Suppose that 1 customer arrives each minute in a Poisson distribution. Is it more likely that 2 customers or 0 customers will arrive each minute?
What will be an ideal response?
P(0 ) = e^(-1 )(1^0 )/0! = .368
P(2 ) = e^(-1 )(1^2 )/2! = .184
Thus it is more likely that 0 customers will arrive each minute.
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