Answer to Question #334821 in Statistics and Probability for bea villamar

Question

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 average 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 :$X\sim N(\mu, \sigma^2).$

Then

$$P(X>2)=1-P(X\le2)=1-P(Z\le\dfrac{2-4}{1})$$

$$=1-P(Z\le-2)\approx0.97725$$

2. Suppose that $X$ has a Poisson distribution with $\lambda=4:X\sim Po(\lambda).$

Then

$$P(X>2)=1-P(X=0)-P(X=1)$$

$$-P(X=2)=1-\dfrac{e^{-4}(4)^0}{0!}-\dfrac{e^{-4}(4)^1}{1!}$$

$$-\dfrac{e^{-4}(4)^2}{2!}\approx0.76190$$