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Suppose that the number of calls coming per minute into an airline reservation center follows a Poisson distribution. Assume that the mean is 3 calls per minute. The probability that at least two calls are received in a given two-minute period is _______.

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Answer: 0.9826

Explanation:

Given : Mean :
\lambda =3\text{ calls per minute}

For two minutes period the new mean would be :


\lambda_1=2*3=6\text{ calls per two minutes}

The formula to calculate the Poisson distribution is given by :_


P(X=x)=(e^(-\lambda_1)\lambda_1^x)/(x!)

Then ,the required probability is given by :-


P(X\geq2)=1-(P(X\leq1))\\\\=1-(P(0)+P(1))\\\\=1-((e^(-6)6^0)/(0!)+(e^(-6)6^1)/(1!))\\\\=1-0.0173512652367\\\\=0.982648734763\approx0.9826

Hence, the probability that at least two calls are received in a given two-minute period is 0.9826.

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