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 stoppages in a 20 hour period.
A).0986
B).1025
C).0878
D).1226
E).1353

Question

Quizplus · Accepted Answer

The Poisson probability function for having exactly $k$ events in a time period is given by $P(k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the average rate of occurrence for the given time period, and $k$ is the number of occurrences. In this case, we want the probability of having no stoppages ($k=0$) in a 20-hour period. The average rate of stoppages per hour is .1, so for a 20-hour period, $\lambda = 20 \times .1 = 2$. Plugging these values into the formula gives $P(0) = \frac{e^{-2} 2^0}{0!} = e^{-2} \approx 0.1353$.

On Average Stoppages on the Assembly Line at Jamari Industries