.025 = 0.2$

This probability is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.2}*(0.2)^{0}}{(0)!} = 0.8187$

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 $\mu = 24*0.025 = 0.6$

Either we have at least one failure, or we have no failures. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.6}*(0.2)^{0}}{(0)!} = 0.5488$

So

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5488 = 0.4512$

There is a 45.12% probability of at least one failure in a 24-hour day.