On average stoppages on the assembly line at Jamari Industries have a Poisson distribution with a mean of .1 per hour.Use the Poisson probability function to determine the probability of having no more than one stoppage in a 30-hour period.
A).1992
B).1689
C).1364
D).2567
E).1043

Question

Quizplus · Accepted Answer

The Poisson probability function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the average rate (mean) of occurrences in a fixed interval of time or space, $k$ is the number of occurrences, and $e$ is the base of the natural logarithm (approximately equal to 2.71828). Given a mean of .1 stoppages per hour, for a 30-hour period, $\lambda = 0.1 \times 30 = 3$. To find the probability of having no more than one stoppage, we calculate the probability of having 0 stoppages and 1 stoppage, then sum these probabilities.For 0 stoppages ($k=0$): $P(X=0) = \frac{e^{-3} 3^0}{0!} = e^{-3}$For 1 stoppage ($k=1$): $P(X=1) = \frac{e^{-3} 3^1}{1!} = 3e^{-3}$Summing these probabilities: $P(X \leq 1) = e^{-3} + 3e^{-3} = 4e^{-3}$Using the value of $e^{-3} \approx 0.0498$, the total probability is approximately $4 \times 0.0498 = 0.1992$ or 19.92%, which corresponds to choice A when expressed as a percentage to four decimal places.

On Average Stoppages on the Assembly Line at Jamari Industries