Answer to Question #235371 in Statistics and Probability for ntachi

Question

The average demand on a factory store for a certain electric motor is 8 per week.

When the storeman places an order for these motors, delivery takes one week.

If the demand for motors has a Poisson distribution, how low can the storeman

allow his stock to fall before ordering a new supply if he wants to be at least

95% sure of meeting all requirements while waiting for his new supply to arrive?

Answer 1

From the Poisson distribution

$P(x=k)= \frac{e^{-\lambda} \lambda^k}{k!}\\ \therefore P(x=k)≥0.95$

Then find the value of n

$\frac{e^{-8} 8^0}{0!}+\frac{e^{-8} 8^1}{1!}+...+\frac{e^{-8} 8^n}{n!}≥0.95\\ e^{-8}[\frac{ 8^0}{0!}+\frac{ 8^1}{1!}+...+\frac{8^n}{n!}]≥0.95\\ [\frac{ 8^0}{0!}+\frac{ 8^1}{1!}+...+\frac{8^n}{n!}]≥2831.9\\ \implies n= 13$