Question

(3 points) Buchtal, a manufacturer of ceramic tiles, reports on average 3.1 job-related accidents per year. Accident categories include trip, fall, struck by equipment, transportation, and handling. The number of accidents is approximately Poisson. Please upload your work for all of the parts at the end. (0.5 pts.) a) What is the probability that more than one accident occurs per year

Answer 1

Answer:

0.8743 = 87.43% probability that more than one accident occurs per year

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 interval.

Buchtal, a manufacturer of ceramic tiles, reports on average 3.1 job-related accidents per year.

This means that $\mu = 3.1$

What is the probability that more than one accident occurs per year?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

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

Then

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

$P(X = 0) = \frac{e^{-3.6}*(3.6)^{0}}{(0)!} = 0.0273$

$P(X = 1) = \frac{e^{-3.6}*(3.6)^{1}}{(1)!} = 0.0984$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0273 + 0.0984 = 0.1257$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1257 = 0.8743$

0.8743 = 87.43% probability that more than one accident occurs per year