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Question
The average number of phone calls per minute into the switchboard of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be no phone at all
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Solution
The average number of phone cells per minute into the switchboard of a company is λ = 2.5
x follows poisson distribution with
P(x) = `("e"^(-lambda) lambda^x)/(x!)`
= `((2^(-2.5) (2.5)^x)/(x!))`
P(no phone at all) = P(X = 0)
= `("e"^(-2.5) (2.5)^0)/(0!)`
= `"e"^(-2.5)`
= 0.08208
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