the average number of automobiles per minute stopping for a gas at a particular service station along the coastal road is 3. what is a probability that in any given minute more than two automobile will stop for gas?
We have that
"\\mu=3"
x = 2
This follows Poisson distribution
The Poisson probability can be calculated by the formula:
"P(x,\\mu)=\\frac{e^{-\\mu}\\mu^x}{x!}"
Need to find "P(x>2,3) = 1 - P(x\\le2,3)"
where "P(x\\le2,3)= P(0,3)+P(1,3)+P(2,3)"
"P(0,3)=\\frac{e^{-3}3^0}{0!}=0.05"
"P(1,3)=\\frac{e^{-3}3^1}{1!}=0.15"
"P(2,3)=\\frac{e^{-3}3^2}{2!}=0.22"
"P(x>2,3) =1-0.05-0.15-0.22=0.58"
Answer: 0.58
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