Question #330730

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?

Expert's answer

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

  1. Suppose that XX has a normal distribution with parameters μ=4\mu=4 and σ\sigma . Then, P(X>2)=2+f(x)dxP(X>2)=\int_{2}^{+\infty}f(x)dx, where f(x)=1σ2πe12(xμσ)2f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12\left(\frac{x-\mu}{\sigma}\right)^2}. For σ=1\sigma=1 we get: P(X>2)0.977.P(X>2)\approx0.977.
  2. Suppose that XX has a Poisson distribution with λ=4.\lambda=4. Remind that for Poisson distribution P(X=k)=λkeλk!P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}. Thus, P(X>2)=P(X=3)+P(X=4)+...0.762P(X>2)=P(X=3)+P(X=4)+...\approx0.762

Thus, for the normal distribution we receive 0.9770.977 and for the Poisson distribution 0.7620.762.


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