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Find $P(3 < X)$, where $X$ is a Poisson variable with parameter $\lambda = 4.33$.
Hint: For a Poisson distribution, $P(X = r) = \mathrm{e}^{-\lambda} \dfrac{\lambda^r}{r!}$.
$P(X = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c0}}\hspace{35px}}~}) = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c2}}\hspace{100px}}~}}{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c3}}\hspace{100px}}~}} = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c4R}}\hspace{100px}}~}$ $P(X = 0) = 0.01316755 \times \dfrac{1}{1} = 0.01316755$
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c2}}\hspace{100px}}~}}{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c3}}\hspace{100px}}~}} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c4R}}\hspace{100px}}~}$ $P(X = 1) = 0.01316755 \times \dfrac{4.33}{1} = 0.05701548$
$P(X = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c0}}\hspace{35px}}~}) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c2}}\hspace{100px}}~}}{\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c3}}\hspace{100px}}~}} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c4R}}\hspace{100px}}~}$ $P(X = 2) = 0.01316755 \times \dfrac{18.7489}{2} = 0.12343852$
$P(X = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c0}}\hspace{35px}}~}) = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c2}}\hspace{100px}}~}}{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c3}}\hspace{100px}}~}} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c4R}}\hspace{100px}}~}$ $P(X = 3) = 0.01316755 \times \dfrac{81.182737}{6} = 0.17816292$
Hint: Add together the individual probabilities.
$P(X \le 3) = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BR}}\hspace{100px}}~}$ $P(X \le 3) = 0.37178447$
Therefore,
$P(3 < X) = 1 - \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{100px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{100px}}~}$ $P(3 < X) = 1 - 0.37178447 = 0.62821553$