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Competitive Stats
Q142Section III (Statistics)142 of 145

F(1.5)=P(X≤1)=(1+λ)e−λF(1.5)=P(X\le1)=(1+\lambda)e^{-\lambda}

Let X be Poisson variate with mean λ\lambda. Then distribution function of X, evaluated at 1.5 is:

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