Answer to Question #227613 in Statistics and Probability for Teddy

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

Given the average demand on a factory store for a certain electric motor = 8 per week

x=8

Demand for follows the Poisson distribution

From the Poisson distribution

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

Finding the value of n

$\frac{e^{-8}8^0}{0!}+\frac{e^{-8}8^1}{1!}+\frac{e^{-8}8^2}{2!}+...+\frac{e^{-8}8^n}{n!}≥0.95\\ e^{-8}(\frac{8^0}{0!}+\frac{8^1}{1!}+...+\frac{8^n}{n!})≥0.95\\ \frac{8^0}{0!}+\frac{8^1}{1!}+...+\frac{8^n}{n!}≥ 283191009$

From the above series, the minimum n= 13 satisfies to meet 95% sure of meeting all requirements while waiting for his new supply to arrive

So the storeman can allow the minimum of 13.