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
We have that
"\\mu=4"
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>2,4) = 1 - P(x\\le2,4)"
where "P(x\\le2,4)= P(0,4)+P(1,4)+P(2,4)"
"P(0,4)=\\frac{e^{-4}4^0}{0!}=0.0183"
"P(1,4)=\\frac{e^{-4}4^1}{1!}=0.0733"
"P(2,4)=\\frac{e^{-4}4^2}{2!}=0.147"
"P(x>2,4) =1-0.0183-0.0733-0.147=0.7614"
Answer: 0.7614
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