Answer to Question #330730 in Statistics and Probability for almen

Question

The average number of automobiles per minute stopping for gas at a particular service station along the Coastal Road is 4. What is the probability that in any given minute more than two will stop for gas?

Answer 1

It has to be pointed out that the distribution is not specified. Suppose that $X$ is a random variable that corresponds to the number of automobiles per minute. The aim is to compute $P(X>2)$. Consider the most typical distributions:

1. Suppose that $X$ has a normal distribution with parameters $\mu=4$ and $\sigma$ . Then, $P(X>2)=\int_{2}^{+\infty}f(x)dx$, where $f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12\left(\frac{x-\mu}{\sigma}\right)^2}$. For $\sigma=1$ we get: $P(X>2)\approx0.977.$
2. Suppose that $X$ has a Poisson distribution with $\lambda=4.$ Remind that for Poisson distribution $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$. Thus, $P(X>2)=P(X=3)+P(X=4)+...\approx0.762$

Thus, for the normal distribution we receive $0.977$ and for the Poisson distribution $0.762$.