Question

The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a mean of 0.018 failures 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 1 failure in a 24-hour day? Round your answers to four decimal places (e.g. 98.7654).

Answer 1

Answer:

The probability that the instrument does not fail in an 8-hour shift is $P(X=0) \approx 0.8659$

The probability of at least 1 failure in a 24-hour day is $P(X\geq 1 )\approx 0.3508$

Step-by-step explanation:

The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

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

Let X be the number of failures of a testing instrument.

We know that the mean $\mu = 0.018$ failures per hour.

(a) To find the probability that the instrument does not fail in an 8-hour shift, you need to:

For an 8-hour shift, the mean is $\mu=8\cdot 0.018=0.144$

$P(X=0)=\frac{e^{-0.144}0.144^0}{0!}\\\\P(X=0) \approx 0.8659$

(b) To find the probability of at least 1 failure in a 24-hour day, you need to:

For a 24-hour day, the mean is $\mu=24\cdot 0.018=0.432$

$P(X\geq 1 )=1-P(X=0)\\\\P(X\geq 1 )=1-\frac{e^{-0.432}0.432^0}{0!}\\\\P(X\geq 1 )\approx 0.3508$